Houghton Mifflin Harcourt

Algebra 2, Volume 1 1st Edition Module 1 Analyzing Function

Page 4 Problem 1 Answer

In the question, it is given that A is (1,5).

It is required to determine the location of A′ when A is rotated 90∘ clockwise.

To solve this, multiply the x-value with −1 and switch it with the y-coordinate to get the location of A′.

Multiply the x-coordinate of the point A(1,5) with −1 and simplify using mathematical operations.

A(1⋅(−1),5)=A(−1,5)

Switch the x-value of the point A(−1,5) with the y-coordinate to get the location of A′.

A(−1,5)=A′(5,−1)

The location of A′ when A(1,5) is rotated 90∘ clockwise is A′(5,−1).

Page 4 Problem 2 Answer

In the question, it is given that A is (1,5).

It is required to determine the location of A′ when A is translated  1 unit left.

To solve this, subtract the x-value by 1 as it is translating to the left to get the location of A′.

Subtract the x-coordinate of the point A(1,5) by 1 and simplify using subtraction operations.

A(1−1,5)=A′(0,5)

The location of A′ when A(1,5) is translated 1 unit left is A′(0,5).

Page 4 Problem 3 Answer

In the question, it is given that A is (1,5).

It is required to determine the location of A′ when A is reflected across the x-axis.

To solve this, multiply the y-value with −1 and simplify to get the location of A′.

When the point is reflected across the x-axis, the y-coordinate will become negative and the x-coordinate will remain the same.

Multiply the y-coordinate of the point A(1,5) with −1 and simplify using mathematical operations.

A(1,5⋅(−1))=A′(1,−5)

The location of A′ when A(1,5) is reflected across the x-axis is A′(1,−5).

Page 6 Problem 4 Answer

In the question intervals [0,5], [0,5) and (0,5) are given.

It is required to show the common interval and the difference between them.

To solve this, check the interval notations of the given intervals and evaluate the answer.

Write the given intervals,[0,5], [0,5) and (0,5)And check their interval notations.

Upon visualization, it is evident that numbers between 0 and 5 are common in the interval [0,5], [0,5) and (0,5).

The difference between the intervals [0,5], [0,5) and (0,5) are,

In interval [0,5], 0 and 5 both are included in the interval.

In interval [0,5), 0 is included but the 5 is not.

In interval (0,5), the numbers between 0 and 5 are included, but the value 0 and 5 are not.

In the intervals [0,5], [0,5) and (0,5), the numbers between 0 and 5 are common.

The difference is that, in interval [0,5], 0 and 5 both are included in the interval, in interval [0,5), 0 is included but the 5 is not and in interval (0,5), the numbers between 0 and 5 are included, but the value 0 and 5 are not.

Page 4 Problem 5 Answer

It is given in the question that the graph of the quadratic function f(x)=−x² is, It is required to write the domain and range of function using an inequality, set notation and using interval notation.

To solve this, use the concept of intervals along with the knowledge of domain and range to write the notations in the form of inequality, set and intervals.

The graph of the function f(x)=−x² is

By analyzing the function and the graph the domain and range of the function can be known.

The domain of function f(x)=−x² in inequality, set notation and using interval notation is as follows,

Inequality notation: −∞<x<∞

Set notation: {x∣​−∞<x<∞}

Interval notation: (−∞,∞)

The range of function f(x)=−x² in inequality, set notation and using interval notation is as follows,

Inequality notation: −∞<x⩽0

Set notation: {x∣​−∞<x⩽0}

Interval notation: (−∞,0]

The end behaior of the function f(x)=−x² is,

As x→∞

, f(x)→−∞

As x→−∞

, f(x)→−∞

The domain and range of function are as follows,

Domain:

Inequality notation: −∞<x<∞

Set notation: {x∣​−∞<x<∞}

Interval notation: (−∞,∞)

Range:

Inequality notation: −∞<x⩽0

Set notation: {x∣​−∞<x⩽0}

Interval notation: (−∞,0]

The end behavior of the function f(x)=−x2 is, As x→∞, f(x)→−∞

As x→−∞

, f(x)→−∞

The function f(x)=−x² will always lead to a non positive real number.

Page 4 Problem 6 Answer

It is given that the domain of the function f(x)=3/4x+2 is changed to (−4,4) instead of [−4,4].

It is required to depict the change in the graph due to change in the domain of the function.

To solve this, insert the end point values of the domain in the function and analyze the result.

Earlier the domain of the function f(x)=3/4x+2 was [−4,4].

Calculate the f(x) for the endpoints of domain.

Substitute x=−4 in function f(x) and simplify using the division and addition properties.

f(−4)=3/4

(−4)+2

f(−4)=(−3)+2

f(−4)=−1

​Substitute x=4 in function f(x) and simplify using the division and addition properties.

f(4)=3/4

(4)+2

f(4)=(3)+2

f(4)=5

​The domain of the function  changes from  to .

Since, and , then the new domain , will not contain  and  in its range.

Therefore, the endpoints  and  will not be included in the graph for the domain .

The graph does not include the endpoints (4,5) and (−4,−1) when the domain is (−4,4) instead of [−4,4].

Page 9 Problem 7 Answer

In the question, the given function and domain is f(x)=2/3x−1, (−∞,3].

It is required to draw the graph for the function and determine its range.

To solve this, first plot the graph of the given function, then substitute the value of x

from the domain to the given function and simplify to determine its range.

Plot the graph of the function f(x)=2/3x−1 with the domain (−∞,3]

Substitute the value x=3 in the function f(x)=2/3x−1 and simplify using simple Mathematical operations.

Calculation when x=3 is substituted gives

f(x)=2/3

(3)−1

f(3)=2−1

f(3)=1

​Substitute the value x=0 in the function f(x)=2/3x−1 and simplify using simple Mathematical operations.

Calculation when x=0 is substituted gives

f(x)=2/3

(0)−1

f(0)=0−1

f(0)=−1

​As x→−∞,f(x)→−∞

Range: (−∞,1]

The range of the function f(x)=2/3x−1 is (−∞,1].

Page 11 Problem 8 Answer

In the question, it is given that Joyce jogs for 30min.

It is required to determine how the domain changes.

To solve this, domain is the time interval in minutes during which Joyce jogs.

Since Joyce jogs for 30 min and time is always a positive integer.Therefore, Domain: 0⩽t⩽30

The domain of the function changes as 0⩽t⩽30.

Page 11 Problem 9 Answer

In the question, it is given that Joyce jogs for 30min.

It is required to determine how the graph changes.

To solve this, plot the graph with intervals 0⩽t⩽30 and 0⩽d⩽3 to observe the changes.

Plot the graph with the intervals 0⩽t⩽30 and 0⩽d⩽3.

The graph becomes steeper as the interval of time and distance change which is directly proportional to speed.

The graph can be plotted with 0⩽t⩽30 and 0⩽d⩽3

we observe the change that the graph becomes steeper as the interval of time and distance change which is directly proportional to speed.

Page 11 Problem 10 Answer

In the question, it is given that Joyce jogs for 30min.

It is required to determine how the range changes.

To solve this, range is the distance (in miles) covered by Joyce in the time t in the domain.

Since Joyce jogging rate is 0.1miles/min.

Her jogging distance (in miles) at any time t (in minutes) is given by d(t)=0.1t

Therefore, Range: 0⩽d⩽3.

The range of the function changes as 0⩽d⩽3.

Page 11 Problem 11 Answer

In the question it is given that on a moving walkway one is carried 25 feet every 15sec for 1min.

It is required to write a function to model the situation and determine domain, range and plot the graph.

To solve this, model the function using distance in terms of time to form a linear function, then obtain the domain and range. Plot the graph using the obtained domain and range.

The function modelling the distance in terms of time is a linear function:

d(t)=mt+n

Since at time t=0, the distance will be zero, simplify using simple addition properties to find n:

d(0)=0

0+n=0

n=0

​The rate mcan be computed as m=d/t where d is 25feet and t is 15sec

Substitute the respective values in the equation for rate and simplify using simple division.

m=d/t

m=25/15

m=5/3

Substituting the value of min the linear function of time d(t)=5/3t

For the domain, use the time interval in seconds.

t=1min

t=60sec

​The domain: {t∣0⩽t⩽60}=[0,60]

For the range, it is the distance for the given time interval:

For t=0

d(0)=0

Similarly for t=60

d(60)=100

The range: {d∣0⩽d⩽100}=[0,100]

Plot the graph with the interval {t∣0⩽t⩽60}=[0,60]and {d∣0⩽d⩽100}=[0,100]

The function is d(t)=5/3t, the domain for the function is {t∣0⩽t⩽60}=[0,60] and range is {d∣0⩽d⩽100}=[0,100].

Page 11 Problem 12 Answer

In the question it is given aand bare real numbers and a<b

It is required to write four different intervals with aand bas endpoints

To solve this, consider the four intervals as (a,b), (a,b], [a,b), [a,b]

(a,b): This interval includes real numbers from a to b excluding a and b.

(a,b]: This interval includes real numbers from a  to b excluding a .

[a,b): This interval includes real numbers from a  to b excluding b.

[a,b]: This interval includes real numbers from a  to b.

The four different intervals with aand bas endpoints are (a,b), (a,b], [a,b), [a,b].

Page 12 Exercise 1 Answer

In the question, the following number line is given:

It is required to write the interval as given in the number line as an inequality, set notation and interval notation.

To solve this, use the number line to determine the line’s interval and express it as an inequality, set notation and interval notation.

For the given number line, the interval is all real numbers from 5 to infinity, including 5.

The finite interval for the number line is x≥5.

Thus, the set notation for the number line is {x∣x≥5}and the interval notation is [5,+∞).

The number line’s interval as an inequality is x≥5.

The set notation for the number line is {x∣x≥5} and the interval notation is [5,+∞).

Page 12 Exercise 2 Answer

In the question, the given interval is (5,100].

It is required to write the interval as an inequality and using set notation.

To solve this, use the numbers and the parenthesis of the interval to express it as an inequality and as a set notation.

The interval (5,100] is all real numbers from 5 to 100, including 100 but not including 5.

The interval expressed as an inequality is 5<x≤100.

Thus, the interval using set notation is {x∣5<x≤100}.

The interval written as an inequality is 5<x≤100. The interval expressed as a set notation is {x∣5<x≤100}.

Page 12 Exercise 3 Answer

In the question, the given interval is −25≤x<30.

It is required to write the interval using set notation and interval notation.

To solve this, use the numbers and inequality signs in the interval to express it as a set notation and interval notation.

The inequality −25≤x<30 means that the interval is all real numbers from −25 to 30, including −25 but not including 30.

The interval expressed as a set notation is {x∣−25≤x<30}.

Thus, the interval expressed as an interval notation is [−25,30).

The interval written as a set notation is {x∣−25≤x<30}. The interval expressed as an interval notation is [−25,30).

Page 12 Exercise 4 Answer

In the question, the given interval is {x∣−3<x<5}.

It is required to write the interval as an inequality and using interval notation.

To solve this, use the numbers and inequality signs in the set notation to express it as an inequality and interval notation.

The interval is all real numbers from −3 to 5, but not including −3 and 5.

The interval expressed as an inequality from the set notation {x∣−3<x<5} is −3<x<5.

Thus, the interval expressed as an interval notation is (−3,5).

The interval written as an inequality is −3<x<5. The interval expressed as an interval notation is (−3,5).

Page 12 Exercise 5 Answer

In the question, the graph of quadratic function f(x)=x²+2 is given as shown below:

It is required to write the function’s domain and range as an inequality, set notation and interval notation.

It is also required to describe the end behaviour of the function.

To solve this, use the graph to determine the function’s domain and range.

Write the domain and range as an inequality, set notation and interval notation. Describe the end behaviour of the function from the curve in the graph.

The function’s f(x)=x²+2 domain will be the set of all real numbers. The domain written as an interval is −∞<x<+∞.

The domain expressed as a set notation is {x∣−∞<x<+∞}.

Thus, the domain expressed as an interval notation is (−∞,+∞).

The function’s f(x)=x²+2 range will be the set of all real numbers from 2 to infinity, including 2.

The range written as an interval is 2≤f(x)<+∞\.

The range expressed as a set notation is {f(x)∣2≤f(x)<+∞}.

Thus, the range expressed as an interval notation is [2,+∞).

As the value of x tends towards +∞ or −∞, the end behaviour of function f(x) tends towards +∞.

The domain written as an interval is −∞<x<+∞ and as a set notation is {x∣−∞<x<+∞}.

The domain expressed as an interval notation is (−∞,+∞).

The range written as an interval is 2≤f(x)<+∞ and as a set notation is {f(x)∣2≤f(x)<+∞}. The range expressed as an interval notation is [2,+∞).

As the value of x tends towards+∞ or −∞, the end behaviour of functionf(x)tends towards +∞.

Page 13 Exercise 6 Answer

In the question, the given function is f(x)=−x+5 and the domain is [−3,2]. The below diagram has been given:

It is required to draw the function’s graph and state the function’s range using the same notation as the domain.

To solve this, draw the graph using the given function and the domain.

Determine the function’s range from the graph and write it in the same notation as the domain.

The function’s domain is given as [−3,2]. Hence, the x-coordinates of two endpoints of the line are −3 and 2.

The y-coordinates of two endpoints will be f(−3) and f(2).

Substitute x as −3 in the equation f(x)=−x+5 and simplify using mathematical operations to determine the y-coordinate.

​f(x)=−(−3)+5

=3+5

=8

​Hence, one endpoint of the line given by f(x)=−x+5 is (−3,8).

Substitute x as 2 in the equation f(x)=−x+5 and simplify using mathematical operations to determine the y-coordinate.

​f(x)=−2+5

=3

​Hence, another endpoint of the line given by f(x)=−x+5 is (2,3).

Draw the function’s f(x)=−x+5 graph in the domain [3,8].

Page 12 Exercise 7 Answer

From the graph, the function’s range is [3,8].

The function’s f(x)=−x+5 graph in the domain [−3,2] is shown below:

The function’s range is [3,8].

Page 13 Exercise 8 Answer

In the question, the given function is f(x)=3x/2+1 and the domain is {x∣x>−2}. The below diagram has been given:

It is required to draw the function’s graph and state the function’s range using the same notation as the domain.

To solve this, draw the graph using the given function and the domain. Determine the function’s range from the graph and write it in the same notation as the domain.

The function’s domain is given as {x∣x>−2}.

The domain of the function is all real numbers from −2 to +∞ but not including x.

Hence, the x-coordinate of one endpoint of the ray is −2.

The y-coordinate of endpoint will be f(−2).

Substitute x as −2 in the equation f(x)=3x/2+1 and simplify using mathematical operations to determine the y-coordinate.

​f(x)=3⋅(−2)/2+1

=−3+1

=−2

​Hence, one endpoint of the line given by f(x)=3x/2+1 is (−2,−2).

Take another point x=2 to find another point for drawing the graph.

Substitute x as 2 in the equation f(x)=3x/2+1 and simplify using mathematical operations to determine the y-coordinate.

​f(x)=3⋅2/2+1

=3+1

=4

​

Hence, another endpoint of the line given by f(x)=3x/2+1 is (2,4).

Draw the function’s f(x)=3x/2+1 graph in the domain {x∣x>−2}.

From the graph, the function’s range is {f(x)∣f(x)>−2}.

The function’s f(x)=3x/2+1

graph in the domain{x∣x>−2} is shown below:

The function’s range is {f(x)∣f(x)>−2}.

Page 14 Exercise 9 Answer

In the question, it is given that a building is 90 metres above ground and an elevator starts at a floor, and descends 2 metres every 0.5 seconds for 6 seconds.

The below diagram has been given:

It is required to write a function modelling the situation and draw the function’s graph after stating its domain.

It is also required to identify the function’s range using inequality.

To solve this, use the given data to write a function that models the situation. Find the domain and draw the graph using the domain.

Determine the function’s range from the graph.

The elevator travels at 2 metres every 0.5 seconds, or 4 metres every second.

Let t be the total time the elevator descends. The total distance f(t) travelled by the elevator descending from 90 metres can be written as:

f(t)=90−4t

Since the total time travelled by the elevator has been given as 6seconds, the function’s domain is 0≤t≤6.

Substitute t as 6 in the equation f(t)=90−4t and simplify using mathematical operations to determine the range.

​f(t)=90−4⋅6

=90−24

=66

​Hence, the range of the function is 66≤f(t)≤90.

Draw the function’s f(t)=90−4t graph in the domain 0≤t≤6.

The given situation can be written as the function’s f(t)=90−4t with the domain 0≤t≤6  and range 66≤f(t)≤90.

The function’s graph has been shown below:

Page 15 Exercise 10 Answer

It is given in the question that the height from the bridge deck to the top of the tower, where a cable is anchored is 500 feet, and the length of the cable is 1200 feet.

It is required to sketch the cable on a coordinate plane, where x-axis represents the bridge deck and y-axis represents the tower.

It is required to write the linear function for the model graph and identify the domain and range, writing each as an inequality, using set notation, and using interval notation.

To solve this, draw a graph where x axis represents the bridge deck and y-axis represents the tower.

Find the coordinates of the end points where the cables are attached using the Pythagoras theorem.

Write the linear function for the cable line obtained and identify the domain and range of the function.

Draw the graph for the required situation and mark the points accordingly.

Use the Pythagoras theorem to get the distance OB.

It is given that OA=500and ABthe length of the cable is 1200.

According to Pythagoras theorem,OA²+OB²=AB²

Subtract OA² from both the sides and simplify the equation OA²+OB²=AB²,

OB²=AB²−OA²

Square root each sides of the equation OB²=AB²−OA²,

OB=√AB²−OA²

Substitute the value of OA=500 and AB=1200 in the equation OB=√AB²−OA²,

OB=√(1200)²−(500)²

Square the terms within the bracket in the equation OB=√(1200)²−(500)², and simplify using the subtraction properties.

​OB=√1440000−250000

OB=√1190000

OB=1090.87

OB≈1091

​Determine the coordinates of points A and B.

It is known that OA=500 and OB=1091 .

So the coordinates of points A and B are,A(0,500) B(1091,0)​

Write the equation for the line AB  using the coordinates of points A and B.

The equation of a line with two known points is,

y−y₁=y₂−y₁

x₂−x₁

(x−x₁)  where, x, y are the variables and (x₁,y₁) and (x₂,y₂) are the two points on the line.

Let point A be (x₁,y₁) and point B be (x₂,y₂).

Substitute the value A(0,500) and B(1091,0) in the equation of line, y−y₁

=y₂−y₁/x₂−x₁(x−x₁)

y−500=0−500/1091−0(x−0)

Add 500 on both sides of the equation y−500=0−500/1091−0

(x−0),  and simplify using the addition property.

y=0−500/1091−0

(x−0)+500

y=−500/1091

x+500​

Thus, the linear equation of line is y=−500

1091

x+500

By visualizing the graph, it is evident that input values of x can vary from 0 to 1091.

Therefore, the domain of function is 0⩽x⩽1091.

Set notation: {x∣0⩽x⩽1091}

Interval notation: [0,1091]

Substitute the endpoints of the domain in the function, the required results obtained will be the Range of function.

Put x=0 in the equation

y=−500/1091

x+500, and simplify using addition properties.

y=−500/1091

x+500

y=−500/1091

(0)+500

y=500​

Put x=1091in the equation y=−500/1091 x+500, and simplify using addition properties.

y=−500/1091

x+500

y=−500/1091 (1091)+500

y=−500+500

y=0

Therefore, the range of function is0⩽y⩽500.

Set notation:{x∣0⩽y⩽500}

Interval notation:[0,500]

The linear function whose graph models the cable is y=−500/1091

x+500.

Domain of the function:

Inequality: 0⩽x⩽1091

Set notation: {x∣0⩽x⩽1091}

Interval notation: [0,1091]

Range of the function:

Inequality: 0⩽y⩽500

Set notation:{x∣0⩽y⩽500}

Interval notation:[0,500]

The model graph is as shown below,

Algebra 2, Volume 1 1st Edition Module 2 Absolute Value, Functions, Equations And Inequalities

Page 87 Problem 1 Answer

The absolute value expression in an absolute value inequality is generally of the form ∣ax+b∣.

An absolute value inequality can be solved either graphically or algebraically.

To solve an absolute value inequality graphically, the left and right sides of the inequality are assumed to be the functions f(x) and g(x) respectively.

Then, the two functions are graphed on the same coordinate grid. Finally, the intervals on the x-axis which satisfy the given inequality are observed to find the required solution.

To solve an absolute value inequality algebraically, first the operations of multiplication, division, addition, and subtraction are used to isolate the absolute value expression of the form ∣ax+b∣ on one side of the inequality.

Then, the constant on the other side of the inequality and the sign of inequality are used to rewrite the inequality as one or two linear inequalities.

Finally, the obtained inequalities are solved for x to find the required solution.

The solutions of an absolute value inequality can be obtained either graphically or algebraically.

The graphic method of solving an absolute value inequality involves assuming the left and right sides of the inequality to be the functions f(x) and g(x) respectively, graphing the two functions on the same coordinate grid, and finally, observing the intervals on the x-axis which satisfy the given inequality to find the required solution.

The algebraic method of solving an absolute value inequality involves isolating the absolute value expression on one side of the inequality, then using the constant on the other side of the inequality and the sign of inequality to rewrite the inequality as one or two linear inequalities, and finally, solving the obtained inequalities for x to find the required solution.

Page 87 Problem 2 Answer

The given table contains the possible solutions of the given absolute value inequality ∣x∣+2<5.

The question requires to determine which of the integers between −5 and 5 are solutions of the given inequality, and complete the given table by writing ‘yes’ or ‘no’ if the number is a solution of the given inequality.

Then, it is required to plot the solutions on the number line.

To complete the table and plot the solutions, use the given integers, substitute them in the given absolute value inequality, and checking whether the number satisfies the inequality or not.

Then, write ‘yes’ against the numbers that satisfy the inequality and are a solution, or write ‘no’ otherwise. Finally, plot the solutions on the given number line.

Substitute −5 for x in the absolute value inequality ∣x∣+2<5, and simplify the left-hand side.

​∣−5∣+2<5

5+2<5

7<5

This inequality is not true. Therefore, x=−5 is not a solution of the given absolute value inequality.

Substitute −4 for x in the absolute value inequality ∣x∣+2<5, and simplify the left-hand side.

​​∣−4∣+2<5

4+2<5

6<5

This inequality is not true. Therefore, x=−4 is not a solution of the given absolute value inequality.

Substitute −3 for x in the absolute value inequality ∣x∣+2<5, and simplify the left-hand side.

​​∣−3∣+2<5

3+2<5

5<5

This inequality is not true. Therefore, x=−3 is not a solution of the given absolute value inequality.

Substitute −2 for x in the absolute value inequality ∣x∣+2<5, and simplify the left-hand side.

​​∣−2∣+2<5

2+2<5

4<5

This inequality is true. Therefore, x=−2 is a solution of the given absolute value inequality.

Substitute −1 for x in the absolute value inequality ∣x∣+2<5, and simplify the left-hand side.

​​∣−1∣+2<5

1+2<5

3<5

​This inequality is true. Therefore, x=−1 is a solution of the given absolute value inequality.

Substitute 0 for x in the absolute value inequality ∣x∣+2<5, and simplify the left-hand side.

​​∣0∣+2<5

0+2<5

2<5

​This inequality is true. Therefore, x=0 is a solution of the given absolute value inequality.

Substitute 1 for x in the absolute value inequality ∣x∣+2<5, and simplify the left-hand side.

​∣1∣+2<5

1+2<5

3<5

This inequality is true. Therefore, x=1 is a solution of the given absolute value inequality.

Substitute 2 for x in the absolute value inequality ∣x∣+2<5, and simplify the left-hand side.

​​∣2∣+2<5

2+2<5

4<5

​This inequality is true. Therefore, x=2 is a solution of the given absolute value inequality.

Substitute 3 for x in the absolute value inequality ∣x∣+2<5, and simplify the left-hand side.

​∣3∣+2<5

3+2<5

5<5

This inequality is not true. Therefore, x=3 is not a solution of the given absolute value inequality.

Substitute 4 for x in the absolute value inequality ∣x∣+2<5, and simplify the left-hand side.

​∣4∣+2<5

4+2<5

6<5

​This inequality is not true. Therefore, x=4 is not a solution of the given absolute value inequality.

Substitute 5 for x in the absolute value inequality ∣x∣+2<5, and simplify the left-hand side.

​∣5∣+2<5

5+2<5

7<5

This inequality is not true. Therefore, x=5 is not a solution of the given absolute value inequality.

Complete the given table by writing ‘yes’ against the numbers that are solutions of the given absolute value inequality, and ‘no’ against the numbers that are not solutions of the given absolute value inequality.

Plot the obtained solutions on the given number line.

The required solutions are the integers −2,−1,0,1,2.

The completed table is:

The required plot of the solutions on the number line is:

Page 87 Problem 3 Answer

The given table contains the possible solutions of the given absolute value inequality ∣x∣+2>5.

The question requires to determine which of the integers between −5 and 5

are solutions of the given inequality, and complete the given table by writing ‘yes’ or ‘no’ if the number is a solution of the given inequality.

Then, it is required to plot the solutions on the number line.

To complete the table and plot the solutions, use the given integers, substitute them in the given absolute value inequality, and checking whether the number satisfies the inequality or not.

Then, write ‘yes’ against the numbers that satisfy the inequality and are a solution, or write ‘no’ otherwise.

Finally, plot the solutions on the given number line.

Substitute −5 for x in the absolute value inequality ∣x∣+2>5, and simplify the left-hand side.

​​∣−5∣+2>5

5+2>5

7>5

​This inequality is true. Therefore, x=−5 is a solution of the given absolute value inequality.

Substitute −4 for x in the absolute value inequality ∣x∣+2>5, and simplify the left-hand side.

​​∣−4∣+2>5

4+2>5

6>5

​This inequality is true. Therefore, x=−4 is a solution of the given absolute value inequality.

Substitute −3 for x in the absolute value inequality ∣x∣+2>5, and simplify the left-hand side.

​∣−3∣+2>5

3+2>5

5>5

​This inequality is not true. Therefore, x=−3 is not a solution of the given absolute value inequality.

Substitute −2 for x in the absolute value inequality ∣x∣+2>5, and simplify the left-hand side.

​​∣−2∣+2>5

2+2>5

4>5

​This inequality is not true. Therefore, x=−2 is not a solution of the given absolute value inequality.

Substitute −1 for x in the absolute value inequality ∣x∣+2>5, and simplify the left-hand side.

​​∣−1∣+2>5

1+2>5

3>5

​This inequality is not true. Therefore, x=−1 is not a solution of the given absolute value inequality.

Substitute 0 for x in the absolute value inequality ∣x∣+2>5, and simplify the left-hand side.

​​∣0∣+2>5

0+2>5

2>5

​This inequality is not true. Therefore, x=0 is not a solution of the given absolute value inequality.

Substitute 1 for x in the absolute value inequality ∣x∣+2>5, and simplify the left-hand side.

​​∣1∣+2>5

1+2>5

3>5

​This inequality is not true. Therefore, x=1 is not a solution of the given absolute value inequality.

Substitute 2 for x in the absolute value inequality ∣x∣+2>5, and simplify the left-hand side.

​​∣2∣+2>5

2+2>5

4>5

​This inequality is not true. Therefore, x=2 is not a solution of the given absolute value inequality.

Substitute 3 for x in the absolute value inequality ∣x∣+2>5, and simplify the left-hand side.

​∣3∣+2>5

3+2>5

5>5

​This inequality is not true. Therefore, x=3 is not a solution of the given absolute value inequality.

Substitute 4 for x in the absolute value inequality ∣x∣+2>5, and simplify the left-hand side.

​​∣4∣+2>5

4+2>5

6>5

​This inequality is true. Therefore, x=4 is a solution of the given absolute value inequality.

Substitute 5 for x in the absolute value inequality ∣x∣+2>5, and simplify the left-hand side.

​​∣5∣+2>5

5+2>5

7>5

​This inequality is true. Therefore, x=5 is a solution of the given absolute value inequality.

Complete the given table by writing ‘yes’ against the numbers that are solutions of the given absolute value inequality, and ‘no’ against the numbers that are not solutions of the given absolute value inequality.

Plot the obtained solutions on the given number line.

The required solutions are the integers −5,−4,4,5.

The completed table is:

The required plot of the solutions on the number line is:

Page 87 Problem 4 Answer

The given equation is ∣x∣+2=5.

The question requires to write the solutions of the given equation, and then relate them to the solutions obtained in the previous two parts of this exercise.

Given that the integers between −5 and 5 that are the solutions of the absolute value inequality ∣x∣+2<5 are −2,−1,0,1,2, as calculated in part (a) of this exercise.

Given that the integers between −5 and 5 that are the solutions of the absolute value inequality ∣x∣+2>5 are −5,−4,4,5, as calculated in the previous part of this exercise.

To solve the equation, isolate the absolute value on one side of the equation, rewrite the equation as two equations, and solve the equations for x to find the solutions.

Next, observe the solutions obtained in the previous two parts of this exercise, and relate the solutions of the given absolute value equation.

Subtract 2 from both sides of ∣x∣+2=5.

∣x∣=3

Rewriting the equation, the two equations obtained are x=−3 or x=3.

Thus, the two solutions of the given absolute value equation are x=−3 or x=3.

The integers solutions between −5 and 5 of the absolute value inequality ∣x∣+2<5 are −2,−1,0,1,2.

The integers solutions between −5 and 5 of the absolute value inequality ∣x∣+2>5 are −5,−4,4,5.

It can be observed that solutions contain all the integers between −5 and 5, except −3 and 3.

These remaining integers are the solutions of the absolute value equation ∣x∣+2=5.

Thus, the solutions of ∣x∣+2<5, ∣x∣+2>5, and ∣x∣+2=5 are distinct and not repeated, and each integer between −5 and 5 is the solution of either one of the two inequalities, or the equation.

The required solutions are x=−3 or x=3.

The solutions of ∣x∣+2<5 and ∣x∣+2>5 contain all the integers between −5 and 5, except −3 and 3, which are the solutions of the absolute value equation ∣x∣+2=5.

The solutions of ∣x∣+2<5, ∣x∣+2>5, and ∣x∣+2=5 are different and not repeated. Also, each integer between −5 and 5 is the solution of either one of the two inequalities, or the equation.

Page 87 Problem 5 Answer

It is given that x is a real number.

The given absolute value inequalities are ∣x∣+2<5 and ∣x∣+2>5.

The question requires to graph all real solutions of the two absolute value inequalities on the given number lines.

Given that the solutions of the absolute value equation ∣x∣+2=5 are −3,3, as calculated in the previous part of this exercise.

To graph the solutions, observe the given solutions of the absolute value equation ∣x∣+2=5, form intervals using the two solutions, and check the value of the left-hand side of the inequalities for all real values.

Then, graph the solutions on the given number lines for both the inequalities.

The solutions of the absolute value equation ∣x∣+2=5 are −3,3.

When x<−3, the value of ∣x∣+2 is greater than 5, which does not satisfy the inequality ∣x∣+2<5.

When x=−3, the value of ∣x∣+2 is equal to 5, which does not satisfy the inequality ∣x∣+2<5.

When x is between −3 and 3, both not included, the value of ∣x∣+2 is less than 5, which satisfies the inequality ∣x∣+2<5.

When x=3, the value of ∣x∣+2 is equal to 5, which does not satisfy the inequality ∣x∣+2<5.

When x>3, the value of ∣x∣+2 is greater than 5, which does not satisfy the inequality ∣x∣+2<5.

Thus, the solution of the absolute value inequality ∣x∣+2<5 is −3<x<3.

Graph the obtained solutions of ∣x∣+2<5 on the given number line.

Consider the absolute value inequality ∣x∣+2>5.

When x<−3, the value of ∣x∣+2 is greater than 5, which satisfies the inequality ∣x∣+2>5.

When x=−3, the value of ∣x∣+2 is equal to 5, which does not satisfy the inequality ∣x∣+2>5.

When x is between −3 and 3, both not included, the value of ∣x∣+2 is less than 5, which does not satisfy the inequality ∣x∣+2<5.

When x=3, the value of ∣x∣+2 is equal to 5, which does not satisfy the inequality ∣x∣+2>5.

When x>3, the value of ∣x∣+2 is greater than 5, which satisfies the inequality ∣x∣+2>5.

Thus, the solution of the absolute value inequality ∣x∣+2>5 is x<−3 or x>3.

Graph the obtained solutions of ∣x∣+2>5 on the given number line.

The graph of all real solutions of ∣x∣+2<5 is:

The graph of all real solutions of ∣x∣+2>5 is:

Page 88 Problem 6 Answer

The absolute value inequality ∣x∣+2>5 can be simplified as ∣x∣>3.

This equation is true for all values of x greater than 3, that is x>3, and for all values of x less than −3, that is x<−3.

There is no value of x that is within both the intervals x<−3 and x>3.

Therefore, the solution for the given inequality will be written as a disjunction, that is using the word “or”.

Thus, the solution for the inequality ∣x∣+2>5 is the disjunction x<−3 or x>3.

The solution for the given inequality will be written as a disjunction using the word “or” because there is no value of x lying within both the intervals x<−3 and x>3.

Therefore, the required solution for the inequality ∣x∣+2>5 is the disjunction x<−3 or x>3.

Page 88 Problem 7 Answer

The given absolute value inequalities are ∣x∣+2≤5 and ∣x∣+2≥5.

The question requires to describe the solutions of the given absolute value inequalities using inequalities that do not include an absolute value.

Given that the solutions of the absolute value equation ∣x∣+2=5 are −3,3, as calculated in the previous part of this exercise.

To graph the solutions, simplify the given absolute value inequalities, and check the value of the left-hand side of the inequalities for all real values using intervals.

Then, write the solutions for which the inequalities are satisfied as intervals which do not contain any absolute values.

The first given absolute value inequality ∣x∣+2≤5 can be simplified as ∣x∣≤3.

When x<−3, the value of ∣x∣ is greater than 3, which does not satisfy the inequality ∣x∣≤3.

When x=−3, the value of ∣x∣ is equal to 3, which satisfies the inequality ∣x∣≤3.

When x is between −3 and 3, both not included, the value of ∣x∣ is less than 3, which satisfies the inequality ∣x∣≤3.

When x=3, the value of ∣x∣ is equal to 3, which satisfies the inequality ∣x∣≤3.

When x>3, the value of ∣x∣ is greater than 3, which does not satisfy the inequality ∣x∣≤3.

Therefore, the solutions of the absolute value inequality ∣x∣+2≤5 are all the values between the integers −3 and 3, both included.

Thus, the solution of the absolute value inequality ∣x∣+2≤5 is the interval −3≤x≤3.

The second given absolute value inequality ∣x∣+2≥5 can be simplified as ∣x∣≥3.

When x<−3, the value of ∣x∣ is greater than 3, which satisfies the inequality ∣x∣≥3.

When x=−3, the value of ∣x∣ is equal to 3, which satisfies the inequality ∣x∣≥3.

When x is between −3 and 3, both not included, the value of ∣x∣ is less than 3, which does not satisfy the inequality ∣x∣≥3.

When x=3, the value of ∣x∣ is equal to 3, which satisfies the inequality ∣x∣≥3.

When x>3, the value of ∣x∣ is greater than 3, which satisfies the inequality ∣x∣≥3.

Therefore, the solutions of the absolute value inequality ∣x∣+2≥5 are all the values lesser than or equal to −3, or greater than and equal to 3.

Thus, the solution of the absolute value inequality ∣x∣+2≥5 is the disjunction x≤−3 or x≥3

Page 88 Problem 8 Answer

The given absolute value inequality is ∣x−2∣−3<1.

The question requires to solve the given inequality graphically.

To solve the given inequality, let f(x) and g(x) be functions equal to left and right side of the given inequality respectively.

Then, write the translations used to form the function f(x) from the parent absolute value function h(x)=∣x∣, and use them to determine the vertex of f(x).

Next, write the transformations used to form f(x) from the parent absolute value function, take two points lying on the parent absolute value function, and use them to determine two points lying on f(x).

Next, plot the points and vertex, and use them to graph the function f(x) along with the constant function g(x).

Finally, write the intervals of x where the graph of f(x) is below the graph of g(x).

The given inequality is of the form f(x)<g(x).

Let f(x)=∣x−2∣−3 and g(x)=1.

The required solutions are the intervals on the x-axis where the graph of f(x) lies below the graph of g(x).

The vertex of the parent absolute value function h(x)=∣x∣ is (0,0).

The vertex of f(x) is the point where (0,0) will be mapped using translations.

It can be observed that f(x) involves translations of h(x) by 2 units to the right and 3 units down.

Therefore, the vertex of f(x) is the point (0+2,0−3)=(2,−3).

The points (2,2) and (−2,2) lie on the parent absolute value function h(x)=∣x∣.

The transformations used to form f(x)=∣x−2∣−3 from the parent absolute value function can be used to determine the two points lying on f(x).

The x-coordinate of the points lying on h(x) will be shifted 2 units to the right when mapped to f(x).

The y-coordinate of the points lying on h(x) will be moved down 3 units.

Therefore, the point (2,2) moves to (2+2,∣2∣−3)=(4,−1).

The point (−2,2) moves to (−2+2,∣2∣−3)=(0,−1).

Plot the vertex (2,−3) and the points (0,−1) and (4,−1), and then draw straight lines to plot the graph of f(x)=∣x−2∣−3.

Plot the function g(x)=1 as a horizontal line on the grid where f(x) is plotted.

The graph of f(x) lies below the graph of g(x) between x=−2 and x=6.

Thus, the solution of the inequality ∣x−2∣−3<1 is x>−2 and x<6.

The solution of the inequality ∣x−2∣−3<1 is x>−2 and x<6.

The solution is obtained using the graph:

Page 89 Problem 9 Answer

The inequality in part (A) changes from ∣x+3∣+1>4 to ∣x+3∣+1≥4.

Let f(x) be equal to ∣x+3∣+1, and g(x) be equal to 4.

The given inequality is of the form f(x)≥g(x).

The required solutions are the intervals on the x-axis where the graph of f(x) lies above the graph of g(x), or intersects it.From the graph, it can be observed that the graph of f(x)

lies above the graph of g(x) when x<−6 and x>0.

Also, the graph of f(x) intersects the graph of g(x) when x is equal to −6 or 0.

Thus, the solution of the inequality ∣x+3∣+1≥4 is x≤−6 or x≥0.

It can be observed that the solution x<−6 or x>0 of ∣x+3∣+1>4 changes to x≤−6 or x≥0 when the inequality changes to ∣x+3∣+1≥4.

The signs of inequality of the solution changed to include equal to.

The solution of the inequality ∣x+3∣+1≥4 is x≤−6 or x≥0.

The solution x<−6 or x>0 of ∣x+3∣+1>4 changes to x≤−6 or x≥0 when the inequality changes to ∣x+3∣+1≥4.

The signs of inequality of the solution are changed to include equal to in the solution of the new inequality.

Page 89 Problem 10 Answer

The graph of f(x) lies entirely above the graph of g(x).

This means that the graph of f(x) lies above the graph of g(x) for all values of x.

The solution of the inequality f(x)>g(x) is the interval where the graph of f(x) lies above the graph of g(x).

Therefore, the inequality f(x)>g(x) has infinitely many solutions lying in the interval (−∞,∞).

The solution of the inequality f(x)<g(x) is the interval where the graph of f(x) lies below the graph of g(x).

Since the graph of f(x) lies above the graph of g(x) for all values of x, the inequality f(x)<g(x) has no solution.

The inequality f(x)>g(x) has infinitely many solutions lying in the interval (−∞,∞).

The inequality f(x)<g(x) has no solution.

 

Algebra 2 Volume 1 1st Edition Module 2 Absolute Value, Functions, Equations, And Inequalities

Page 77 Problem 1 Answer

The absolute value expression in an absolute value equation is generally of the form ∣ax+b∣.

To solve an absolute value equation, first, the operations of multiplication, division, addition, and subtraction are used to isolate the absolute value expression ∣ax+b∣  on one side of the equation.

If the absolute value expression ∣ax+b∣  is equal to a negative number, then the equation cannot be solved and there is no solution.

If the absolute value expression ∣ax+b∣ is equal to 0, then the equation can be written as the linear equation ax+b=0, which can then be solved to obtain the only solution of the absolute value equation.

If the absolute value expression ∣ax+b∣ is equal to a positive number, then the equation can be written as a disjunction of two linear equations, which can then be solved to obtain the two solutions of the absolute value equation.

An absolute value equation can be solved by isolating the absolute value expression in the equation on one side of the equation, and determining whether the equation can be solved.

Then, the equation can be written as either one linear equation or a disjunction of two linear equations, which can be solved to obtain the required solutions.

Page 77 Problem 2 Answer

The given absolute value equation is 2∣x−5∣−4=2. The steps to solve the equation graphically are given.

The question requires to solve the given equation.

To solve the given equation, let f(x) and g(x) be functions equal to left and right side of the given equation respectively.

Then, write the translations used to form the function f(x) from the parent absolute value function h(x)=∣x∣, and use them on the vertex to determine the vertex of f(x).

Next, write the coefficient of the expression with the independent variable in the given function to write the stretch or compress required.

Then, take two points lying on the parent absolute value function, and determine the location where they will be mapped on f(x) using the transformations used to form f(x) from h(x).

Next, plot the points and vertex, and use them to graph the function f(x)  along with the constant function g(x), and mark the points of intersection.

Finally, write the x-coordinates of those points to obtain the required solutions.

Let f(x)=2∣x−5∣−4 and g(x)=2.

The vertex of the parent absolute value function h(x)=∣x∣ is (0,0).

The vertex of f(x) is the point where (0,0) will be mapped using translations.

It can be observed that f(x) involves translations of h(x) by 5 units to the right and 4 units down.

Therefore, the vertex of f(x) is the point (0+5,0−4)=(5,−4).

The points (2,2) and (−2,2) lie on the parent absolute value function h(x)=∣x∣.

Compare the function f(x) to f(x)=a∣x−h∣−k, and write the value of the constant a.

a=2

Since ∣a∣=2, that is ∣a∣>1, the graph of f(x) will involve a vertical stretch of the parent absolute value function by a factor 2.

The transformations used to form f(x)=2∣x−5∣−4 from the parent absolute value function can be used to determine the two points lying on f(x).

The x-coordinate of the points lying on h(x) will be shifted 5 units to the right when mapped to f(x).

The y-coordinate of the points lying on h(x) will be vertically stretched by a factor of 2, and then moved down 4 units.

Therefore, the point (2,2) moves to (2+5,2⋅∣2∣−4)=(7,0).

The point (−2,2) moves to (−2+5,2⋅∣2∣−4)=(3,0).

Plot the vertex (5,−4) and the points (7,0) and (3,0), and then draw straight lines to plot the graph of f(x)=2∣x−5∣−4.

Plot the function g(x)=2 as a horizontal line on the grid where f(x) is plotted and mark the points of intersection of f(x) and g(x).

It can be observed that the points of intersection are (2,2) and (8,2).

Thus, the solutions of the given equation are the x-coordinates of the points of intersection, that is 2 and 8.

The solutions of the given equation are the x-coordinates of the points of intersection, that is 2 and 8.

Page 77 Problem 3 Answer

The given function is 2∣x−5∣−4=2. Let f(x) be equal to 2∣x−5∣−4, and g(x) be equal to 2.

The points of intersection of the graphs of the two functions are (2,2) and (8,2), as calculated in the previous part of this exercise.

Thus, the solutions of the given equation are the x-coordinates of the points of intersection, that is 2 and 8.

This can be written as the disjunction x equal to 2 or x equal to 8.

The required solution of the given equation is the disjunction x=2 or x=8.

Page 77 Problem 4 Answer

Assume an absolute value equation ∣ax+b∣=c.The two equations formed by rewriting the equation are ax+b equal to −c or ax+b equal to c.

The two equations formed have one solution each because they are linear equations.

Therefore, the absolute value equation has two solutions.

The absolute value equation ∣ax+b∣=c cannot have three or four solutions because the expression inside the absolute value parentheses is a linear expression, that is an expression with degree 1.

Therefore, the two equations formed are linear equations which one possible solution each.

Since most absolute value equations have a linear expression of the form ax+b inside the absolute value parentheses, most absolute value equations are expected to have two solutions.

Most absolute value equations are expected to have two solutions because most absolute value equations have a linear expression of the form ax+b inside the absolute value parentheses.

Most absolute value equations generally do not have three or four solutions because the two equations formed by rewriting the common absolute value equation ∣ax+b∣=c are linear equations with only one solution each.

Page 77 Problem 5 Answer

Assume an absolute value equation ∣ax+b∣=c.If c is equal to a negative number, then the equation cannot be solved because the absolute value expression ∣ax+b∣ cannot be equal to the negative number c.

Therefore, an absolute value equation will have no solution if the equation is of the form ∣ax+b∣=c, where c<0.

The graph of an absolute value equation with no solution does not contain any points because there is no point which satisfies the equation.

If c is equal to 0, then the equation can be written as the linear equation ax+b=c, which has only one solution.

Therefore, an absolute value equation will have only one solution if the equation is of the form ∣ax+b∣=0.

The graph of an absolute value equation with one solution is a straight line because it can be written as a linear equation.

An absolute value equation will have no solution if the equation is of the form ∣ax+b∣=c, where c<0.

The graph of such an absolute value equation does not contain any points.

An absolute value equation will have only one solution if the equation is of the form ∣ax+b∣=0.

The graph of such an absolute value equation is a straight line.

Page 78 Problem 6 Answer

The given equation is 1/2∣x+2∣=10.

The question requires to solve the given equation algebraically, and then graph its solutions on the number line.

To solve the equation and graph the solutions, isolate the absolute value on one side of the equation, rewrite the equation as two equations, and solve the equations for x to find the solutions.

Next, mark the solutions on the given number line.

Multiply both sides of 1/2∣x+2∣=10 by 2.

∣x+2∣=20

Rewriting the equation, the two equations obtained are x+2=20 or x+2=−20.

Subtract 2 from both sides of x+2=20.

x=18 Subtract 2 from both sides of x+2=−20.

x=−22

Thus, the two solutions are x=−22 or x=18.

Graph the obtained solutions on the given number line.

The required solutions are x=−22 or x=18.

The required graph of the solutions on the number line is:

Page 77 Problem 7 Answer

The given equation is −2∣3x−6∣+5=1.

The question requires to solve the given equation algebraically, and then graph its solutions on the number line.

To solve the equation and graph the solutions, isolate the absolute value on one side of the equation, rewrite the equation as two equations, and solve the equations for x to find the solutions.

Next, mark the solutions on the given number line.

Subtract 5 from both sides of the equation, and divide both sides by −2.​

−2∣3x−6∣=−4

∣3x−6∣=−4/−2

∣3x−6∣=2​

Rewriting the equation, the two equations obtained are 3x−6=2 or 3x−6=−2.

Add 6 on both sides of 3x−6=2 and divide both sides by 3.​

3x=8

x=8/3

Add 6 on both sides of 3x−6=−2 and divide both sides by 3.​

3x=4

x=4/3

Thus, the two solutions are x=4/3 or x=8/3.

Graph the obtained solutions on the given number line.

The required solutions are x=4/3 or x=8/3.

The required graph of the solutions on the number line is:

Page 78 Problem 8 Answer

The given equation is 3∣4x−5∣−2=19.

The question requires to solve the given equation algebraically, and then graph its solutions on the number line.

To solve the equation and graph the solutions, isolate the absolute value on one side of the equation, rewrite the equation as two equations, and solve the equations for x to find the solutions.

Next, mark the solutions on the given number line.

Add 2 on both sides of the equation, and divide both sides by 3.

​3∣4x−5∣=21

∣4x−5∣=7

Rewriting the equation, the two equations obtained are 4x−5=7 or 4x−5=−7.

Add 5 on both sides of 4x−5=7 and divide both sides by 4.

​4x=12

x=3

​Add 5 on both sides of 4x−5=−7 and divide both sides by 4.

​4x=−2

x=−2/4

x=−1/2

Thus, the two solutions are x=−1/2 or x=3.

Graph the obtained solutions on the given number line.

The required solutions are x=−1/2 or x=3.

The required graph of the solutions on the number line is:

Page 79 Problem 9 Answer

The given equation is 9∣4/3 x−2∣+7=7.

The question requires to isolate the absolute value expression on one side of the equation and determine if the given equation has a solution.

Then, it is required to write “no solution” if there is no solution, or solve the equation if it has a solution.

To answer the question, isolate the absolute value expression on one side of the equation, and check whether the right-hand side of the equation is non-negative or not.

Then, rewrite the equation, and solve the equation for x to find the solution.

Subtract 7 from both sides of the equation, and divide both sides by 9.​

9∣4/3x−2∣=0

∣4/3x−2∣=0​

The absolute value expression ∣4/3x−2∣ is equal to the non-negative number 0.

Thus, the equation can be solved.

Rewriting the equation, the equation obtained is 4/3

x−2=0.

Add 2 on both sides of 4/3x−2=0 and multiply both sides by 3/4.​

4/3

x=2

x=6/4

x=3/2

Thus, the solution is x=3/2.

Yes, the given equation can be solved. The required solution is x=3/2.

Page 79 Problem 10 Answer

The given equation is 3/5∣2x−4∣−3=−3.

The question requires to isolate the absolute value expression on one side of the equation and determine if the given equation has a solution.

Then, it is required to write “no solution” if there is no solution, or solve the equation if it has a solution.

To answer the question, isolate the absolute value expression on one side of the equation, and check whether the right-hand side of the equation is non-negative or not.

Then, rewrite the equation, and solve the equation for x to find the solution.

Add 3 on both sides of the equation, and multiply both sides by 5/3.​

3/5

∣2x−4∣=0

∣2x−4∣=0

The absolute value expression ∣2x−4∣ is equal to the non-negative number 0.

Thus, the equation can be solved.

Rewriting the equation, the equation obtained is 2x−4=0.

Add 4 on both sides of 2x−4=0 and divide both sides by 2.

2x=4

x=2​

Thus, the solution is x=2.

Yes, the given equation can be solved. The required solution is x=2.

Page 81 Exercise 1 Answer

The given absolute value equation is 2∣x+1∣+5=9.

The question requires to solve the given absolute value equation graphically.

To solve the given equation graphically, let f(x) and g(x) be functions equal to left and right side of the given equation respectively.

Then, write the translations used to form the function f(x) from the parent absolute value function h(x)=∣x∣, and use them to determine the vertex of f(x).

Next, write the transformations used to form the points on the function f(x) from the points on the parent absolute value function, and use two points lying on the parent absolute value function to determine two points on f(x).

Next, plot the points and vertex, and use them to graph the function f(x)  along with the constant function g(x), and mark the points of intersection.

Finally, write the x-coordinates of those points as a disjunction to obtain the required solution.

Let f(x)=2∣x+1∣+5 and g(x)=9.

The vertex of the parent absolute value function h(x)=∣x∣ is (0,0).

The vertex of f(x) is the point where (0,0) will be mapped using translations.

It can be observed that f(x) involves translations of h(x) by 1 unit to the left and 5 units up.

Therefore, the vertex of f(x) is the point (0−1,0+5)=(−1,5).

The points (2,2) and (−2,2) lie on the parent absolute value function h(x)=∣x∣. Compare the function f(x) to f(x)=a∣x+h∣+k, and write the value of the constant a.

a=2

Since ∣a∣=2, that is ∣a∣>1, the graph of f(x) will involve a vertical stretch of the parent absolute value function by a factor 2.

The transformations used to form f(x)=2∣x+1∣+5 from the parent absolute value function can be used to determine the two points lying on f(x).

The x-coordinate of the points lying on h(x) will be shifted 1 unit to the left when mapped to f(x).

The y-coordinate of the points lying on h(x) will be vertically stretched by a factor of 2, and then moved up 5 units.

Therefore, the point (2,2) moves to (2−1,2⋅∣2∣+5)=(1,9).

The point (−2,2) moves to (−2−1,2⋅∣2∣+5)=(−3,9).

Plot the vertex (−1,5) and the points (−3,9) and (1,9), and then draw straight lines to plot the graph of f(x)=2∣x+1∣+5.

Plot the function g(x)=9 as a horizontal line on the grid where f(x) is plotted and mark the points of intersection of f(x) and g(x).

It can be observed that the points of intersection are (−3,9) and (1,9).

Thus, the solutions of the given equation are the x-coordinates of the points of intersection, that is −3 and 1. This can be written as the disjunction x=−3 or x=1.

The required solution of the given equation is the disjunction x=−3 or x=1.

Page 81 Exercise 2 Answer

The given absolute value equation is −2∣x+5∣+4=2.

The question requires to solve the given absolute value equation graphically.

To solve the given equation graphically, let f(x) and g(x) be functions equal to left and right side of the given equation respectively.

Then, write the translations used to form the function f(x) from the parent absolute value function h(x)=∣x∣, and use them to determine the vertex of f(x).

Next, write the transformations used to form the points on the function f(x) from the points on the parent absolute value function, and use two points lying on the parent absolute value function to determine two points on f(x).

Next, plot the points and vertex, and use them to graph the function f(x) along with the constant function g(x), and mark the points of intersection.

Finally, write the x-coordinates of those points as a disjunction to obtain the required solution.

Let f(x)=−2∣x+5∣+4 and g(x)=2.

The vertex of the parent absolute value function h(x)=∣x∣ is (0,0).

The vertex of f(x) is the point where (0,0) will be mapped using translations.

It can be observed that f(x) involves translations of h(x) by 5 units to the left and 4 units up.

Therefore, the vertex of f(x) is the point (0−5,0+4)=(−5,4).

The points (2,2) and (−2,2) lie on the parent absolute value function h(x)=∣x∣.

Compare the function f(x) to f(x)=a∣x+h∣+k, and write the value of the constant a.

a=−2

Since ∣a∣=2, that is ∣a∣>1, the graph of f(x) will involve a vertical stretch of the parent absolute value function by a factor 2.

Since a=−2, that is a<0, the graph of f(x) will involve a reflection of the parent absolute value function about the x-axis.

The transformations used to form f(x)=−2∣x+5∣+4 from the parent absolute value function can be used to determine the two points lying on f(x).

The x-coordinate of the points lying on h(x) will be shifted 5 units to the left when mapped to f(x).

The y-coordinate of the points lying on h(x) will be reflected about the x-axis, vertically stretched by a factor of 2, and then moved up 4 units.

Therefore, the point (2,2) moves to (2−5,−2⋅∣2∣+4)=(−3,0).

The point (−2,2) moves to (−2−5,−2⋅∣2∣+4)=(−7,0).

Plot the vertex (−5,4) and the points (−3,0) and (−7,0), and then draw straight lines to plot the graph of f(x)=−2∣x+5∣+4.

Plot the function g(x)=2 as a horizontal line on the grid where f(x) is plotted and mark the points of intersection of f(x) and g(x).

It can be observed that the points of intersection are (−6,2) and (−4,2).

Thus, the solutions of the given equation are the x-coordinates of the points of intersection, that is −6 and −4. This can be written as the disjunction x=−6 or x=−4.

The required solution of the given equation is the disjunction x=−6 or x=−4.

Page 81 Exercise 3 Answer

The given absolute value equation is ∣3/2(x−2)∣+3=2.

The question requires to solve the given absolute value equation graphically.

To solve the given equation graphically, let f(x) and g(x) be functions equal to left and right side of the given equation respectively.

Then, write the translations used to form the function f(x) from the parent absolute value function h(x)=∣x∣, and use them to determine the vertex of f(x).

Next, write the transformations used to form the points on the function f(x) from the points on the parent absolute value function, and use two points lying on the parent absolute value function to determine two points on f(x).

Finally, plot the points and vertex, use them to graph the function f(x) along with the constant function g(x), and check the points of intersection to determine the solution if any.

Let f(x)=∣3/2(x−2)∣+3 and g(x)=2.

The vertex of the parent absolute value function h(x)=∣x∣ is (0,0).

The vertex of f(x) is the point where (0,0) will be mapped using translations.

It can be observed that f(x) involves translations of h(x) by 2 units to the right and 3 units up.

Therefore, the vertex of f(x) is the point (0+2,0+3)=(2,3).

The points (2,2) and (−2,2) lie on the parent absolute value function h(x)=∣x∣.

Compare the function f(x) to f(x)=∣1/b(x−h)∣+k, and write the value of the constant b.

​1/b=3/2

b=2/3

Since ∣b∣=2/3, that is 0<∣b∣<1, the graph of f(x) will involve a horizontal stretch of the parent absolute value function by a factor 2/3.

The transformations used to form f(x)=∣3/2(x−2)∣+3 from the parent absolute value function can be used to determine the two points lying on f(x).

The x-coordinate of the points lying on h(x) will be horizontally stretched by a factor of 2/3, and then shifted 2 units to the right when mapped to f(x).

The y-coordinate of the points lying on h(x) will be moved up 3 units.

Therefore, the point (2,2) moves to (2/3⋅2+2,∣2∣+3)=(10/3,5).

The point (−2,2) moves to (2/3⋅(−2)+2,∣2∣+3)=(2/3,5).

Plot the vertex (2,3) and the points (2/3,5) and (10/3,5), and then draw straight lines to plot the graph of f(x)=∣3/2(x−2)∣+3.

Plot the function g(x)=2 as a horizontal line on the grid where f(x) is plotted.

It can be observed that the graphs do not intersect at any point.

Therefore, there is no value of x for which f(x) is equal to g(x), that is for which ∣3/2(x−2)∣+3=2.

Thus, the given equation has no solution.

The given absolute value equation has no solution.

Page 81 Exercise 4 Answer

The given equation is ∣2x∣=3.

The question requires to solve the given equation algebraically, and then graph its solutions on the number line.

To solve the equation and graph the solutions, rewrite the equation as two equations, and solve the equations for x to find the solutions.

Next, mark the solutions on the given number line.

Rewriting the given equation, the two equations obtained are 2x=3 or 2x=−3.

Divide both sides of 2x=3 by 2.

​2x/2=3/2

x=1.5

​Divide both sides of 2x=−3 by 2.

​2x/2=−3/2

x=−1.5

Thus, the two solutions are x=−1.5 or x=1.5.

Graph the obtained solutions on the given number line.

The required solutions are x=−1.5 or x=1.5.

The required graph of the solutions on the number line is:

Page 81 Exercise 5 Answer

The given equation is ∣1/3x+4∣=3.

The question requires to solve the given equation algebraically, and then graph its solutions on the number line.

To solve the equation and graph the solutions, rewrite the equation as two equations, and solve the equations for x to find the solutions.

Next, mark the solutions on the given number line.

Rewriting the given equation, the two equations obtained are 1/3x+4=3 or 1/3x+4=−3.

Subtract 4 from both sides of 1/3x+4=3, and multiply both sides by 3.​

1/3

x=−1

x=−3​

Subtract 4 from both sides of 1/3x+4=−3, and multiply both sides by 3.

1/3

x=−7

x=−21

Thus, the two solutions are x=−21 or x=−3.

Graph the obtained solutions on the given number line.

The required solutions are x=−21 or x=−3.

The required graph of the solutions on the number line is:

Page 82 Exercise 6 Answer

The given equation is 3∣2x−3∣+2=3.

The question requires to solve the given equation algebraically, and then graph its solutions on the number line.

To solve the equation and graph the solutions, isolate the absolute value on one side of the equation, rewrite the equation as two equations, and solve the equations for x to find the solutions.

Next, mark the solutions on the given number line.

Subtract 2 from both sides of the equation, and divide both sides by 3.

​3∣2x−3∣=1

∣2x−3∣=1/3

Rewriting the equation, the two equations obtained are 2x−3=1/3 or 2x−3=−1/3.

Add 3 on both sides of 2x−3=1/3 and divide both sides by 2.

​2x=1/3+3

2x=10/3

x=5/3

Add 3 on both sides of 2x−3=−1/3 and divide both sides by 2.

​2x=−1/3+3

2x=8/3

x=4/3

Thus, the two solutions are x=4/3 or x=5/3.

Graph the obtained solutions on the given number line.

The required solutions are x=4/3 or x=5/3.

The required graph of the solutions on the number line is:

Page 82 Exercise 7 Answer

The given equation is −8∣−x−6∣+10=2.

The question requires to solve the given equation algebraically, and then graph its solutions on the number line.

To solve the equation and graph the solutions, isolate the absolute value on one side of the equation, rewrite the equation as two equations, and solve the equations for x to find the solutions.

Next, mark the solutions on the given number line.

Subtract 10 from both sides of the equation, and divide both sides by −8.

​−8∣−x−6∣=−8

∣−x−6∣=−8/−8

∣−x−6∣=1​

Rewriting the equation, the two equations obtained are −x−6=1 or −x−6=−1.

Add 6 on both sides of −x−6=1, and multiply both sides by −1.

−x=7

−x(−1)=7(−1)

x=−7

​Add 6 on both sides of −x−6=−1, and multiply both sides by −1.

​−x=5

−x(−1)=5(−1)

x=−5

Thus, the two solutions are x=−5 or x=−7.

Graph the obtained solutions on the given number line.

The required solutions are x=−5 or x=−7.

The required graph of the solutions on the number line is:

Page 82 Exercise 8 Answer

The given equation is −3∣x−3∣+3=6.

The question requires to isolate the absolute value expression on one side of the equation and determine if the given equation has a solution.

Then, it is required to write “no solution” if it cannot be solved, or solve the equation and graph the solution(s) if it can be solved.

To answer the question, isolate the absolute value expression on one side of the equation, and check whether the right-hand side of the equation is non-negative or not to determine if the equation can be solved.

Subtract 3 from both sides of the equation, and divide both sides by −3.

​−3∣x−3∣=3

∣x−3∣=3/−3

∣x−3∣=−1

The absolute value expression ∣x−3∣ cannot be equal to the negative number −1.

Thus, the equation cannot be solved, and has no solution.

No, the equation cannot be solved, and has no solution.

Page 82 Exercise 9 Answer

The given equation is 2(∣x+4∣+3)=6.

The question requires to isolate the absolute value expression on one side of the equation and determine if the given equation has a solution.

Then, it is required to write “no solution” if it cannot be solved, or solve the equation and graph the solution(s) if it can be solved.

To answer the question, isolate the absolute value expression on one side of the equation, and check whether the right-hand side of the equation is non-negative or not.

Then, rewrite the equation, and solve the equation for x to find the solution. Finally, graph the solution on the given number line.

Divide both sides of the equation 2(∣x+4∣+3)=6 by 2, and then subtract 3 from both sides.

​∣x+4∣+3=3

∣x+4∣=0

The absolute value expression ∣x+4∣ is equal to the non-negative number 0.

Thus, the equation can be solved.

Rewriting the equation, the equation obtained is x+4=0.

Subtract 4 from both sides of x+4=0.

x=−4

Thus, the solution is x=−4.

Graph the obtained solution on the given number line.

Yes, the given equation can be solved. The required solution is x=−4.

The required graph of the solution on the number line is:

Page 82 Exercise 10 Answer

The given equation is 7∣1/2x+31/2∣−2=5.

The question requires to solve the given absolute value equation.

To solve the equation, isolate the absolute value on one side of the equation, rewrite the equation as two equations, and solve the equations for x to find the solutions.

Next, mark the solutions on the given number line.

Add 2 on both sides of the equation 7∣1/2x+31/2∣−2=5, rewrite the mixed fraction as an improper fraction, and divide both sides by 7.

​7∣1/2x+31/2∣=7

7∣1/2x+7/2∣=7

∣1/2x+7/2∣=1

​Rewriting the equation, the two equations obtained are 1/2x+7/2=1 or 1/2x+7/2=−1.

Subtract 7/2 from both sides of 1/2x+7/2=1 and multiply both sides by 2.

​1/2x=1−7/2

1/2x=−5/2

x=−5​

Subtract 7/2 from both sides of 1/2x+7/2=−1 and multiply both sides by 2.

​1/2x=−1−7/2

1/2x=−9/2

x=−9

Thus, the two solutions are x=−9 or x=−5.

Graph the obtained solutions on the given number line.

The required solutions are x=−9 or x=−5.

The required graph of the solutions on the number line is:

Page 83 Exercise 11 Answer

The given equation is −5∣−3x+2∣−2=−2.

The question requires to solve the given absolute value equation.

To solve the equation, isolate the absolute value on one side of the equation, rewrite the equation, and solve the equation for x to find the solution.

Next, mark the solution on the given number line.

Add 2 on both sides of the equation −5∣−3x+2∣−2=−2, and then divide both sides by −5.

​−5∣−3x+2∣=0

∣−3x+2∣=0

Rewriting the equation, the equation obtained is −3x+2=0.

Add 3x on both sides of −3x+2=0, and divide both sides by 3.

2=3x

2/3=x

Thus, the solution is x=2/3.

Graph the obtained solution on the given number

The required solution is x=2/3.

The solution can be represented on the given number line as:

Page 81 Exercise 12 Answer

It is given that the depth of the river bottom in feet is given by the absolute value function d(h)=1/5

∣h−240∣−48, where h is the horizontal distance from the left-hand shore in feet. Also, it is given that buoys are to be placed where the river bottom is 30 ft below the surface.

The question requires to determine the horizontal distance from the left-hand shore where the buoys should be placed.

To determine the horizontal distances, substitute the depth of the river bottom as −30

ft in the given absolute value function, isolate the absolute value on one side of the equation, rewrite the equation as two equations, and solve the equations for x

to find the solutions, and thus, the horizontal distances from the left-hand shore where the buoys should be placed.

The depth of the river bottom is 30 ft when d(h)=−30.

Substitute −30 for d(h) in d(h)=1/5∣h−240∣−48, add 48 on both sides, and multiply both sides by 5.

​−30=1/5∣h−240∣−48

18=1/5∣h−240∣

90=∣h−240∣

Rewriting the equation, the two equations obtained are h−240=90 or h−240=−90.

Add 240 on both sides of h−240=90.

h=330 Add 240 on both sides of h−240=−90.

h=150

Thus, the buoys should be placed 150 feet and 330 feet away from the left-hand shore.

The buoys should be placed 150 feet and 330 feet away from the left-hand shore.

Page 84 Exercise 13 Answer

Subtract 2 from both sides of the equation −1/2∣3x−3∣+2=1, and multiply both sides by −2.

−1/2

∣3x−3∣=−1

∣3x−3∣=2

​Rewriting the equation, the two equations obtained are 3x−3=−2 or 3x−3=2.

Add 3 on both sides of 3x−3=−2 and divide both sides by 3.

​3x=1

x=1/3

Add 3 on both sides of 3x−3=2 and divide both sides by 3.

​3x=5

x=5/3

Therefore, the two solutions are x=1/3 or x=5/3.Thus, the values of x satisfying the given equation are option (A) x=5/3and option (C) x=1/3.

The given equation is written as ∣3x−3∣=2 where the absolute value expression is isolated.

When x is equal to −5/3, the value of ∣3(−5/3)−3∣ is ∣−5−3∣=8 and not 2. Thus, option (B) is incorrect.

When x is equal to −1/3, the value of ∣3(−1/3)−3∣ is ∣−1−3∣=4 and not 2. Thus, option (D) is incorrect.

When x is equal to 3, the value of ∣3(3)−3∣ is ∣9−3∣=6 and not 2. Thus, option (E) is incorrect.

When x is equal to −3, the value of ∣3(−3)−3∣ is ∣−9−3∣=12 and not 2. Thus, option (F) is incorrect.

When x is equal to 1, the value of ∣3(1)−3∣ is ∣3−3∣=0 and not 2. Thus, option (G) is incorrect.

When x is equal to −1, the value of ∣3(−1)−3∣ is ∣−3−3∣=6 and not 2. Thus, option (H) is incorrect.

The values of x satisfying the given equation are option (A) x=5/3 and option (C) x=1/3.

Page 84 Exercise 14 Answer

The given equation is −3∣x−4∣−4=3. The incorrect solution for solving the given equation is given.The given solution involves the equation ∣x−4∣=−7/3.

The absolute value ∣x−4∣ cannot be equal to the negative number −7/3.

Therefore, the equation cannot be solved and has no solution.

Thus, the error in the solution is that the equation ∣x−4∣=−7/3 cannot be rewritten as the given disjunction of the two linear equations.

The correct answer is that the given equation cannot be solved and has no solution.

The error in the solution is that the equation ∣x−4∣=−7/3  is never true, and cannot be rewritten as the given disjunction of the two linear equations.

There is no correct solution because the given equation cannot be solved, and thus, has no solution.

Algebra 2, Volume 1 1st Edition Module 2 Absolute Value, Functions, Equations And Inequalities

Page 64 Problem 1 Answer

The given equation is r+9=7.

The question requires to solve the given equation.

To find the required value, use subtraction in the given equation to isolate the variable r on one side of the equation.

Subtract 9 from both sides of the given equation, and combine the terms.

​r+9−9=7−9

r=−2

Thus, the solution is r=−2.

The required solution is r=−2.

Page 64 Problem 2 Answer

The given equation is w/4=−3.

The question requires to solve the given equation.

To find the required value, use multiplication in the given equation to isolate the variable w on one side of the equation.

Multiply both sides of w/4=−3 by 4.

​W/4(4)=−3(4)

w=−12

Thus, the solution is w=−12.

The required solution is w=−12.

Page 64 Problem 3 Answer

The given equation is 10b=14.

The question requires to solve the given equation.

To find the required value, use division in the given equation to isolate the variable b on one side of the equation.

Divide both sides of 10b=14 by 10.

​10b/10=14/10

b=1.4

Thus, the solution is b=1.4.

The required solution is b=1.4.

Page 64 Problem 4 Answer

The given equation is 3y=2(x−3).

The question requires to determine the slope and y-intercept.

To determine the required slope and y-intercept, isolate the variable y on one side of the equation using division to rewrite it in slope-intercept form.

Then, compare the obtained equation with the slope-intercept form y=mx+b to find the values of m and b, and thus, the required slope and y-intercept.

Multiply the terms in the equation 3y=2(x−3), and then divide both sides by 3.

​3y=2x−6

3y/3=2x−6/3

y=2/3

x−2…………(1)

​Compare equation (1) with the slope-intercept form y=mx+b, and write the values of m and b.​

m=2/3

b=−2

Thus, the slope and y-intercept of the given equation are 2/3 and −2respectively.

The required slope and y-intercept are 2/3 and −2 respectively.

Page 64 Problem 5 Answer

The given inequality is y≥−x+2.

The question requires to graph and label the given inequality on the coordinate plane.

To graph, the inequality, write the equation for the boundary line of the given inequality, and compare it to the slope-intercept form y=mx+b to find the values of m and b, and thus, find the slope and y-intercept of the boundary line.

Then, use the slope and y-intercept to find two points lying on the boundary line.

Next, use an arbitrary point to determine the direction in which the graph of the inequality will be shaded.

Finally, use the obtained points to graph and label the given inequality on the coordinate plane.

The equation of the boundary line for the inequality y≥−x+2 is: y=−x+2

Compare the equation of the boundary line with the slope-intercept form y=mx+b, and write the values of m and b.​

m=−1

b=2

Therefore, the slope and y-intercept are −1 and 2 respectively.

Thus, the boundary line passes through (0,2).

The slope of the boundary line is −1, that is −1/1.

The point (0,2) lies on the boundary line.

Therefore, a second point lying on the boundary line is 0+1,2−1)=(1,1).

Assume the arbitrary point (0,0).

Substitute 0 for x and 0 for y in the inequality y≥−x+2.

0≥−0+2

0≥2

Therefore, the point (0,0) does not satisfy the inequality y≥−x+2.

Thus, the inequality y≥−x+2 will be shaded away from the  the origin (0,0).

Graph the y-intercept of (0,2), plot the second point (1,1), draw a solid line through the points, and shade the region above the line, that is away from the origin (0,0).

Then, label the inequality graphed.

The required labelled graph is

 

Page 64 Problem 6 Answer

The given inequality is y<x−1.

The question requires to graph and label the given inequality on the coordinate plane.

To graph the inequality, write the equation for the boundary line of the given inequality, and compare it to the slope-intercept form y=mx+b to find the values of m and b, and thus, find the slope and y-intercept of the boundary line.

Then, use the slope and y-intercept to find two points lying on the boundary line.

Next, use an arbitrary point to determine the direction in which the graph of the inequality will be shaded.

Finally, use the obtained points to graph and label the given inequality on the coordinate plane.

The equation of the boundary line for the inequality y<x−1 is: y=x−1

Compare the equation of the boundary line with the slope-intercept form y=mx+b, and write the values of m and b.

​m=1

b=−1

Therefore, the slope and y-intercept are 1 and −1 respectively.

Thus, the boundary line passes through (0,−1).

The slope of the boundary line is 1, that is 1/1.

The point (0,−1) lies on the boundary line.

Therefore, a second point lying on the boundary line is 0+1,−1+1)=(1,0).

Assume the arbitrary point (0,0).

Substitute 0 for x and 0 for y in the inequality y<x−1.

​0<0−1

0<−1

Therefore, the point (0,0) does not satisfy the inequality y<x−1.

Thus, the inequality y<x−1 will be shaded away from the origin (0,0).

Graph the y-intercept of (0,−1), plot the second point (1,0), draw a dashed line through the points, and shade the region below the line, that is away from the origin (0,0).

Then, label the inequality graphed.

The required labelled graph is:

The features of the graph of an absolute value function can be identified by sketching its graph and observing the features.

First, the input-output table for the absolute value function has to be constructed.

Next, the coordinates of a few points lying on the graph are written and plotted.

Then, the graph of the absolute value function is graphed using the points.

The drawn graph can be observed to find the features of the absolute value function such as its shape, symmetry, domain, and range.

The features of the graph of an absolute value function can be identified by sketching its graph and observing the features such as its shape, symmetry, domain, and range.

Page 65 Problem 7 Answer

The given table shows the values of some inputs of the absolute value function f(x)=∣x∣, which can be written as the piecewise function:

f(x)={x x≥0−x x<0

The question requires to complete the given input-output table.

To complete the table, use the given piecewise function to determine the value of output for each input, and then use the outputs to complete the input-output table.

Substitute −8 for x in the absolute value function f(x)=∣x∣, and simplify using the piecewise definition of the absolute value function.

f(−8)=∣−8∣

=−(−8)

=8

​Therefore, when the input is −8, the output is 8.

Substitute −4 for x in the absolute value function f(x)=∣x∣, and simplify using the piecewise definition of the absolute value function.

​f(−4)=∣−4∣

=−(−4)

=4

Therefore, when the input is −4, the output is 4.

Substitute 0 for x in the absolute value function f(x)=∣x∣, and simplify using the piecewise definition of the absolute value function.

​f(0)=∣0∣

=0

Therefore, when the input is 0, the output is 0.

Substitute 4 for x in the absolute value function f(x)=∣x∣, and simplify using the piecewise definition of the absolute value function.

​f(4)=∣4∣

=4

​Therefore, when the input is 4, the output is 4.

Substitute 8 for x in the absolute value function f(x)=∣x∣, and simplify using the piecewise definition of the absolute value function.

​f(8)=∣8∣

=8

Therefore, when the input is 8, the output is 8.

Complete the given input-output table using the calculated outputs.

The completed input-output table is:

Page 65 Problem 8 Answer

It is given that the absolute value function f(x)=∣x∣  can be written as the piecewise function:

f(x)={ x x≥0−x x<0​

The question requires to plot the points obtained from the previous part of this exercise on the coordinate grid, and use the points to complete the graph of the absolute value function.

Given the input-output table completed in the previous part of this exercise:

To plot the points and graph the function, use the input-output table to write the coordinates of points lying on the graph of f(x)=∣x∣, and connect the points to sketch the graph of the function.

The points lying on the graph of f(x)=∣x∣ are of the form (x,f(x)).

From the input-output table, it can be determined that the points (−8,8),(−4,4),(0,0),(4,4),(8,8) lie on the graph of f(x)=∣x∣.

Plot the points on the coordinate grid.

Join the points and draw straight lines to form the graph of the function f(x)=∣x∣.

The required plot of the points, and the graph of the function f(x)=∣x∣ is:

Page 65 Problem 9 Answer

The given incomplete statements state the features of the graph of the absolute value function f(x)=∣x∣.

The question requires to complete the given statements about the features of the absolute value function, using the graph sketched in the previous part of this exercise.

Given the graph of the absolute value function f(x)=∣x∣ sketched in the previous part of this exercise:

To complete the statements, observe the graph of the function and use it to determine the line about which the function is symmetric, its domain, and its range.

From the graph of the absolute value function, it can be observed that f(x)=∣x∣ is symmetric about the y-axis.

This means that the absolute value function f(x)=∣x∣ is an even function.

It can be observed that the graph of the function f(x)=∣x∣ stretches infinitely towards the left and right. Therefore, the absolute value function is defined for all values of x.

Hence, the domain of f(x)=∣x∣ is all real numbers.

It can be observed that the points lying on the graph of the function f(x)=∣x∣lie either on, or above the x-axis. Therefore, the values of the outputs f(x) can be either 0, or greater than 0.

Thus, the range of f(x)=∣x∣ is all non-negative real numbers.

The function f(x)=∣x∣ is symmetric about the y-axis, and therefore is an even function.

The domain of f(x)=∣x∣ is all real numbers.

The range of f(x)=∣x∣ is all non-negative real numbers.

Page 66 Problem 10 Answer

The given function is g(x)= −1/2(x+3)+1 . The incomplete solution to graph the function is given.

The question requires to predict how the graph of the given function will look, and verify this using a graphing calculator.

Then, it is required to sketch the graph of the given function.

To sketch the graph, write the translations used to form the given function from the parent absolute value function f(x)=∣x∣, and use them on the vertex to determine the vertex of g(x).

Next, write the coefficient of the expression with the independent variable in the given function to write the stretch or compress required.

Then, take two points lying on the parent absolute value function, and determine the location where they will be mapped on g(x) using the transformations used to form g(x) from f(x).

Finally, plot the points and vertex and graph the given function.

The vertex of the parent absolute value function f(x)=∣x∣ is (0,0).

The vertex of g(x) is the point where (0,0) will be mapped using translations.

It can be observed that g(x) involves translations of f(x) by 3 units to the left and 1 unit up.

Therefore, the vertex of g(x) is the point (0−3,0+1)=(−3,1).

The points (2,2) and (−2,2) lie on the parent absolute value function f(x)=∣x∣.

Compare the function g(x) to g(x)=∣−1/b(x+h)+k|, and write the value of the constant b.b=2

Since ∣b∣=2, that is ∣b∣>1, the graph of g(x) will involve a horizontal stretch of the parent absolute value function by a factor 2.

Since b is negative, the graph of g(x) will involve a reflection of the parent absolute value function about the y-axis.

The transformations used to form g(x)=∣−1/2(x+3)+1∣ from the parent absolute value function can be used to determine the two points lying on g(x).

The x-coordinate of the points lying on f(x) will be reflected in the y-axis, horizontally stretched by a factor of 2, and shifted 3 units to the left when mapped to g(x).

The y-coordinate of the points lying on f(x) will be moved up 1 unit.

Therefore, the point (2,2) moves to (−2⋅2−3,∣2∣+1)=(−7,3).

The point (−2,2) moves to (−2⋅(−2)−3,∣2∣+1)=(1,3).

Plot the vertex (−3,1) and the points (−7,3) and (1,3), and then draw straight lines to sketch the graph of g(x)=∣−1/2(x+3)+1∣.

The vertex of g(x) is (−3,1).

Two points lying on g(x) are (−7,3) and (1,3).

The required grap

Page 67 Problem 11 Answer

The given graph shows the graph of an absolute value function. The incomplete steps used to find the function are given.

The question requires to write the function in the form g(x)=a∣1/b(x−h)+k| for the given graph, when b=1.

To determine the function, let the given function be g(x)=a∣1/b(x−h)+k|, use the graph to find the vertex, and thus, the values of h,k.

Then, rewrite the function using the obtained values.

Next, find a point other than vertex lying on the graph, and substitute its coordinates in the function to find the value of a. Finally, use the value of a in the function to obtain the function in the required form.

Let the given function be g(x)=a∣1/b(x−h)+k|.

From the graph, it can be observed that the vertex of g(x) is at (1,6).

This means that the values of h and k are 1 and 6 respectively.

Substitute 1 for b, 1 for h, and 6 for k in g(x)=a∣1/b(x−h)+k|.

​g(x)=a∣1/1(x−1)+6|

g(x)=a∣x−1∣+6………………(1)

​From the graph, it can be observed that the point with the integer coordinates (0,3) lies on the graph.

Substitute 0 for x, and 3 for g(x) in equation (1), subtract 6 from both sides, and solve for a.

​g(x)=a∣x−1∣+6

3=a∣0−1∣+6

3=a∣−1∣+6

−3=a(1)

−3=a​

Substitute −3 for a in g(x)=a∣x−1∣+6.

g(x)=−3∣x−1∣+6

Thus, the required form of the absolute value function is g(x)=−3∣x−1∣+6.

The required form of the graphed function is g(x)=−3∣x−1∣+6.

Page 67 Problem 12 Answer

The given function is g(x)=−1/5∣(x+6)∣+4.

The question requires to find the vertex and two points of the given function, and use them to graph g(x).

To graph the function, compare the given function to the parent absolute value function, and use the translations on its vertex to determine the vertex of g(x).

Then, take two points lying on the parent absolute value function, and determine the location where they will be mapped on g(x) using the transformations used to form g(x)

from f(x).

Finally, plot the points and vertex and graph the given function.

The vertex of the parent absolute value function f(x)=∣x∣ is (0,0).

The vertex of g(x) is the point where (0,0) will be mapped using translations.

It can be observed that g(x) involves a translation of f(x) by 6 units to the left and 4 units up.

Therefore, the vertex of g(x) is the point (0−6,0+4)=(−6,4).

The points (8,8) and (−8,8) lie on the parent absolute value function f(x)=∣x∣.

The transformations used to form g(x)=−1/5∣(x+6)∣+4 from the parent absolute value function can be used to determine the two points lying on g(x).

The coefficient of the absolute value in the given function g(x)=−1/5∣(x+6)∣+4 is −1/5, which is less than 1.

The x-coordinate of the points lying on f(x) will be shifted 6 units to the left when mapped to g(x).

The y-coordinate of the points lying on f(x) will be reflected about the x-axis by multiplying by −1, vertically compressed by the factor 1/5, and then shifted 4 units up when mapped to g(x).

Calculate the location where the point (8,8) will be mapped.​

(8−6,−1/5⋅∣8∣+4)=(2,−1/5⋅8+4)

=(2,−8/5+4)

=(2,12/5)

​Thus, (2,12/5) lies on the graph of g(x).

Calculate the location where the point (−8,8) will be mapped.

(−8−6,−1/5⋅∣8∣+4)=(−14,−1/5⋅8+4)

=(−14,−8/5+4)

=(−14,12/5)

Thus, (−14,12/5) lies on the graph of g(x).

Plot the vertex (−6,4) and the points (2,12/5) and (−14,12/5), and then draw straight lines to sketch the graph of g(x)=−1/5∣(x+6)∣+4.

The vertex of g(x) is (−6,4).

Two points lying on g(x) are (−14,12/5) and (2,12/5).

The required graph of g(x) is:

Page 68 Problem 13 Answer

The given graph shows the graph of an absolute value function.

The question requires to write the function graphed in the form g(x)=a∣1/b(x−h)+k|.

To determine the function, assume b=1, use the graph to find the vertex, and thus, the values of h,k. Then, rewrite the function using the obtained values.

Next, find a point other than vertex lying on the graph, and substitute its coordinates in the function to find the value of a.

Finally, use the value of a in the function to obtain the function in the required form.

Let b=1.

From the graph, it can be observed that the vertex of the graphed absolute value function is (−5,−1).

This means that the values of h and k are −5 and −1 respectively.

Substitute 1 for b, −5 for h, and −1 for k in the equation g(x)=a∣1/b(x−h)+k|.​

g(x)=a∣1/1(x−(−5))+(−1)|

g(x)=a∣x+5∣−1………………(1)​

From the graph, it can be observed that the point (−8,5) lies on the graph.

Substitute −8 for x, and then 5 for g(−8) in equation (1), add 1

on both sides, and then divide both sides by 2.​

g(−8)=a∣−8+5∣−1

5=a∣−3∣−1

6=a(3)

2=a

Substitute 2 for a in g(x)=a∣x+5∣−1.

g(x)=2∣x+5∣−1

Thus, the required form is g(x)=2∣x+5∣−1.

The required form of the graphed function is g(x)=2∣x+5∣−1.

Page 69 Problem 14 Answer

It is given that a beam of light that originates from a point 10 ft above the floor, reflects off a mirror 15 ft away from the wall containing the source of the light.

Also, it is given that after reflecting, it hits the opposite wall which is 8.5 ft away from the mirror.

The question requires to determine the height from the floor where the light hits the other wall after reflecting off the mirror.

To determine the height, draw the diagram representing the given situation, label the diagram, and let the base of the first wall be the origin.

Then, observe that the path of the ball is the graph of an absolute value function, and write a general function for the path.

Assume b=1, use the graph to find the vertex, and thus, the values of h,k.

Then, rewrite the function using the obtained values. Next, find a point other than vertex lying on the graph, and substitute its coordinates in the function to find the value of a.

Then, use the value of a in the function to obtain the function representing the path of the light.

Finally, use the distance of the second wall from the origin in the function, simplify to find the required height, and justify the answer.

Label the diagram representing the given sit

Here, A and B represent the points at which the light originates from and hits the other wall respectively.

The model of the path of the light will be of the form g(x)=a∣1/b(x−h)+k|.

Let the base of the first wall be the origin (0,0).

Therefore, the vertex of g(x) is the point at which the mirror is placed, that is (15,0).

Assume that b=1.

The vertex of the path of the ball is (15,0).

This means that the values of h and k are 15 and 0 respectively.

Substitute 1 for b, 15 for h, and 0 for k in the equation g(x)=a∣1/b(x−h)+k|.

g(x)=a∣/1/1(x−15)+0|

g(x)=a∣x−15∣………………(1)​

Another point on g(x) is the point from which the light originates from, that is (0,10).

Substitute 0 for x, and then 10 for g(0) in equation (1), and then divide both sides by 15.

​g(0)=a∣0−15∣

10=a∣−15∣

10=a(15)2/3=a

​Substitute 2/3 for a in g(x)=a∣x−15∣.

g(x)=2/3∣x−15∣

Thus, the function representing the path of the light is g(x)=2/3∣x−15∣.

The value of g(x) at x=15+8.5, that is 23.5 ft will give the height of the light beam on the opposite wall.

Substitute 23.5 for x in g(x)=2/3∣x−15∣, and simplify the expression.

​g(23.5)=2/3∣23.5−15∣

=2/3∣8.5∣

=2/3(8.5)

=17/3

≈5.67

​Therefore, the value of g(23.5) is about 5.67.

Thus, the beam hits the opposite wall at a height of about 5.67 ft.

The answer of 5.67 ft makes sense because the function is symmetric with respect to the vertical line passing through the vertex (15,0), that is x=15.

The distance from this vertical line to the second wall is a little more than half the distance from the line to the first wall from where the beam originates.

Since the beam originates at a height of 10 off the floor, the beam, after reflection, should hit the second wall at a height of a little over 1/2(10)=5 ft.

Thus, the answer 5.67 ft is justified.

The light beam hits the other wall at a height of about 5.67 ft.

Page 70 Problem 15 Answer

It is given that a student bounces a ball 3 m away from himself from a height of 1.4 m, to pass it to the other student standing.

The question requires to determine the position where the second student must stand to catch the ball at the height 1.2 m.

To determine the position, draw the diagram representing the given situation, label the diagram, and assume the point at which the first student stands as the origin.

Then, observe that the path of the ball is the graph of an absolute value function, and write a general function for the path.

Assume b=1, use the graph to find the vertex, and thus, the values of h,k. Then, rewrite the function using the obtained values.

Next, find a point other than vertex lying on the graph, and substitute its coordinates in the function to find the value of a.

Then, use the value of a in the function to obtain the function representing the path of the ball.

Finally, use the height at which the second student has to catch the ball in the function, and solve for x to find the required position.

Draw the diagram representing the given situation.

Here, A and B represent the points at which first and second student catch the ball respectively.

Point C represents the point where the ball bounces.

The graph of the path of the ball can be represented by the function g(x)=a∣1/b(x−h)+k|.

Let the point O where the first student stands be the origin (0,0).

Therefore, the vertex of the graph of the function is (3,0).

Assume that b=1.

The vertex of the path of the ball is (3,0).

This means that the values of h and k are 3 and 0 respectively.

Substitute 1 for b, 3 for h, and 0 for k in the equation g(x)=a∣1/b(x−h)+k|.

​g(x)=a∣1/1(x−3)+0|

g(x)=a∣x−3∣………………(1)

​

From the graph, it can be observed that the point (0,1.4) represents the point where the first student catches the ball.

Substitute 0 for x, and then 1.4 for g(0) in equation (1), and then divide both sides by 3.

​g(0)=a∣0−3∣

1.4=a∣−3∣

1.4=a(3)

1.4/3=a

​Substitute 1.4/3 for a in g(x)=a∣x−3∣.

g(x)=1.4/3∣x−3∣

Thus, the function representing the path of the ball is g(x)=1.4/3∣x−3∣.

The second student catches the ball at a height of 1.2 m.

Substitute 1.2 for g(x) in g(x)=1.4/3∣x−3∣, and multiply both sides by 3/1.4.

​1.2=1.4/3∣x−3∣1.2(3/1.4)=1.4/3(3/1.4)∣x−3∣

6(3/7)=∣x−3∣18/7=∣x−3∣………………(2)

​Form two equations by rewriting the absolute value in equation (2).

​18/7=x−3

18/7=−(x−3)

Simplify the right-hand side of 18/7=−(x−3), and solve for x.

​18/7=−x+3

x=3−18/7

x=3/7

x≈0.43

This value of x implies that the student must stand 0.43 m to the right of the origin.

However, the ball bounces 3 m away from the origin, and between the two students.

Thus, the value of x cannot be 0.43.

Add 3 on both sides of the other equation 18/7=x−3.

​18/7+3=x−3+3

39/7=x

5.57≈x

This value of x implies that the student must stand 5.57 m to the right of the origin.

Thus, the second student must stand about 5.57 m away from the first student to catch the ball at a height of 1.2 m.

The answer of 5.57 m makes sense because the function is symmetric with respect to the vertical line passing through the vertex (3,0), that is x=3.

The height at which the second student catches the ball is about 1.2

1.4≈0.857 times the height at which the first student catches the ball.

Therefore, the beam bounces at a distance of 3 m away from the first wall, it should hit the second wall when the distance of the vertical line from the second wall is about 0.857(3)=2.57 m.

This means that the second student should stand at about 3+2.57=5.57 m from the first student.

Thus, the answer 5.57 m is justified.

The second student must stand about 5.57 m away from the first student to catch the ball at a height of 1.2 m.

Page 70 Problem 16 Answer

The parameters in the general form g(x)=a∣1/b(x−h)+k| of the absolute value function are a,b,h,k.

The parameters h and k represent the x-coordinate and y-coordinate of the vertex of the function.

Parameter b determines the factor by which the x-coordinates of points lying on the parent absolute value function are stretched, compressed, and/or reflected to form the x-coordinates of the points lying on g(x).

The parameter a determines the factor by which the y-coordinates of points lying on the parent absolute value function are stretched, compressed, and/or reflected to form the y-coordinates of the points lying on g(x).

In the general form g(x)=a∣1/b(x−h)+k| of the absolute value function, the parameters h and k represent the x-coordinate and y-coordinate of the vertex of the function, the parameter b determines the factor by which the x-coordinates of points lying on the parent absolute value function are stretched, compressed, and/or reflected to form the x-coordinates of the points lying on g(x), and finally, the parameter a determines the factor by which the y-coordinates of points lying on the parent absolute value function are stretched, compressed, and/or reflected to form the y-coordinates of the points lying on g(x).

Page 70 Problem 17 Answer

The vertex of f(x)=∣x∣ is the origin (0,0).When the function f(x)=∣x∣ is vertically stretched or compressed, the y-coordinates of the points lying on it are multiplied by a factor.

When the function is horizontally stretched or compressed, the x-coordinates of the points lying on it are multiplied by a factor.

Since the product of 0 and the factor is always 0, the coordinates of the vertex remain (0,0) after multiplication by the factor.

Therefore, the vertex remains same when the function is stretched or compressed.

When the function f(x)=∣x∣ is vertically translated up or down, a constant is added or subtracted from the y-coordinates of the points lying on it.

When the function is horizontally translated to the left or right, a constant is added or subtracted from the x-coordinates of the points lying on it.

When a constant is added or subtracted from the coordinates of the vertex (0,0), the resulting vertex is a point other than (0,0).

Therefore, the vertex does not remain the same when the function is translated.

The vertex remains same when the function is stretched or compressed because the coordinates of the vertex remain (0,0) after multiplication by any factor.

The vertex does not remain the same when the function is translated because the vertex is not (0,0) after a constant is added to or subtracted from either of the coordinates.

Page 70 Problem 18 Answer

Every absolute value function is of the form g(x)=a∣1/b(x−h)+k|.

The graph of an absolute value function is formed using two inclined straight lines with the vertex as the common initial point.

The graph of an absolute value function is symmetric about the vertical line passing through the vertex.The domain of graph of an absolute value function is all real numbers.

The shape of the graph of an absolute value function is similar to the letter V or an inverted V, formed using two inclined straight lines with the vertex as the common initial point.

The graph of an absolute value function is symmetric about the straight vertical line that passes through its vertex.

The domain of the graph of every absolute value function is all real numbers.

Page 71 Exercise 1 Answer

The given function is g(x)=5∣x−3∣.

The question requires to predict how the graph of the given function will look, and verify this using a graphing calculator.

Then, it is required to sketch the graph of the given function.

To answer the question, compare the function to g(x)=a∣1/b(x−h)+k|to find the values of the constants, and then predict the shape and position of the graph in comparison to the graph of the parent absolute value function.

Next, graph the function on a graphing calculator and verify the prediction.

To sketch the graph, compare the given function to the parent absolute value function, and use the translations on its vertex to determine the vertex of g(x).

Then, take two points lying on the parent absolute value function, and determine the location where they will be mapped on g(x) using the transformations used to form g(x) from f(x).

Finally, plot the points and vertex and graph the given function.

Rewrite the given function in the form g(x)=a∣1/b(x−h)+k|.

g(x)=5∣1/1(x−3)+0|

Compare the function g(x) to g(x)=a∣1/b(x−h)+k|, and write the values of the constants.

​a=5

b=1

h=3

k=0

The value of a is positive. Therefore, the graph of g(x) will open upwards.

Since a=5, the graph of g(x) will be a vertically stretch of the parent absolute value function by a factor 5.

Since h=3, the graph of g(x) will be a horizontal translation of the parent absolute value function by 3 units to the right.

Since k=0, the graph of g(x) will not be involve a vertical translation of the parent absolute value function.

Therefore, it is predicted that the graph of g(x) will open upwards.

It will be a vertically stretch of the parent absolute value function by a factor 5, will involve a horizontal translation of the parent absolute value function by 3 units to the right, and will not be involve a vertical translation of the parent absolute value function.

On the graphing calculator, press the Y= key, and assign 5∣x−3∣ to Y₁.

Next, set the viewing rectangle to [−5,5] by [−5,5], and press the GRAPH key to obtain the required graph of the function.

From the graph, it can be observed that the prediction is verified.

The vertex of the parent absolute value function f(x)=∣x∣ is (0,0).

The vertex of g(x) is the point where (0,0) will be mapped using translations.

Therefore, the vertex of g(x) is the point (0+3,0)=(3,0).

The points (1,1) and (−1,1) lie on the parent absolute value function f(x)=∣x∣.

The transformations used to form g(x)=5∣x−3∣ from the parent absolute value function can be used to determine the two points lying on g(x).

The x-coordinate of the points lying on f(x) will be shifted 3 units to the right when mapped to g(x).

The y-coordinate of the points lying on f(x) will be vertically stretched by the factor 5 when mapped to g(x).

Calculate the location where the point (1,1) will be mapped.

​(1+3,5⋅∣1∣)=(1+3,5⋅1)

=(4,5)

Thus, (4,5) lies on the graph of g(x).

Calculate the location where the point (−1,1) will be mapped.

​(−1+3,5⋅∣1∣)=(−1+3,5⋅1)

=(2,5)

Thus, (2,5) lies on the graph of g(x).

Plot the vertex (3,0) and the points (4,5) and (2,5), and then draw straight lines to sketch the graph of g(x)=5∣x−3∣.

It is predicted that the graph of g(x) will open upwards, will be a vertically stretch of the parent absolute value function by a factor of 5, will involve a horizontal translation of the parent absolute value function by 3 units to the right, and will not be involve a vertical translation of the parent absolute value function.

The prediction is verified using the following graph made using a graphing calculator:

The required sketched graph of g(x) is:

Page 71 Exercise 2 Answer

The given function is g(x)=−4∣x+2∣+5.

The question requires to predict how the graph of the given function will look, and verify this using a graphing calculator.

Then, it is required to sketch the graph of the given function.

To answer the question, compare the function to g(x)=a∣1/b(x−h)+k| to find the values of the constants, and then predict the shape and position of the graph in comparison to the graph of the parent absolute value function.

Next, graph the function on a graphing calculator and verify the prediction.

To sketch the graph, compare the given function to the parent absolute value function, and use the translations on its vertex to determine the vertex of g(x).

Then, take two points lying on the parent absolute value function, and determine the location where they will be mapped on g(x) using the transformations used to form g(x)

from f(x).

Finally, plot the points and vertex and graph the given function.

Rewrite the given function in the form g(x)=a∣1/b(x−h)+k|.

g(x)=−4∣1/1(x+2)+5|

Compare the function g(x) to g(x)=a∣1/b(x−h)+k|, and write the values of the constants.

​a=−4

b=1

h=−2

k=5

​The value of a is negative. Therefore, the graph of g(x) will open downwards.

Since a=−4, the graph of g(x) will be a vertically stretch of the parent absolute value function by a factor 4, and a reflection about the x-axis.

Since h=−2, the graph of g(x) will be a horizontal translation of the parent absolute value function by 2 units to the left.

Since k=5, the graph of g(x) will involve a vertical translation of the parent absolute value function by 5 units up.

Therefore, it is predicted that the graph of g(x) will open downwards.

It will be a reflection of the parent absolute value function about the x-axis, a vertically stretch by a factor 4, a horizontal translation by 2 units to the left, and finally, a vertical translation by 5 units up.

On the graphing calculator, press the Y= key, and assign −4∣x+2∣+5 to Y₁.

Next, set the viewing rectangle to [−5,5] by [−5,5], and press the GRAPH key to obtain the required graph of the function.

From the graph, it can be observed that the prediction is verified.

The vertex of the parent absolute value function f(x)=∣x∣ is (0,0).

The vertex of g(x) is the point where (0,0) will be mapped using translations.

Therefore, the vertex of g(x) is the point (0−2,0+5)=(−2,5).

The points (1,1) and (−1,1) lie on the parent absolute value function f(x)=∣x∣.

The transformations used to form g(x)=−4∣x+2∣+5 from the parent absolute value function can be used to determine the two points lying on g(x).

The x-coordinate of the points lying on f(x) will be shifted 2 units to the left when mapped to g(x).

The y-coordinate of the points lying on f(x) will be reflected about the x-axis by multiplying by −1, vertically stretched by the factor 4, and translated 5 units up when mapped to g(x).

Calculate the location where the point (1,1) will be mapped.

(1−2,−4⋅∣1∣+5)=(1−2,−4⋅1+5)

=(−1,−4+5)

=(−1,1)

Thus, (−1,1) lies on the graph of g(x).

Calculate the location where the point (−1,1) will be mapped.

​(−1−2,−4⋅∣1∣+5)=(−1−2,−4⋅1+5)

=(−3,−4+5)

=(−3,1)

​Thus, (−3,1) lies on the graph of g(x).

Plot the vertex (−2,5) and the points (−1,1) and (−3,1), and then draw straight lines to sketch the graph of g(x)=−4∣x+2∣+5.

It is predicted that the graph of g(x) will open downwards, will be a reflection of the parent absolute value function about the x-axis, a vertical stretch by a factor 4, a horizontal translation by 2 units to the left, and finally, a vertical translation by 5 units up.

The prediction is verified using the following graph made using a graphing calculator:

The required sketched graph of g(x) is:

Page 71 Exercise 3 Answer

The given function is g(x)=∣7/5(x−6)+4|.

The question requires to predict how the graph of the given function will look, and verify this using a graphing calculator.

Then, it is required to sketch the graph of the given function.

To answer the question, rewrite and compare the function to g(x)=a∣(x−h)∣+k to find the values of the constants, and then predict the shape and position of the graph in comparison to the graph of the parent absolute value function.

Next, graph the function on a graphing calculator and verify the prediction.

To sketch the graph, compare the given function to the parent absolute value function, and use the translations on its vertex to determine the vertex of g(x).

Then, take two points lying on the parent absolute value function, and determine the location where they will be mapped on g(x) using the transformations used to form g(x)

from f(x).

Finally, plot the points and vertex and graph the given function.

Rewrite the given function in the form g(x)=a∣(x−h)∣+k.

g(x)=7/5∣x−6∣+4

Compare the function g(x) to g(x)=a∣(x−h)∣+k, and write the values of the constants.

​a=7/5

h=6

k=4

​The value of a is positive. Therefore, the graph of g(x) will open upwards.

Since a=7/5, the graph of g(x) will be a vertically stretch of the parent absolute value function by a factor 7/5.

Since h=6, the graph of g(x) will be a horizontal translation of the parent absolute value function by 6 units to the right.

Since k=4, the graph of g(x) will involve a vertical translation of the parent absolute value function by 4 units up.

Therefore, it is predicted that the graph of g(x) will open upwards. It will be a vertically stretch of the parent absolute value function by a factor 7/5, a horizontal translation by 6 units to the right, and finally, a vertical translation by 4 units up.

On the graphing calculator, press the Y= key, and assign |7/5(x−6)+4| to Y₁.

Next, set the viewing rectangle to [−10,10] by [−10,10], and press the GRAPH key to obtain the required graph of the function.

From the graph, it can be observed that the prediction is verified.

The vertex of the parent absolute value function f(x)=∣x∣ is (0,0).

The vertex of g(x) is the point where (0,0) will be mapped using translations.

Therefore, the vertex of g(x) is the point (0+6,0+4)=(6,4).

The points (2,2) and (−2,2) lie on the parent absolute value function f(x)=∣x∣.

The transformations used to form g(x)=7/5∣x−6∣+4 from the parent absolute value function can be used to determine the two points lying on g(x).

The x-coordinate of the points lying on f(x) will be shifted 6 units to the right when mapped to g(x).

The y-coordinate of the points lying on f(x) will be vertically stretched by the factor 7/5, and translated 4 units up when mapped to g(x).

Calculate the location where the point (2,2) will be mapped.

(2+6,7/5⋅∣2∣+4)=(2+6,7/5⋅2+4)

=(8,14/5+4)

=(8,34/5)

=(8,6.8)

Thus, (8,6.8) lies on the graph of g(x).

Calculate the location where the point (−2,2) will be mapped.

​(−2+6,7/5⋅∣2∣+4)=(−2+6,7/5⋅2+4)

=(4,14/5+4)

=(4,34/5)

=(4,6.8)

Thus, (4,6.8) lies on the graph of g(x).

Plot the vertex (6,4) and the points (4,6.8) and (8,6.8), and then draw straight lines to sketch the graph of g(x)=∣7/5(x−6)+4|.

It is predicted that the graph of g(x) will open upwards, will be vertically stretch of the parent absolute value function by a factor 7/5, a horizontal translation by 6 units to the right, and finally, a vertical translation by 4 units up.

The prediction is verified using the following graph made using a graphing calculator:

The required sketched graph of g(x) is:

Page 71 Exercise 4 Answer

The given function is g(x)=7/4∣(x−2)∣−3.

The question requires to predict how the graph of the given function will look, and verify this using a graphing calculator.

Then, it is required to sketch the graph of the given function.

To answer the question, compare the function to g(x)=a∣(x−h)∣−k to find the values of the constants, and then predict the shape and position of the graph in comparison to the graph of the parent absolute value function.

Next, graph the function on a graphing calculator and verify the prediction.

To sketch the graph, compare the given function to the parent absolute value function, and use the translations on its vertex to determine the vertex of g(x).

Then, take two points lying on the parent absolute value function, and determine the location where they will be mapped on g(x) using the transformations used to form g(x)

from f(x).

Finally, plot the points and vertex and graph the given function.

Compare the function g(x) to g(x)=a∣(x−h)∣−k, and write the values of the constants.

​a=7/4

h=2

k=3

The value of a is positive. Therefore, the graph of g(x) will open upwards.

Since a=7/4, the graph of g(x) will be a vertically stretch of the parent absolute value function by a factor 7/4.

Since h=2, the graph of g(x) will be a horizontal translation of the parent absolute value function by 2 units to the right.

Since k=3, the graph of g(x) will involve a vertical translation of the parent absolute value function by 3 units down.

Therefore, it is predicted that the graph of g(x) will open upwards. It will be a vertical stretch of the parent absolute value function by a factor 7/4, a horizontal translation by 2 units to the right, and finally, a vertical translation by 3 units down.

On the graphing calculator, press the Y= key, and assign 7/4∣(x−2)∣−3 to Y₁.

Next, set the viewing rectangle to [−5,5] by [−5,5], and press the GRAPH key to obtain the required graph of the function.

From the graph, it can be observed that the prediction is verified.

The vertex of the parent absolute value function f(x)=∣x∣ is (0,0).

The vertex of g(x) is the point where (0,0) will be mapped using translations.

Therefore, the vertex of g(x) is the point (0+2,0−3)=(2,−3).

The points (2,2) and (−2,2) lie on the parent absolute value function f(x)=∣x∣.

The transformations used to form g(x)=7/4∣(x−2)∣−3 from the parent absolute value function can be used to determine the two points lying on g(x).

The x-coordinate of the points lying on f(x) will be shifted 2 units to the right when mapped to g(x).

The y-coordinate of the points lying on f(x) will be vertically stretched by the factor 7/4, and translated 3 units down when mapped to g(x).

Calculate the location where the point (2,2) will be mapped.

​(2+2,7/4⋅∣2∣−3)=(2+2,7/4⋅2−3)

=(4,7/2−3)

=(4,1/2)

=(4,0.5)

Thus, (4,0.5) lies on the graph of g(x).

Calculate the location where the point (−2,2) will be mapped.

​(−2+2,7/4⋅∣2∣−3)=(−2+2,7/4⋅2−3)

=(0,7/2−3)

=(0,1/2)

=(0,0.5)

Thus, (0,0.5) lies on the graph of g(x).

Plot the vertex (2,−3) and the points (4,0.5) and (0,0.5), and then draw straight lines to sketch the graph of g(x)=7/4∣(x−2)∣−3.

It is predicted that the graph of g(x) will open upwards, will be vertical stretch of the parent absolute value function by a factor 7/4, a horizontal translation by 2 units to the right, and finally, a vertical translation by 3 units down.

The prediction is verified using the following graph made using a graphing calculator:

The required sketched graph of g(x) is:

Page 72 Exercise 5 Answer

The given function is g(x)=∣x∣.

The question requires to graph the given function, and determine its domain and range.

To graph the function, use the piecewise definition of the absolute value function g(x)=∣x∣ to form an input-output table.

Then, use the input-output table to write the coordinates of points lying on the graph of g(x)=∣x∣, and connect the points to sketch the graph of the function.

Finally, use the graph to determine the domain and range of the function.

The absolute value function g(x)=∣x∣ can be defined as the piecewise function:g(x)={x x≥0−x x<0

Substitute −8 for x in the absolute value function g(x)=∣x∣, and simplify using the piecewise definition of the absolute value function.

g(−8)=∣−8∣

=−(−8)

=8

Therefore, when the input is −8, the output is 8.

Substitute −4 for x in the absolute value function g(x)=∣x∣, and simplify using the piecewise definition of the absolute value function.

​g(−4)=∣−4∣

=−(−4)

=4

​Therefore, when the input is −4, the output is 4.

Substitute 0 for x in the absolute value function g(x)=∣x∣, and simplify using the piecewise definition of the absolute value function.​

g(0)=∣0∣

=0​

Therefore, when the input is 0, the output is 0.

Substitute 4 for x in the absolute value function g(x)=∣x∣, and simplify using the piecewise definition of the absolute value function.

​g(4)=∣4∣

=4

​Therefore, when the input is 4, the output is 4.

Substitute 8 for x in the absolute value function g(x)=∣x∣, and simplify using the piecewise definition of the absolute value function.

g(8)=∣8∣

=8

Therefore, when the input is 8, the output is 8.

Construct an input-output table using the calculated inputs and outputs.

The points lying on the graph of g(x)=∣x∣ are of the form (x,g(x)).

From the input-output table, it can be determined that the points (−8,8),(−4,4),(0,0),(4,4),(8,8) lie on the graph of g(x)=∣x∣.

Plot the points, join them, and draw straight lines to form the graph of the function g(x)=∣x∣.

From the graph, it can be observed that the graph of the function g(x)=∣x∣ stretches infinitely towards the left and right.

Therefore, the given absolute value function is defined for all values of x.

Hence, the domain of g(x)=∣x∣ is (−∞,∞).

Also, it can be observed that the points lying on the graph of the function g(x)=∣x∣ lie either on, or above the x-axis.

Therefore, the values of the outputs g(x) can be either 0, or greater than 0.

Thus, the range of g(x)=∣x∣ is [0,∞).

The required graph is:

The domain of g(x)=∣x∣ is (−∞,∞).

The range of g(x)=∣x∣ is [0,∞).

Page 72 Exercise 6 Answer

The given function is g(x)=−7/6∣(x−2)∣.

The question requires to predict how the graph of the given function will look, and verify this using a graphing calculator.

Then, it is required to sketch the graph of the given function.

To sketch the graph, write the translations used to form the given function from the parent absolute value function f(x)=∣x∣, and use them on the vertex to determine the vertex of g(x).

Next, write the coefficient of the absolute value in the given function to find the factor by which the parent absolute function will be vertically stretched.

Then, take two points lying on the parent absolute value function, and determine the location where they will be mapped on g(x)

using the transformations used to form g(x) from f(x). Next, plot the points and vertex and graph the given function.

Finally, use the drawn graph to find and write the domain and range of the function.

The vertex of the parent absolute value function f(x)=∣x∣ is (0,0).

The vertex of g(x) is the point where (0,0) will be mapped using translations.

It can be observed that g(x) involves a translation of f(x) by 2 units to the right.

Therefore, the vertex of g(x) is the point (0+2,0)=(2,0).

Compare the function g(x) to g(x)=−a∣(x−h)∣, and write the value of the constant a.a=7/6

Since a=7/6, that is ∣a∣>1, the graph of g(x) will involve a vertical stretch of the parent absolute value function by a factor 7/6.

Since the coefficient of the absolute value ∣(x−2)∣ in the given function is −7/6, that is negative, the graph of g(x) will involve reflection of the parent absolute value function about the x-axis.

The points (6,6) and (−6,6) lie on the parent absolute value function f(x)=∣x∣.

The transformations used to form g(x)=−7/6∣(x−2)∣ from the parent absolute value function can be used to determine the two points lying on g(x).

The x-coordinate of the points lying on f(x) will be shifted 2 units to the right when mapped to g(x).

The y-coordinate of the points lying on f(x) will be reflected about the x-axis, and then vertically stretched by the factor 7/6.

Calculate the location where the point (6,6) will be mapped.

​(6+2,−7/6⋅∣6∣)=(6+2,−7/6⋅6)=(8,−7)

Calculate the location where the point (−6,6) will be mapped.​

(−6+2,−7/6⋅∣6∣)=(−6+2,−7/6⋅6)

=(−4,−7)

​Therefore, the two points (−4,−7) and (8,−7) lie on the graph of g(x).

Plot the vertex (2,0) and the points (−4,−7) and (8,−7), and then draw straight lines to sketch the graph of g(x)=−7/6∣(x−2)∣.

From the graph, it can be observed that the graph of the function g(x)=−7/6∣(x−2)∣ stretches infinitely towards the left and right. Therefore, the given absolute value function is defined for all values of x.

Hence, the domain of g(x)=−7/6∣(x−2)∣ is (−∞,∞).

Also, it can be observed that the y-coordinates of the points lying on the graph of the function g(x)=−7/6∣(x−2)∣ are either 0, or less than 0.

Thus, the range of g(x)=−7/6∣(x−2)∣ is (−∞,0].

The required graph is:

The domain of g(x) is (−∞,∞).

The range of g(x) is (−∞,0].

Page 71 Exercise 7 Answer

The given function is g(x)=∣5/7(x−4)|.

The question requires to predict how the graph of the given function will look, and verify this using a graphing calculator.

Then, it is required to sketch the graph of the given function.

To sketch the graph, write the translations used to form the given function from the parent absolute value function f(x)=∣x∣, and use them on the vertex to determine the vertex of g(x).

Next, write the coefficient of the absolute value in the given function to find the factor by which the parent absolute function will be vertically stretched.

Then, take two points lying on the parent absolute value function, and determine the location where they will be mapped on g(x) using the transformations used to form g(x) from f(x).

Next, plot the points and vertex and graph the given function.

Finally, use the drawn graph to find and write the domain and range of the function.

The vertex of the parent absolute value function f(x)=∣x∣ is (0,0).

The vertex of g(x) is the point where (0,0) will be mapped using translations.

It can be observed that g(x) involves a translation of f(x) by 4 units to the right.

Therefore, the vertex of g(x) is the point (0+4,0)=(4,0).

Rewrite the given function.

g(x)=5/7∣(x−4)∣

Compare the function g(x) to g(x)=a∣(x−h)∣−k, and write the value of the constant a.

a=5/7

Since a=5/7, that is 0<a<1, the graph of g(x) will involve a vertical compression of the parent absolute value function by a factor 5/7.

The points (4,4) and (−4,4) lie on the parent absolute value function f(x)=∣x∣.

The transformations used to form g(x)=5/7∣(x−4)∣ from the parent absolute value function can be used to determine the two points lying on g(x).

The x-coordinate of the points lying on f(x) will be shifted 4 units to the right when mapped to g(x).

The y-coordinate of the points lying on f(x) will be vertically compressed by the factor 5/7 when mapped to g(x).

Calculate the location where the point (4,4) will be mapped.

(4+4,5/7⋅∣4∣)=(4+4,5/7⋅4)

=(8,20/7)

Calculate the location where the point (−4,4) will be mapped.

​(−4+4,5/7⋅∣4∣)=(−4+4,5/7⋅4)

=(0,20/7)

Therefore, the two points (0,20/7) and (8,20/7) lie on the graph of g(x).

Plot the vertex (4,0) and the points (0,20/7) and (8,20/7), and then draw straight lines to sketch the graph of g(x)=∣5/7(x−4)∣.

From the graph, it can be observed that the graph of the function g(x)=∣5/7(x−4)∣ stretches infinitely towards the left and right.

Therefore, the given absolute value function is defined for all values of x.

Hence, the domain of g(x)=∣5/7(x−4)∣ is (−∞,∞).

Also, it can be observed that the y-coordinates of the points lying on the graph of the function g(x)=∣5/7(x−4)∣ are either 0 or greater than 0.

Thus, the range of g(x)=∣5/7(x−4)∣is [0,∞).

The required The domain of g(x) is (−∞,∞).

The range of g(x) is [0,∞).

Page 72 Exercise 8 Answer

The given function is g(x)=∣−7/3(x+5)−4|.

The question requires to predict how the graph of the given function will look, and verify this using a graphing calculator.

Then, it is required to sketch the graph of the given function.

To sketch the graph, write the translations used to form the given function from the parent absolute value function f(x)=∣x∣, and use them on the vertex to determine the vertex of g(x).

Next, write the coefficient of the absolute value in the given function to find the factor by which the parent absolute function will be vertically stretched.

Then, take two points lying on the parent absolute value function, and determine the location where they will be mapped on g(x) using the transformations used to form g(x) from f(x).

Next, plot the points and vertex and graph the given function. Finally, use the drawn graph to find and write the domain and range of the function.

The vertex of the parent absolute value function f(x)=∣x∣ is (0,0).

The vertex of g(x) is the point where (0,0) will be mapped using translations.

It can be observed that g(x) involves translations of f(x) by 5 units to the left and 4 units down.

Therefore, the vertex of g(x) is the point (0−5,0−4)=(−5,−4).

Rewrite the given function.​

g(x)=7/3∣−(x+5)∣−4

g(x)=7/3∣(x+5)∣−4

Compare the function g(x) to g(x)=a∣(x+h)∣−k, and write the value of the constant a.

a=7/3

Since a=7/3, that is a>1, the graph of g(x) will involve a vertical stretch of the parent absolute value function by a factor 7/3.

The points (3,3) and (−3,3) lie on the parent absolute value function f(x)=∣x∣.

The transformations used to form g(x)=7/3∣(x+5)∣−4 from the parent absolute value function can be used to determine the two points lying on g(x).

The x-coordinate of the points lying on f(x) will be shifted 5 units to the left when mapped to g(x).

The y-coordinate of the points lying on f(x) will be vertically stretched by the factor 7/3, and shifted down by 4 units.

Calculate the location where the point (3,3) will be mapped.

(3−5,7/3⋅∣3∣−4)=(3−5,7/3⋅3−4)

=(−2,7−4)

=(−2,3)

​Calculate the location where the point (−3,3) will be mapped.

​(−3−5,7/3⋅∣3∣−4)=(−3−5,7/3⋅3−4)

=(−8,7−4)

=(−8,3)

​Therefore, the two points (−8,3) and (−2,3) lie on the graph of g(x).

Plot the vertex (−5,−4) and the points (−8,3) and (−2,3), and then draw straight lines to sketch the graph of g(x)=∣−7/3(x+5)−4|.

From the graph, it can be observed that the graph of the function g(x)=∣−7/3(x+5)−4| stretches infinitely towards the left and right.

Therefore, the given absolute value function is defined for all values of x.

Hence, the domain of g(x)=∣−7/3(x+5)−4| is (−∞,∞).

Also, it can be observed that the y-coordinates of the points lying on the graph of the function g(x)=∣−7/3(x+5)−4| are either −4, or greater than −4.

Thus, the range of g(x)=∣−7/3(x+5)−4| is [−4,∞).

The required graph is:

The domain of g(x) is (−∞,∞).

The range of g(x) is [−4,∞).

Page 72 Exercise 9 Answer

The given graph shows the graph of an absolute value function.

The question requires to write the function in standard form for the given graph, when a=1.

To determine the function, let the given function be g(x)=a∣1/b(x−h)+k|, use the graph to find the vertex, and thus, the values of h,k.

Then, rewrite the function using the obtained values.

Next, find a point other than vertex lying on the graph, and substitute its coordinates in the function to find the value of b.

Finally, use the value of b in the function to obtain the function in the required form.

Let the given function be g(x)=a∣1/b(x−h)+k|, that is an absolute value function in standard form.

From the graph, it can be observed that the vertex of the graphed absolute value function is (−7,4).

This means that the values of h and k are −7 and 4 respectively.

Substitute 1 for a, −7 for h, and 4 for k in g(x)=a∣1/b(x−h)+k|.

g(x)=1∣1/b(x−(−7))+4|

g(x)=∣1/b(x+7)+4|………………(1)

From the graph, it can be observed that the point (−6,6) lies on the graph.

Substitute −6 for x, and then 6 for g(−6) in equation (1), and subtract 4 from both sides.

​g(−6)=∣1/b(−6+7)+4|

6=∣1/b(1)+4|

2=∣1/b|……………(2)

​Form two equations by rewriting the absolute value in equation (2).

​2=1/b

2=−1/b

If 2 is equal to 1/b, then b=1/2.

If 2 is equal to −1/b, then b=−1/2.

Since the absolute value in g(x)=∣1/b(x+7)+4| remains the same for both values of b, consider the value b=1/2.

Substitute 1/2 for b in g(x)=∣1/b(x+7)+4|, and simplify the equation.

​g(x)=∣1/12(x+7)+4|

g(x)=∣2(x+7)∣+4

g(x)=2∣(x+7)∣+4

Thus, the graphed absolute value function is g(x)=2∣(x+7)∣+4.

The required absolute value function is g(x)=2∣(x+7)∣+4.

Page 72 Exercise 10 Answer

The given graph shows the graph of an absolute value function.

The question requires to write the function in standard form for the given graph, when b=1.

To determine the function, let the given function be g(x)=a∣1/b(x−h)+k|, use the graph to find the vertex, and thus, the values of h,k.

Then, rewrite the function using the obtained values

. Next, find a point other than vertex lying on the graph, and substitute its coordinates in the function to find the value of a.

Finally, use the value of a in the function to obtain the function in the required form.

Let the given function be g(x)=a∣1/b(x−h)+k|, that is an absolute value function in standard form.

From the graph, it can be observed that the vertex of the graphed absolute value function is (4,3).

This means that the values of h and k are 4 and 3 respectively.

Substitute 1 for b, 4 for h, and 3 for k in g(x)=a∣1/b(x−h)+k|.​

g(x)=a∣1/1(x−4)+3|

g(x)=a∣x−4∣+3………………(1)​

From the graph, it can be observed that the point (0,2) lies on the graph.

Substitute 0 for x, and then 2 for g(0) in equation (1), subtract 3 from both sides, and then divide both sides by 4.

​g(0)=a∣0−4∣+3

2=a∣−4∣+3

−1=a(4)

−1/4=a​

Substitute −1/4 for a in g(x)=a∣x−4∣+3.

g(x)=−1/4∣x−4∣+3

Thus, the graphed absolute value function is g(x)=−1/4∣x−4∣+3.

The required absolute value function is g(x)=−1/4∣x−4∣+3.

Page 74 Exercise 11 Answer

The vertex of f(x)=∣x∣, that is the parent absolute value function, is (0,0).

The vertex of y=∣x+6∣−4 is the point where (0,0) will be mapped using translations.

It can be observed that y=∣x+6∣−4 involves translations of f(x) by 6 units to the left and 4 units down.

Therefore, the vertex of the function y=∣x+6∣−4 is the point:

(0−6,0−4)=(−6,−4)

From the given graphs, only the graph in option (C) has the vertex at (−6,−4).

Thus, the graph of the function y=∣x+6∣−4 is the graph in option (C).

The graph in option (A) has the vertex at (6,−4), and not at (−6,−4).

Thus, the graph in option (A) is not the graph of the function y=∣x+6∣−4.

The graph in option (B) has the vertex at (6,4), and not at (−6,−4).

Thus, the graph in option (B) is not the graph of the function =∣x+6∣−4.

The graph of the function y=∣x+6∣−4 is the graph in option (C).

Page 74 Exercise 12 Answer

The vertex of f(x)=∣x∣, that is the parent absolute value function, is (0,0).

The vertex of y=∣x−6∣−4 is the point where (0,0) will be mapped using translations.

It can be observed that y=∣x−6∣−4 involves translations of f(x) by 6 units to the right and 4 units down.

Therefore, the vertex of the function y=∣x−6∣−4 is the point:

(0+6,0−4)=(6,−4)

From the given graphs, only the graph in option (A) has the vertex at (6,−4).

Thus, the graph of the function y=∣x−6∣−4 is the graph in option (A).

The graph in option (B) has the vertex at (6,4), and not at (6,−4).

Thus, the graph in option (B) is not the graph of the function y=∣x−6∣−4.

The graph in option (C) has the vertex at (−6,−4), and not at (6,−4).

Thus, the graph in option (C) is not the graph of the function y=∣x−6∣−4.

The graph of the function y=∣x−6∣−4 is the graph in option (A).

Page 74 Exercise 13 Answer

The vertex of f(x)=∣x∣, that is the parent absolute value function, is (0,0).

The vertex of y=∣x−6∣+4 is the point where (0,0) will be mapped using translations.

It can be observed that y=∣x−6∣+4 involves translations of f(x)

by 6 units to the right and 4 units up.

Therefore, the vertex of the function y=∣x−6∣+4 is the point: (0+6,0+4)=(6,4)

From the given graphs, only the graph in option (B) has the vertex at (6,4).

Thus, the graph of the function y=∣x−6∣+4 is the graph in option (B).

The graph in option (A) has the vertex at (6,−4), and not at (6,4).

Thus, the graph in option (A) is not the graph of the function y=∣x−6∣+4.

The graph in option (C) has the vertex at (−6,−4), and not at (6,4).

Thus, the graph in option (C) is not the graph of the function y=∣x−6∣+4.

The graph of the function y=∣x−6∣+4 is the graph in option (B).

Page 75 Exercise 14 Answer

The vertex of f(x)=∣x∣, that is the parent absolute value function, is (0,0).

The vertex of y=∣x+3∣+2 is the point where (0,0) will be mapped using translations.

It can be observed that y=∣x+3∣+2 involves translations of f(x) by 3 units to the left and 2 units up.

Therefore, the vertex of the given function is the point (0−3,0+2)=(−3,2).

The given graph is not the graph of the function y=∣x+3∣+2  because the vertex of the function in the graph is (3,2) and not (−3,2).

The function in the graph involves translations of the parent absolute value function f(x)=∣x∣ by 3 units to the right and 2 units up.

Thus, the correct equation shown in the graph is y=∣x−3∣+2.

The graphed function is not the graph of y=∣x+3∣+2 because the vertex of the function in the graph is (3,2), and the vertex of the given function is (−3,2).

The correct equation shown in the graph is y=∣x−3∣+2.

Page 71 Exercise 15 Answer

It is given that the golf ball is at the point (2.5,2), the hole where the golf ball is to be shot is at (9.5,2), and the point where the golf ball banks off the side wall is (6,8).

The question requires to equation of the path of the ball.

To determine the equation, draw the graph representing the given situation.

Then, observe that the path of the ball is the graph of an absolute value function, and write a general function for the path.

Assume b=1, use the graph to find the vertex, and thus, the values of h,k. Then, rewrite the function using the obtained values.

Next, find a point other than vertex lying on the graph, and substitute its coordinates in the function to find the value of a.

Finally, use the value of a in the function to obtain the function representing the path of the ball.

Draw the graph representing the given points, and join them to show the path of the ball.

It can be observed that the path of the ball is the graph of an absolute value function.

Let the equation for the path of the ball be represented by the function g(x)=a∣1/b(x−h)+k|.

Assume that b=1.

From the graph, it can be determined that the vertex of the graph of the function is (6,8).

This means that the values of h and k are 6 and 8 respectively.

Substitute 1 for b, 6 for h, and 8 for k in the equation g(x)=a∣1/b(x−h)+k|.

​g(x)=a∣1/1(x−6)+8|

g(x)=a∣x−6∣+8………………(1)​

The path of the ball starts from the point (2.5,2).

Substitute 2.5 for x, and then 2 for g(2.5) in equation (1). Then, subtract 8 from both sides, and divide both sides by 3.5.

​g(2.5)=a∣2.5−6∣+8

2=a∣−3.5∣+8

−6=a(3.5)

−6/3.5=a

−12/7=a

​Substitute −12/7 for a in g(x)=a∣x−6∣+8.

g(x)=−12/7∣x−6∣+8

Thus, the equation representing the path of the ball is g(x)=−12/7∣x−6∣+8.

The required equation is g(x)=−12/7∣x−6∣+8.

Page 75 Exercise 16 Answer

It is given that the golf ball is at the point (2.5,2), the hole where the golf ball is to be shot is at (9.5,2), and the point where the golf ball banks off the side wall is (6,8).

The question requires to determine if the player makes the shot.

Given the equation g(x)=−12/7∣x−6∣+8 representing the path of the ball as calculated in the previous part of this exercise.

To answer the question, substitute the point (9.5,2) on the equation for path of the ball, and check whether the point lies on the path.

Substitute 9.5 for x and 2 for g(9.5) in equation g(x)=−12/7∣x−6∣+8, and simplify both sides of the equation.​

g(9.5)=−12/7∣9.5−6∣+8

2=−12/7∣3.5∣+8

2=−12/7(3.5)+8

2=−6+8

2=2

The point (9.5,2) satisfies the equation for the path of the ball because the equation 2=2 is always true.

Therefore, the point (9.5,2)representing the hole where the golf ball is aimed at lies on the path of the ball.

Thus, the player makes the shot.

Yes, the player makes the shot because the point (9.5,2) representing the hole, where the golf ball is to be shot, lies on the path of the ball.

Page 76 Exercise 17 Answer

It is given that the orchestra starts at fortissimo, changes to pianissimo after four measures, and again increases to fortissimo after four more measures.

Also, it is given that a fortissimo and pianissimo are at sound levels 90 and 50 decibels respectively.

The question requires to determine the function representing the sound level s as a function of the number of measures m.

To determine the function, use the points of the form (m,s) and the given information to find three points lying on the graph of the required function.

Plot the points, join them, observe the graph formed, and write the required function as a general absolute value function.

Assume b=1, use the graph to find the vertex, and thus, the values of h,k.

Then, rewrite the function using the obtained values.

Next, find a point other than vertex lying on the graph, and substitute its coordinates in the function to find the value of a.

Then, use the value of a in the function to obtain the function representing the sound level.

The points (m,s) can be used to represent the sound level at a particular number of measures.

The initial sound level at zero number of measures is s=90 decibels.

The sound level after four number of measures is s=50 decibels.

The sound level after eight number of measures is s=90 decibels.

Therefore, the points (0,90),(4,50),(8,90) lie on the graph of the function representing the sound level.

Plot the points, and join them to show the graph of the function representing the sound level.

The graph of the function representing the sound level s as a function of the number of measures m can be represented by the function s=a∣1/b(m−h)+k|.

Assume that b=1.

The vertex of the function is the point is the relative extrema (4,50).

This means that the values of h and k are 4 and 50 respectively.

Substitute 1 for b, 4 for h, and 50 for k in the equation s=a∣1/b(m−h)+k|.

​s=a∣1/1(m−4)+50|

s=a∣m−4∣+50………………(1)

​The point representing the initial sound level at zero numbers of measures is the point (0,90).

Substitute 0 for x and 90 for s in equation (1). Then, subtract 50 from both sides, and then divide both sides by 4.

​90=a∣0−4∣+50

90=a∣−4∣+50

40=a(4)

10=a

​Substitute 10 for a in s=a∣m−4∣+50.

s=10∣m−4∣+50

Thus, the function representing the sound level s as a function of the number of measures m is s=10∣m−4∣+50.

The required function is s=10∣m−4∣+50.

Page 76 Exercise 18 Answer

It is given that the orchestra starts at fortissimo, changes to pianissimo after four measures, and again increases to fortissimo after four more measures.

Also, it is given that a fortissimo and pianissimo are at sound levels 90 and 50 decibels respectively.

The question requires to determine the number of measures at which the orchestra is at the loudness of mezzo forte, that is 70 decibels.

Given the equation s=10∣m−4∣+50 representing the sound level s as a function of the number of measures m, as calculated in the previous part of this exercise.

To determine the number of measures, substitute the sound level as 70 in the function for the sound level, and solve the equation for m by rewriting the absolute value.

Substitute 70 for s in equation s=10∣m−4∣+50, subtract 50 from both sides, and divide both sides by 10.

​70=10∣m−4∣+50

20=10∣m−4∣

2=∣m−4∣………………(1)

​Form two equations by rewriting the absolute value in equation (1).

​2=m−4

2=−(m−4)

Add 4 on both sides of 2=m−4 to solve for m.

​2+4=m−4+4

6=m

​Simplify the right-hand side of 2=−(m−4), and solve for m.

​2=−m+4

m=4−2

m=2

​The two values of m are 2 and 6.

Thus, the orchestra is at the loudness of mezzo forte in 2 measures and in 6 measures.

The required number of measures at which orchestra is at the loudness of mezzo forte are 2 measures and 6 measures.

Page 76 Exercise 19 Answer

It is given that the orchestra starts at fortissimo, changes to pianissimo after four measures, and again increases to fortissimo after four more measures.

Also, it is given that a fortissimo and pianissimo are at sound levels 90 and 50 decibels respectively.

The question requires to describe how the graph of the given function will look.

Given the equation s=10∣m−4∣+50 representing the sound level s as a function of the number of measures m, as calculated in part (a) of this exercise.

To answer the question, compare the given function to the general form s=a∣(m−h)∣+k to find the values of the constants, and then describe the graph using the transformations that are required to form the function from the parent absolute value function.

Compare the function for sound level to the general form s=a∣(m−h)∣+k, and write the values of the constants.

​a=10

h=4

k=50​

The value of a is positive. Therefore, the graph of s=10∣m−4∣+50 will open upwards.

Since ∣a∣=10, that is greater than 1, the graph will be a vertical stretch of the parent absolute value function by a factor 10.

Since h=4, the graph will be a horizontal translation of the parent absolute value function by 4 units to the right.

Since k=50, the graph will involve a vertical translation of the parent absolute value function by 50 units up.

Therefore, the graph of s=10∣m−4∣+50 will open upwards.

It will be a vertical stretch of the parent absolute value function by a factor 10, a horizontal translation by 4 units to the right, and finally, a vertical translation 50 units up.

The graph of s=10∣m−4∣+50 will open upwards and will be a vertical stretch of the parent absolute value function by a factor 10, a horizontal translation by 4 units to the right, and finally, a verical translation 50 units up.

Algebra 2 Volume 1 1st Edition Unit 1 Functions

Page 2 Problem 1 Answer

In the question, the following words are given:
Coefficient
Domain
Function
Inequality
Interval
Quadratic function
Range
Transformation

The diagram of an information wheel is given as shown below:

It is required to complete the information wheel using correct words from the given list.

To solve this, use the appropriate words from the list to replace the terms in the diagram, using the definitions of those words.

From the diagram, f(x)=3x²

is a function of x and hence, it can be termed as a function.

3 is the coefficient of x².

From the definition of domain, the domain of the function f(x)=3x²is the set of all real numbers.

The range of the function is all the outputs and is given as y=3x².

Finally, since x² is always positive, the interval will contain only non-negative real numbers.

Draw the diagram of the information wheel with the correct words:

The complete information wheel using correct words from the given list is shown below:

Page 2 Problem 2 Answer

In the question, the given incomplete sentence is: A ________ is a constant in the equation of a curve that yields a family of similar curves as it changes.

It is required to complete the sentence. To solve this, use the given sentence and the definition of parameter to find an appropriate word to complete the sentence.

A Parameter is a constant in the equation of a curve that yields a family of similar curves as it changes.

A Parameter is a constant in the equation of a curve that yields a family of similar curves as it changes.

Page 2 Problem 3 Answer

In the question, the given incomplete sentence is: A function f(x) such that f(x)=f(−x) is an ________.

It is required to complete the sentence. To solve this, use the given sentence and the definition of an even function to find an appropriate word to complete the sentence.

A function f(x) such that f(x)=f(−x) is an Even Function.

Page 2 Problem 4 Answer

In the question, the given incomplete sentence is: A compound statement that uses the word or is a _________.

It is required to complete the sentence. To solve this, use the given sentence and the definition of disjunction to find an appropriate word to complete the sentence.

A disjunction uses the word or in a compound statement.

Algebra 2 Volume 1 1st Edition Unit 3 Polynomial, Functions, Expressions, And Equations

Page 232 Problem 1 Answer

A table is given with some statements written in one column. It is asked to write the terms for those statements in another column.

Factor is a number or expression that divides a product exactly.

For example, 3x divides 6x² and gives the quotient 2x. So, 3x is the factor of 6x².

A rational or irrational number is always a real number. For example, 1/2 is a rational number.

Here, 1/2=0.5. Thus, it is a real number.

A number, variable, product, or quotient in expression is called as term.

For example: in the expression, 3x+2, 3xand 2are the terms of the binomial 3x+2.

Parameters are the constants in a function or equation that may be changed. For example, in the equation,x=2t+1, t is the parameter.

A change in the size, position, or shape of a figure of graph is called as transformation.

Transformations are of four types mainly, Reflection, Translation, Stretch, Compression.A numerical factor in a term of an algebraic expression is known as coefficients.

For example, in the expression, 3×2+5, 3 is the coefficient of x².

The table with complete statements is given below:

The table with review words is given by,

Page 232 Problem 2 Answer

It is given an incomplete statement, “A polynomial function of degree 3 is a ____.”

It is required to complete the statement.

The degree of a polynomial is the highest of the degrees of the polynomial’s monomials (individual terms) with non-zero coefficients.

The polynomial having degree 3 are cubic polynomial function. For example: 3×3+2×2+1is a cubic polynomial function.

A polynomial function of degree 3 is a cubic polynomial function.

The complete sentence is, A polynomial function of degree 3 is a cubic polynomial function.

Page 232 Problem 3 Answer

It is given an incomplete statement, “A ___________of a polynomial is a zero of the function associated with that polynomial.”

It is required to complete the statement. A factor of a polynomial is a zero of the function associated with that polynomial.

A factor of a polynomial is a zero of the function associated with that polynomial.

Algebra 2 Volume 1 1st Edition Module 3 Quadratic Equations

Page 139 Problem 1 Answer

The complex solutions of a quadratic equation can be calculated either by completing the square, or by using the quadratic formula.

To complete the square and calculate the complex solutions of a quadratic equation, first the equation is rewritten in the form x²+bx=c. Next, the value of b is identified, and (b/2)² is calculated.

The value of (b/2)² is added on both sides of the rewritten equation, and the left-side of the equation is factorised.

Finally, the definition of square root is used, and the equation is solved for the required values of x.

To use the quadratic formula and calculate the complex solutions of a quadratic equation, the equation is rewritten in the form ax²+bx+c=0

and the values of a,b,c are written. Then, the values of a,b,c are substituted in the quadratic formula and the expression is simplified to determine the required values of x.

To calculate the complex solutions of a quadratic equation, the completing the square method or the quadratic formula are used.

To calculate the complex solutions of a quadratic equation by completing the square, rewrite the quadratic equation in the form x²+bx=c, calculate the value of (b/2)², add (b/2)² on both sides of the rewritten equation, factor the left-side of the equation, use the definition of square root, and finally, solve the resulting equation for x.

Page 139 Problem 2 Answer

The given table shows three quadratic equations of the form ax²+bx+c=0.

The question requires to complete the given table.

To complete the given table, rewrite the equations using subtraction to fill the second column.

Then, use the left and right sides of the rewritten equations to determine the expressions for f(x) and g(x), and use them to fill the third and fourth columns of the table.

The first given quadratic equation is 2x²+4x+1=0.

Subtract 1 from both sides of the equation to rewrite it in the form ax²+bx=−c.

​2x²+4x+1−1=0−12x²+4x=−1

The left-hand side of the equation is of the form ax²+bx. Equating the left-hand side of the rewritten equation to f(x), the obtained function is f(x)=2x²+4x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=−1.

The second given quadratic equation is 2x²+4x+2=0                                                                                                                                                                                                                                                     

Subtract 2 from both sides of the equation to rewrite it in the form ax²+bx=−c.

​2x²+4x+2−2=0−2

2x²+4x=−2

The left-hand side of the equation is of the form ax²+bx. Equating the left-hand side of the rewritten equation to f(x), the obtained function is f(x)=2x²+4x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=−2.

The third given quadratic equation is 2x²+4x+3=0.

Subtract 3 from both sides of the equation to rewrite it in the form ax²+bx=−c.

2x²+4x+3−3=0−3

2x²+4x=−3​

The left-hand side of the equation is of the form ax2+bx. Equating the left-hand side of the rewritten equation to f(x), the obtained function is f(x)=2x²+4x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=−3.

Complete the table using the rewritten equations, and the obtained expressions for f(x) and g(x).

The completed table is:

Page 139 Problem 3 Answer

The given graph shows the graph of the function f(x)=2x²+4x.

The question requires to graph each g(x) and then complete the given table.

Given the following table, as completed in the previous part of this exercise

To draw the graph and complete the given table, write the functions g(x)

to be graphed, and then graph them on the given graph using horizontal lines.

Then, use the points of intersection of f(x) and g(x)=−1 to determine the number of points where they are equal.

Then, use the expressions for the two functions, and rewrite the equation with 0 on one side.

Next, use the number of points of intersection to write the number of real solutions of the obtained equation.

Similarly, find the points of intersection, and thus the number of real solutions of the remaining two equations, and complete the table.

The graph of a constant function g(x)=a is a horizontal line having the y-intercept a.

From the completed table from previous part of the exercise, it can be determined that functions to be graphed are:​

g(x)=−1

g(x)=−2

g(x)=−3

Graph the functions on the given graph by drawing horizontal lines having the y-intercept equal to the right-hand side of the equation.

From the graph, it can be observed that the graph of f(x) and g(x)=−1intersect each other at two points.

This means that 2x²+4x=−1 for two real values of x.

Adding 1 on both sides, the equation becomes 2x²+4x+1=0.

Therefore, the equation 2x²+4x+1=0 has two real solutions.

From the graph, it can be observed that the graph of f(x) and g(x)=−2 touch each other at one point.

This means that 2x²+4x=−2 for one real value of x.

Adding 2 on both sides, the equation becomes 2x²+4x+2=0.

Therefore, the equation 2x²+4x+2=0 has one real solution.

From the graph, it can be observed that the graph of f(x) and g(x)=−3

do not touch or intersect each other at any point.This means that 2x²+4x=−3 for no real value of x.

Adding 3 on both sides, the equation becomes 2x²+4x+3=0.

Therefore, the equation 2x²+4x+3=0 has no real solutions.

Complete the table using the determined number of real solutions of each equation.

The required graph of each g(x) on the given graph is:

Page 139 Problem 4 Answer

The given tables show three quadratic equations of the form ax²+bx+c=0. The given graph shows the graph of the function f(x) equal to −2x²+4x.

The question requires to complete the two given tables and graph the functions g(x).

To complete the first given table, rewrite the equations using subtraction to fill the second column.

Then, use the left and right sides of the rewritten equations to determine the expressions for f(x)

and g(x), and use them to fill the third and fourth columns of the table.

To draw the graph and complete the second given table, write the functions g(x) to be graphed, and then graph them on the given graph using horizontal lines.

Then, use the points of intersection of f(x)and g(x)=1 to determine the number of points where they are equal.

Then, use the expressions for the two functions, and rewrite the equation with 0 on one side.

Next, use the number of points of intersection to write the number of real solutions of the obtained equation.

Similarly, find the points of intersection, and thus the number of real solutions of the remaining two equations, and complete the second table.

The first given quadratic equation is −2x²+4x−1=0.

Add 1 on both sides of the equation to rewrite it in the form ax²+bx=−c.

−2x²+4x−1+1=0+1

−2x²+4x=1

The left-hand side of the equation is of the form ax²+bx. Equating the left-hand side of the rewritten equation to f(x), the obtained function is f(x)=−2x²+4x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=1.

The second given quadratic equation is −2x²+4x−2=0.

Add 2 on both sides of the equation to rewrite it in the form ax²+bx=−c.

​−2x²+4x−2+2=0+2

−2x²+4x=2

The left-hand side of the equation is of the form ax²+bx. Equating the left-hand side of the rewritten equation to f(x), the obtained function is f(x)=−2x²+4x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=2.

The third given quadratic equation is −2x²+4x−3=0.

Add 3 on both sides of the equation to rewrite it in the form ax²+bx=−c.

​−2x²+4x−3+3=0+3

−2x²+4x=3

The left-hand side of the equation is of the form ax²+bx. Equating the left-hand side of the rewritten equation to f(x), the obtained function is f(x)=−2x²+4x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=3.

Complete the first table using the rewritten equations, and the obtained expressions for f(x) and g(x).

The graph of a constant function g(x)=a is a horizontal line having the y-intercept a.

From the completed table from previous part of the exercise, it can be determined that functions to be graphed are:

g(x)=1

g(x)=2

g(x)=3

​Graph the functions on the given graph by drawing horizontal lines having the y-intercept equal to the right-hand side of the equations.

From the graph, it can be observed that the graph of f(x) and g(x)=1intersect each other at two points.

This means that −2x²+4x=1 for two real values of x.

Subtracting 1 from both sides, the equation becomes −2x²+4x−1=0.

Therefore, the equation −2x²+4x−1=0 has two real solutions.

From the graph, it can be observed that the graph of f(x) and g(x)=2 touch each other at one point.

This means that −2×2+4x=2 for one real value of x.

Subtracting 2 from both sides, the equation becomes −2x²+4x−2=0.

Therefore, the equation −2x²+4x−2=0 has one real solution.

From the graph, it can be observed that the graph of f(x) and g(x)=3 do not touch or intersect each other at any point.This means that −2x²+4x=3 for no real value of x.

Subtracting 3 from both sides, the equation becomes −2x²+4x−3=0.

Therefore, the equation −2x²+4x−3=0 has no real solutions.

Complete the table using the determined number of real solutions of each equation.

The completed first table is:

The required graph of each g(x) on the given graph is:

Page 140 Problem 5 Answer

It is given that the minimum value of f(x) noticed from steps A and B of Explore 1 is −2.

The given table shows the values of g(x), and the corresponding number of real solutions of f(x)=g(x).

The question requires to complete the given table.

Given the following graph, as drawn in step B of Explore 1:

To complete the given table, use the graph to determine the number of points of intersection of f(x) and g(x)=−2, and write the number of real solutions of the given equation.

Next, use the graph to determine the number of points of intersection of f(x) and the horizontal lines represented by g(x)>−2, and write the number of real solutions of the given equation.

Similarly, find the number of real solutions of the given equation when g(x)<−2.

Finally, use the obtained number of real solutions to complete the given table.

From the graph, it can be observed that the graph of f(x) and g(x)=−2 touch each other at one point.

This means that when g(x) is equal to the minimum value of f(x), that is −2, f(x)=g(x) for one real value of x.

Thus, the number of real solutions of f(x)=g(x) is one when g(x) is equal to −2.

The graph of g(x)>a for each value of a represents a horizontal line above the line g(x)=a.

From the graph, it can be observed that each horizontal line passing through the graph of f(x) above the line g(x)=−2 intersects the parabola at two points.

This means that when g(x) has a value greater than −2, the graph of f(x) and g(x) intersect each other at two points.

Therefore, when g(x) has a value greater than the minimum value of f(x), that is −2, f(x)=g(x) for two real values of x.

Thus, the number of real solutions of f(x)=g(x) is two when g(x)>−2.

The graph of g(x)<a for each value of a represents a horizontal line below the line g(x)=a.

From the graph, it can be observed that each horizontal line below the line g(x)=−2

does not intersect the parabola at any point.

This means that when g(x) has a value less than −2, the graph of f(x) and g(x) do not intersect each other.

Therefore, when g(x) has a value less than the minimum value of f(x), that is −2, f(x) is not equal to g(x) for any real value of x.

Thus, the number of real solutions of f(x)=g(x) is zero when g(x)<−2.

Complete the table using the determined number of real solutions of f(x)=g(x) for each g(x).

The completed table is:

Page 140 Problem 6 Answer

It is given that the maximum value of f(x) noticed from step C of Explore 1 is 2.

The given table shows the values of g(x), and the corresponding number of real solutions of f(x)=g(x).

The question requires to complete the given table.

Given the following graph, as drawn in step C of Explore 1:

To complete the given table, use the graph to determine the number of points of intersection of f(x)

and g(x)=2, and write the number of real solutions of the given equation.

Next, use the graph to determine the number of points of intersection of f(x) and the horizontal lines represented by g(x)>2, and write the number of real solutions of the given equation.

Similarly, find the number of real solutions of the given equation when g(x)<2.

Finally, use the obtained number of real solutions to complete the given table.

From the graph, it can be observed that the graph of f(x) and g(x)=2 touch each other at one point.

This means that when g(x) is equal to the maximum value of f(x), that is 2, f(x)=g(x) for one real value of x.

Thus, the number of real solutions of f(x)=g(x) is one when g(x) is equal to 2.

The graph of g(x)<a for each value of a represents a horizontal line below the line g(x)=a.

From the graph, it can be observed that each horizontal line passing through the graph of f(x) below the line g(x)=2 intersects the parabola at two points.

This means that when g(x) has a value less than 2, the graph of f(x) and g(x) intersect each other at two points.

Therefore, when g(x) has a value less than the maximum value of f(x), that is 2, f(x)=g(x) for two real values of x.

Thus, the number of real solutions of f(x)=g(x) is two when g(x)<2.

The graph of g(x)>a for each value of a represents a horizontal line above the line g(x)=a.

From the graph, it can be observed that each horizontal line above the line g(x)=2 does not intersect the parabola at any point.

This means that when g(x) has a value greater than 2, the graph of f(x) and g(x) do not intersect each other.

Therefore, when g(x) has a value greater than the maximum value of f(x), that is 2, f(x) is not equal to g(x) for any real value of x.

Thus, the number of real solutions of f(x)=g(x) is zero when g(x)>2

Complete the table using the determined number of real solutions of f(x)=g(x) for each g(x).

The completed table is:

Page 140 Problem 7 Answer

It is given that the maximum value of f(x) given in Reflect 2 is 2. Also, it is given that equation ax²+bx+c=0

where a>0 has real solutions when b²−4ac≥0. The steps used to generalize the results of Reflect 1 are given.

The question requires to generalize the results of Reflect 2, and write what is noticed.

Given the following table, as completed in Reflect 2:

To answer the question, use the table to write the range of values of g(x) for which f(x)=g(x) has real solutions.

Then, determine the maximum value of f(x) by substituting −b/2a for x in the function, and simplifying it.

Next, use the obtained maximum value of f(x) to form an inequality in terms of g(x).

Finally, rewrite the inequality to form a general expression, generalize the result by writing the relation for which ax²+bx+c=0

has real solutions, and write what is observed using the given information.

The quadratic equation ax²+bx+c=0, where a<0, can be written as:ax²+bx=−c

The function f(x) used in Reflect 2 is of the form ax²+bx, where a<0.

The function g(x) is of the form −c.

From the table, it can be observed that the equation f(x)=g(x), where g(x) is equal to −c, has real solutions when g(x) is less than or equal to the maximum value of f(x), that is 2.

The maximum value of f(x) occurs at x=−b/2a.

Calculate the maximum value of f(x) by substituting x=−b/2a in the function, and simplifying the expression.

​f(−b/2a)=a(−b/2a)²+b(−b/2a)

f(−b/2a)=a(b^2/4a²)−b^2/2a

f(−b/2a)=b²/4a−b²/2a

f(−b/2a)=−b²/4a

Therefore, the equation f(x)=g(x) has real solutions when g(x)≤−b²/4a.

Substitute −c for g(x) in the inequality g(x)≤−b²/4a, add b²/4a on both sides, and multiply both sides by the negative number 4a.

​−c≤−b²/4a

b²/4a−c≤0

4a(b²/4a−c)≥4a(0)

b²−4ac≥0

​Therefore, the equation ax²+bx+c=0

where a<0 has real solutions when b²−4ac≥0.

It is given that equation ax²+bx+c=0 where a>0 has real solutions when b²−4ac≥0.

Thus, it can be concluded that the equation ax²+bx+c=0 has real solutions when b²−4ac≥0 for all real values of a.

The equation ax²+bx+c=0 where a<0 has real solutions when b²−4ac≥0.

It is observed that equation ax²+bx+c=0 has real solutions when b²−4ac≥0 for all real values of a.

Page 141 Problem 8 Answer

The given equation is x²−2x+7=0.

The question requires to solve the given equation by completing the square, and then state whether the obtained solutions are real or non-real.

To solve the equation by completing the square, rewrite the equation in the form x²+bx=c.

Then, identify the value of b, and calculate (b/2)².

Then, add (b/2)² on both sides of the rewritten equation, factor the left-side expression using algebraic identities, and use the definition of square root.

Finally, solve the obtained equation to determine the required solutions, and observe them to state whether they are real or non-real.

Subtract 7 from both sides of x²−2x+7=0.

​x²−2x+7−7=0−7

x²−2x=−7……………(1)

​Comparing the equation (1) to the equation x²+bx=c, it can be determined that the value of b is −2.

Therefore, the value of (b/2)² is (−2/2)²=4/4, that is 1.

Add the value of (b/2)², that is 1, on both sides of equation (1).

Then, factorise the left-hand side expression using the algebraic identity a²−2ab+b²=(a−b)², and simplify the right-hand side expression.

​x²−2x+1=−7+1

(x−1)²=−7+1

(x−1)²=−6

​Use the definition of square root to simplify (x−1)²=−6, and then add 1 on both sides.

​(x−1)²=−6

x−1=±√−6

x=1±√−6

Therefore, the required solutions are x=1+√−6 and x=1−√−6.

It can be observed that the solutions x=1±√−6 contain the square root of a negative number.

Therefore, the solutions are non-real.

Thus, there are two non-real solutions: x=1+√−6 and x=1−√−6.

The required solutions are x=1+√−6 and x=1−√−6.The solutions are non-real.

Page 142 Problem 9 Answer

From part A of the example, it can be observed that the equation 3x²+9x−6=0 has two real solutions: −3+√17/2 and −3−√17/2.

Every real number a can be written as the complex number a+0i, having the imaginary part equal to 0.

This means that every real number is a complex number.

The real solutions −3+√17/2 and −3−√17/2 can be written as the complex solutions −3+√17/2+0i and −3−√17/2+0i respectively.

This means that both the real solutions of the equation in part A are complex.

Therefore, the equation in part A has two complex solutions, having the imaginary part equal to 0.

From part B of the example, it can be observed that the equation x²−2x+7=0 has two non-real solutions: 1+√−6 and 1−i√6.

The non-real solutions 1+√−6 and 1−√−6 can be written as the complex solutions 1+i√6and 1−i√6 respectively, having non-zero real and imaginary parts.

This means that both the non-real solutions of the equation in part B are complex.

Therefore, the equation in part B has two complex solutions, having non-zero real and imaginary parts.

The equations in part A and part B both have two complex solutions because every real and non-real number is a complex number.

This means that the solutions of the equation in part A are real and complex, and the solutions of the equation in part B are non-real and complex.

Page 142 Problem 10 Answer

The given equation is x²+8x+17=0.

The question requires to solve the given equation by completing the square, and then state whether the obtained solutions are real or non-real.

To solve the equation by completing the square, rewrite the equation in the form x²+bx=c. Then, identify the value of b, and calculate (b/2)².

Then, add (b/2)² on both sides of the rewritten equation, factor the left-side expression using algebraic identities, and use the definition of square root.

Finally, solve the obtained equation to determine the required solutions, and observe them to state whether they are real or non-real.

Subtract 17 from both sides of x²+8x+17=0.​

x²+8x+17−17=0−17

x²+8x=−17……………(1)​

Comparing the equation (1) to the equation x²+bx=c, it can be determined that the value of b is 8.

Therefore, the value of (b/2)² is (8/2)²

=(4)², that is 16.

Add the value of (b/2)², that is 16, on both sides of equation (1).

Then, factorise the left-hand side expression using the algebraic identity a²+2ab+b²=(a+b)², and simplify the right-hand side expression.

​x²+8x+16=−17+16

(x+4)²=−17+16

(x+4)²=−1

​Use the definition of square root to simplify (x+4)²=−1.

Then, subtract 4 from both sides, and rewrite the expression using the imaginary unit i.

​(x+4)²=−1

x+4=±√−1

x=−4±√−1

x=−4±i

Therefore, the required solutions are x=−4+i and x=−4−i.

It can be observed that the solutions x=−4±i contain the imaginary unit i, which is equal to the square root of the negative number −1.

Therefore, the solutions are non-real. Thus, there are two non-real solutions: x=−4+i and x=−4−i.

The required solutions are x=−4+i and x=−4−i. The solutions are non-real.

Page 142 Problem 11 Answer

The given equation is x²+10x−7=0.

The question requires to solve the given equation by completing the square, and then state whether the obtained solutions are real or non-real.

To solve the equation by completing the square, rewrite the equation in the form x²+bx=c.

Then, identify the value of b, and calculate (b/2)².

Then, add (b/2)² on both sides of the rewritten equation, factor the left-side expression using algebraic identities, and use the definition of square root.

Finally, solve the obtained equation to determine the required solutions, and observe them to state whether they are real or non-real.

Add 7 on both sides of x²+10x−7=0.​

x²+10x−7+7=0+7

x²+10x=7……………(1)

Comparing the equation (1) to the equation x²+bx=c, it can be determined that the value of b is 10.

Therefore, the value of (b/2)² is (10/2)²=(5)2, that is 25.

Add the value of (b/2)², that is 25, on both sides of equation (1).

Then, factorise the left-hand side expression using the algebraic identity a²+2ab+b²=(a+b)², and simplify the right-hand side expression.​

x²+10x+25=7+25

(x+5)²=7+25

(x+5)²=32

​Use the definition of square root to simplify (x+5)²=32, and then subtract 5 from both sides.

​(x+5)²=32

x+5=±√32

x=−5±√32

Rewrite the expression √32, use the product property of square roots, and simplify the square root.

x=−5±√16⋅2

x=−5±√16⋅√2

x=−5±4√2​

Therefore, the required solutions are x=−5+4√2 and x=−5−4√2.

It can be observed that the solutions x=−5±4√2 do not contain the square root of a negative number or the imaginary unit i.Therefore, the solutions are real.

The required solutions are x=−5+4√2 and x=−5−4√2.

The solutions are real.

Page 142 Problem 12 Answer

It is given that the function A(w)=w(50−w) gives the area of the given garden in square feet, where w

is the width of the garden in feet.

The question requires to determine whether the garden can have an area equal to 700 square feet.

It is required to answer this by writing an equation for the given situation, and then determining whether the solutions are real or non-real.

To answer the question, let the area A(w) be equal to 700 square feet.

Then, use the given function to form a quadratic equation in terms of w. Next, compare the equation to aw²+bw+c=0 and identify the values of a,b,c.

Finally, determine the discriminant of the equation to determine whether the solutions are real or non-real, and thus, determine whether the area can be equal to 700 square feet or not.

Let the area of the garden be equal to 700 square feet. Therefore, A(w)=700.

Substitute w(50−w) for A(w) in A(w)=700. Then, multiply the terms, and subtract 700 from both sides.

w(50−w)=700

50w−w²=700−w²+50w−700=0……………(1)

Comparing the equation (1) to the equation aw²+bw+c=0, it can be determined that the values of a,b,c are −1,50,−700 respectively.

Substitute −1 for a, 50 for b, and −700 for c in the formula for discriminant b²−4ac. Then, multiply the terms, and subtract the terms.

​b²−4ac=(50)²−4(−1)(−700)

=2500−2800

=−300

The discriminant is equal to −300, which is less than 0.

This means that the equation A(w)=700 has two non-real solutions.

Therefore, there is no real value of w for which the area of the garden is equal to 700 square feet.

Thus, the garden cannot have an area equal to 700 square feet.

No, the garden cannot have an area equal to 700 square feet.

Page 143 Problem 13 Answer

It is given that the height h of a triangular sail is twice the length of its base b, where the height and base are given in inches.

The question requires to determine whether the area of a triangular sail can be equal to 10 square inches.

It is required to answer this by writing an equation for the given situation, and then determining whether the solutions are real or non-real.

To answer the question, use the given information and the formula for area of a triangle to write an equation for the given situation.

Next, let the area A(b) of the sail be equal to 10 square inches. Then, use the obtained function for area to form a quadratic equation in terms of b.

Next, compare the equation to pb²+qb+r=0 and identify the values of p,q, and r.

Finally, determine the discriminant of the equation to determine whether the solutions are real or non-real, and thus, determine whether the area can be equal to 10 square inches or not.

From the given information, it can be determined that h=2b.

Substitute 2b for h in the formula A=1/2bh for area of a triangle, and multiply the terms.

A=1/2b(2b)

A=b²

A(b)=b²

Let the area of the triangular sail be equal to 10

square inches. Therefore, A(b)=10.

Substitute b² for A(b) in A(b)=10, and subtract 10 from both sides.

​b²=10

b²−10=0……………(1)

Comparing the equation (1) to the equation pb²+qb+r=0, it can be determined that the values of p,q,r are 1,0,−10 respectively.

If a quadratic equation is in the form pb²+qb+r=0, then its determinant is equal to q²−4pr.

Substitute 1 for p, 0 for q, and −10 for r in the formula for discriminant q²−4pr. Then, multiply the terms, and add the terms.

​q²−4pr=(0)²

−4(1)(−10)

=0+40

=40

The discriminant is equal to 40, which is greater than 0.

This means that the equation A(b)=10 has two real solutions.

Therefore, there is some real value of b for which the area of the triangular sail is equal to 10

square inches.

Thus, the triangular sail can have an area equal to 10 square inches.

Yes, the area of the triangular sail can be equal to 10 square inches.

Page 143 Problem 14 Answer

The given equation is 7x²+2x+3=−1.

The question requires to solve the given equation using the quadratic formula, and then check one of the solutions using substitution.

To solve the equation, rewrite the equation in the form ax²+bx+c=0. Then, identify the values of a,b,c

, and substitute the values in the quadratic formula. Next, simplify the expression on the right side to obtain the required solutions.

Finally, substitute one of the solutions in the given equation, and simplify using the distributive property of multiplication and the value of i² to prove that both sides of the equation are equal.

Add 1 on both sides of 7x²+2x+3=−1.

​7x²+2x+3+1=−1+1

7x²+2x+4=0……………(1)

Comparing the equation (1) to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are 7,2,4 respectively.

Substitute 7 for a, 2 for b, and 4 for c in the quadratic formula x=−b±√b²−4ac/2a. Then, multiply the terms and combine the terms in the square root.

​x=−2±√(2)²−4(7)(4)/2(7)

x=−2±√4−112/14

x=−2±√−108/14

Rewrite the square root using the imaginary unit i. Then, simplify the square root, and simplify the expression.

​x=−2±i√108/14

x=−2±i√36⋅3/14

x=−2±6i√3/14

x=−1±3i√3/7

Thus, the required solutions are:

x=−1/7−3i√3/7

x=−1/7+3i√3/7Consider the solution x=−1/7−3i√3/7.

Substitute −1/7−3i√3/7 for x in the given equation 7x²+2x+3=−1.

Then, square the term −1−3i√3/7 in the equation, substitute −1 for i², and simplify the expression in the parentheses.

​7(−1/7−3i√3/7)2

+2(−1/7−3i√3/7)+3=?

−1/7(1/49+6i√3/49+27i²/49)+2(−1/7−3i√3/7)+3=?

−1/7(1/49+6i√3/49+27(−1)/49)+2(−1/7−3i√3/7)+3=?

−1/7(1/49+6i√3/49−27/49)+2(−1/7−3i√3/7)+3=?

−1​Apply the distributive property of multiplication, combine the terms, and simplify the expression on the left side.

1/7+6i√3/7−27/7−2/7−6i√3/7+3=?

−1/−28/7+3=?

−1/−4+3=?

−1/−1=−1

Thus, the solution x=−1/7−3i√3/7 is checked.

The required solutions are:x=−1/7−3i√3/7

x=−1/7+3i√3/7

The solution x=−1/7−3i√3/7 is checked.

Page 145 Problem 15 Answer

The given equation is x²+8x+12=2x.

The question requires to solve the given equation using the quadratic formula, and then check one of the solutions using substitution.

To solve the equation, rewrite the equation in the form ax²+bx+c=0. Then, identify the values of a,b,c, and substitute the values in the quadratic formula.

Next, simplify the expression on the right side to obtain the required solutions.

Finally, substitute one of the solutions in the given equation, and simplify using the distributive property of multiplication and the value of i² to prove that both sides of the equation are equal.

Subtract 2x from both sides of x²+8x+12=2x.

​x²+8x+12−2x=2x−2x

x²+6x+12=0……………(1)

Comparing the equation (1) to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are 1,6,12 respectively.

Substitute 1 for a, 6 for b, and 12 for c in the quadratic formula x=−b±√b²−4ac/2a. Then, multiply the terms and combine the terms in the square root.

​x=−6±√(6)²−4(1)(12)/2(1)

x=−6±√36−48/2

x=−6±√−12/2

​Rewrite the square root using the imaginary unit i. Then, simplify the square root, and simplify the expression.

​x=−6±i√12/2

x=−6±i√4⋅3/2

x=−6±2i√3/2

x=−3±i√3

Thus, the required solutions are:

​x=−3−i√3

x=−3+i√3

Consider the solution x=−3−i√3.

Substitute −3−i√3 for x in the given equation x²+8x+12=2x. Then, square the term −3−i√3 in the equation, substitute −1 for i², and simplify the expression in the parentheses.

​(−3−i√3)²+8(−3−i√3)+12=?

2(−3−i√3)(9+6i√3+3i²)+8(−3−i√3)+12=?

2(−3−i√3)(9+6i√3+3(−1))+8(−3−i√3)+12=?

2(−3−i√3)(9+6i√3−3)+8(−3−i√3)+12=?

2(−3−i√3)​

Apply the distributive property of multiplication, combine the terms, and simplify the expression on the left side.

​9+6i√3−3−24−8i√3+12=?

−6−2i√3

18−24−2i√3=?

−6−2i√3−6−2i√3

=−6−2i√3

Thus, the solution x=−3−i√3 is checked.

The required solutions are:

x=−3−i√3

x=−3+i√3​

The solution x=−3−i√3 is checked.

Page 145 Problem 16 Answer

The given equation is ax²+bx+c=0. The given solution is p+qi.

The question requires to what should be the other solution of the equation, and explain how it is determined.

Let the other solution of the given equation be the complex number f+gi.

To answer the question, use the property for sum of roots of a quadratic equation, and the assumed solution and the given solution to write an equation for the sum of the solutions.

Then, observe both sides to determine the value of the imaginary part of the sum, and solve the resulting equation to determine one of the constants in the assumed solution.

Next, use the property for product of roots of a quadratic equation, and the assumed solution and the given solution to write an equation for the product of the solutions.

Then, observe both sides to determine the value of the imaginary part of the product, and solve the resulting equation to determine the other constant in the assumed solution.

Finally, use the constants to determine the other solution, and explain why is it the solution.

The sum of the two solutions of the quadratic equation ax²+bx+c=0 is the quotient −b/a.

Therefore, the sum of f+gi and p+qi is −b/a.

Form an equation for the sum of the solutions, and group the real and imaginary parts.​

(f+gi)+(p+qi)=−b/a

(f+p)+(gi+qi)=−b/a

(f+p)+(g+q)i=−b/a

The right-hand side expression −b/a is a real number.

Therefore, (f+p)+(g+q)i must also be a real number. This means that the imaginary part (g+q) is equal to 0.

Equate g+q and 0, and subtract q from both sides.​

g+q=0

g=−q​

The product of the two solutions of the quadratic equation ax²+bx+c=0 is the quotient c/a.

Therefore, the product of f+gi and p+qi is c/a.

Form an equation for the product of the solutions.

Then, multiply the terms, substitute −1 for i², and group the real and imaginary parts.

(f+gi)(p+qi)=c/a

fp+fqi+gpi+gqi²=c/a

fp+fqi+gpi+gq(−1)=c/a

fp+fqi+gpi−gq=c/a

(fp−gq)+(fq+gp)i=c/a

The right-hand side expression c/a is a real number.

Therefore, (fp−gq)+(fq+gp)i must also be a real number.

This means that the imaginary part fq+gp is equal to 0.

Equate fq+gp and 0, substitute −q for g, divide both sides by q, and add p on both sides.

​fq+gp=0

fq+(−q)p=0

fq−pq=0

q(f−p)=0

f−p=0

f=p

​Substitute −q for g and p for f in the solution f+gi.

​f+gi=p+(−q)i

f+gi=p−qi

​Therefore, the other solution of the given equation is p−qi.

Thus, it can be observed that the other solution must be p−qi

such that the sum and product of the solutions are equal to −b/a and c/a respectively.

The other solution of the equation must be p−qi because the sum and product of the solutions of the given equation are equal to −b/a and c/a respectively only if the two solutions are complex conjugates of each other.

Page 145 Problem 17 Answer

To complete the square, first the equation is rewritten in the form x²+bx=c. Next, the value of b is identified, and (b/2)² is calculated.

The value of (b/2)² is added on both sides of the rewritten equation, and the left-side of the equation is factorized.

Finally, the definition of square root is used, and the equation is solved for the required values of x.

To use the quadratic formula, the equation is rewritten in the form ax²+bx+c=0 and the values of a,b,c are written. Then, the values of a,b,c

are substituted in the quadratic formula and the expression is simplified to determine the required values of x.

It can be observed that using the quadratic formula to solve a quadratic equation is much simpler and faster as compared to solving a quadratic equation by completing the square.

Thus, using the quadratic formula to solve a quadratic equation is much easier than completing the square.

Using the quadratic formula to solve a quadratic equation is much easier than completing the square because quadratic formula only requires to rewrite the equation in the form ax²+bx+c=0, identify the values of a,b,c, and substitute them into the quadratic formula to obtain the values of x.

The completing the square method takes much longer to solve and is more complex because it involves rewriting the equation in the form x²+bx=c, calculating (b/2)², adding (b/2)² on both sides, factoring the left-side, using the definition of square root, and finally, solving the resulting equation.

Page 146 Problem 18 Answer

The given graph shows the graph of the function f(x) equal to x²+6x. The given equations are:

​x²+6x+6=0

x²+6x+9=0

x²+6x+12=0

​The question requires to use the given graph to determine the number of real solutions of the given equations.

To determine the number of real solutions, rewrite the equations using subtraction in the form ax²+bx=−c, and equate the left and right sides of the rewritten equations to f(x) and g(x).

Then, draw the graph of each g(x) on the given graph using horizontal lines. Next, use the points of intersection of f(x)

and g(x) to determine the number of points where they intersect.

Finally, use the points to determine the number of real solutions for each g(x), and thus the number of real solutions for each equation.

The first given quadratic equation is x²+6x+6=0.

Subtract 6 from both sides of the equation to rewrite it in the form ax²+bx=−c.

x²+6x=−6

The left-hand side of the equation is of the form ax²+bx. Equating the left-hand side of the rewritten equation to f(x) gives the function f(x)=x²+6x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=−6.

The second given quadratic equation is x²+6x+9=0.

Subtract 9 from both sides of the equation to rewrite it in the form ax²+bx=−c.

x²+6x=−9

The left-hand side of the equation is of the form ax²+bx. Equating the left-hand side of the rewritten equation to f(x) gives the function f(x)=x²+6x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=−9.

The third given quadratic equation is x²+6x+12=0.

Subtract 12 from both sides of the equation to rewrite it in the form ax²+bx=−c.

x²+6x=−12

The left-hand side of the equation is of the form ax²+bx. Equating the left-hand side of the rewritten equation to f(x) gives the function f(x)=x²+6x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=−12.

Construct a table having columns for the given equations, the rewritten equations, and the obtained expressions for f(x) and g(x).

From the completed table from previous part of the exercise, it can be determined that functions to be graphed are:

​g(x)=−6

g(x)=−9

g(x)=−12

​Graph the functions on the given graph by drawing horizontal lines having the y-intercept equal to the right-hand side of the equations.

From the graph, it can be observed that the graph of f(x) and g(x)=−6

intersect each other at two points.

This means that x²+6x=−6 for two real values of x.

Adding 6 on both sides, the equation becomes x²+6x+6=0.

Therefore, the equation x²+6x+6=0 has two real solutions. This is because the value of g(x)

is greater than the minimum value of f(x), that is −9.

From the graph, it can be observed that the graph of f(x) and g(x)=−9 touch each other at one point.

This means that x²+6x=−9 for one real value of x.

Adding 9 on both sides, the equation becomes x²+6x+9=0.

Therefore, the equation x²+6x+9=0 has one real solution. This is because the value of g(x)

is equal to the minimum value of f(x), that is −9.

From the graph, it can be observed that the graph of f(x) and g(x)=−12 do not touch or intersect each other at any point.

This means that x²+6x=−12 for no real value of x.

Adding 12 on both sides, the equation becomes x²+6x+12=0.

Therefore, the equation x²+6x+12=0 has no real solution.

This is because the value of g(x) is less than the minimum value of f(x), that is −9.

The equation x²+6x+6=0 has two real solutions.

This is because the equation can be written as x²+6x equal to −6, where the left-hand side represents the given function f(x), and the right-hand side represents the function g(x).

The equation has two real solutions because the graphs of f(x) and g(x) intersect at two points.

The equation x²+6x+9=0 has one real solution.

This is because the equation can be written as x²+6x equal to −9, where the left-hand side represents the given function f(x), and the right-hand side represents the function g(x).

The equation has one real solution because the graphs of f(x) and g(x) intersect at one point.

The equation x²+6x+12=0 has no real solution.

This is because the equation can be written as x²+6x equal to −12, where the left-hand side represents the given function f(x), and the right-hand side represents the function g(x).

The equation has no real solution because the graphs of f(x) and g(x) intersect at no points.

Page 146 Problem 19 Answer

The given graph shows the graph of the function f(x) equal to −1/2x²+3x. The given equations are:​

−1/2x²+3x−3=0

−1/2x²+3x−9/2=0

−1/2x²+3x−6=0

​The question requires to use the given graph to determine the number of real solutions of the given equations.

To determine the number of real solutions, rewrite the equations using subtraction in the form ax²+bx=−c, and equate the left and right sides of the rewritten equations to f(x) and g(x).

Then, draw the graph of each g(x) on the given graph using horizontal lines.

Next, use the points of intersection of f(x) and g(x) to determine the number of points where they intersect.

Finally, use the points to determine the number of real solutions for each g(x), and thus the number of real solutions for each equation.

The first given quadratic equation is −1/2x²+3x−3=0.

Add 3 on both sides of the equation to rewrite it in the form ax²+bx=−c.

−1/2x²+3x=3

The left-hand side of the equation is of the form ax²+bx. Equating the left-hand side of the rewritten equation to f(x) gives the function f(x)=−1/2x²+3x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=3.

The second given quadratic equation is −1/2x²+3x−9/2=0.

Add 9/2 on both sides of the equation to rewrite it in the form ax²+bx=−c.

−1/2x²+3x=9/2

The left-hand side of the equation is of the form ax²+bx. Equating the left-hand side of the rewritten equation to f(x) gives the function f(x)=−1/2x²+3x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=9/2.

The third given quadratic equation is −1/2x²+3x−6=0.

Add 6 on both sides of the equation to rewrite it in the form ax²+bx=−c.

−1/2x²+3x=6

The left-hand side of the equation is of the form ax²+bx. Equating the left-hand side of the rewritten equation to f(x) gives the function f(x)=−1/2x²+3x.

The right-hand side of the equation is of the form −c. Equating the right-hand side of the rewritten equation to g(x), the obtained function is g(x)=6.

Construct a table having columns for the given equations, the rewritten equations, and the obtained expressions for f(x) and g(x).

From the completed table from previous part of the exercise, it can be determined that functions to be graphed are:​

g(x)=3

g(x)=9/2

g(x)=6

Graph the functions on the given graph by drawing horizontal lines having the y-intercept equal to the right-hand side of the equations.

From the graph, it can be observed that the graph of f(x) and g(x)=3 intersect each other at two points.

This means that −1/2x²+3x=3 for two real values of x.

Subtracting 3 from both sides, the equation becomes −1/2x²+3x−3=0.

Therefore, the equation −1/2x²+3x−3=0 has two real solutions.

From the graph, it can be observed that the graph of f(x) and g(x)=9/2 touch each other at one point.

This means that −1/2x²+3x=9/2 for one real value of x.

Subtracting 9/2 from both sides, the equation becomes −1/2x²+3x−9/2=0.

Therefore, the equation −1/2x²+3x−9/2=0 has one real solution.

From the graph, it can be observed that the graph of f(x) and g(x)=6 do not touch or intersect each other at any point.

This means that −1/2x²+3x=6 for no real value of x.

Subtracting 6 from both sides, the equation becomes −1/2x²+3x−6=0.

Therefore, the equation −1/2x²+3x−6=0 has no real solution.

The equation −1/2x²+3x−3=0 has two real solutions.

This is because the equation can be written as −1/2x²+3x equal to 3, where the left-hand side represents the given function f(x), and the right-hand side represents the function g(x).

The equation has two real solutions because the graphs of f(x) and g(x) intersect at two points.

The equation −1/2x²+3x−9/2=0 has one real solution.

This is because the equation can be written as −1/2x²+3x equal to 9/2, where the left-hand side represents the given function f(x), and the right-hand side represents the function g(x).

The equation has one real solution because the graphs of f(x) and g(x) intersect at one point.

The equation −1/2x²+3x−6=0 has no real solution.

This is because the equation can be written as −1/2x²+3x equal to 6, where the left-hand side represents the given function f(x), and the right-hand side represents the function g(x).

The equation has no real solution because the graphs of f(x) and g(x) intersect at no points.

Page 146 Problem 20 Answer

The given equation is x²+2x+8=0.

The question requires to solve the given equation by completing the square, and then state whether the obtained solutions are real or non-real.

To solve the equation by completing the square, rewrite the equation in the form x²+bx=c. Then, identify the value of b, and calculate (b/2)².

Then, add (b/2)² on both sides of the rewritten equation, factor the left-side expression using algebraic identities, and use the definition of square root.

Finally, solve the obtained equation to determine the required solutions, and observe them to state whether they are real or non-real.

Subtract 8 from both sides of x²+2x+8=0.

​x²+2x+8−8=0−8

x²+2x=−8……………(1)

Comparing the equation (1) to the equation x²+bx=c, it can be determined that the value of b is 2.

Therefore, the value of (b/2)² is (2/2)²

=(1)², that is 1.

Add the value of (b/2)², that is 1, on both sides of equation (1).

Then, factorise the left-hand side expression using the algebraic identity a²+2ab+b²=(a+b)², and simplify the right-hand side expression.

​x²+2x+1=−8+1

(x+1)²=−8+1

(x+1)²=−7

​Use the definition of square root to simplify (x+1)²=−7. Then, subtract 1 from both sides, and rewrite the expression using the imaginary unit i.

​(x+1)²=−7

x+1=±√−7

x=−1±√−7

x=−1±i√7

Therefore, the required solutions are x=−1+i√7 and x=−1−i√7.

It can be observed that the solutions x=−1±i√7 contain the imaginary unit i, which is equal to the square root of the negative number −1.

Therefore, the solutions are non-real.

The required solutions are x=−1+i√7 and x=−1−i√7.

The solutions are non-real.

Page 147 Problem 21 Answer

The given equation is x²−5x=−20.

The question requires to solve the given equation by completing the square, and then state whether the obtained solutions are real or non-real.

To solve the equation by completing the square, compare the equation to x²+bx=c.

Then, identify the value of b, and calculate (b/2)². Then, add (b/2)² on both sides of the rewritten equation, factor the left-side expression using algebraic identities, and use the definition of square root.

Finally, solve the obtained equation to determine the required solutions, and observe them to state whether they are real or non-real.

Let the equation x²−5x=−20 be equation (1).

Comparing the equation (1) to the equation x²+bx=c, it can be determined that the value of b is −5.

Therefore, the value of (b/2)² is (−5/2)²=25/4.

Add the value of (b/2)², that is 25/4, on both sides of equation (1).

Then, factorise the left-hand side expression using the algebraic identity a²−2ab+b²=(a−b)², and simplify the right-hand side expression.

​x²−5x+25/4

=−20+25/4

(x−5/2)²=−20+25/4

(x−5/2)²=−55/4

Use the definition of square root to simplify (x−5/2)²=−55/4. Then, add 5/2 on both sides, and rewrite the expression using the imaginary unit i.

​(x−5/2)²=−55/4

x−5/2=±√−55/4

x=5/2±√−55/4

x=5/2±i√55/4

Apply the quotient property of square roots, and then simplify the square root in the denominator.

​x=5/2±i√55/√4

x=5/2±i√55/2

x=5±i√55/2

Therefore, the required solutions are x=5+i√55/2 and x=5−i√55/2.

 

It can be observed that the solutions x=5±i√55/2 contain the imaginary unit i, which is equal to the square root of the negative number −1.

Therefore, the solutions are non-real.

The required solutions are x=5+i√55/2 and x=5−i√55/2.

The solutions are non-real.

Page 147 Problem 22 Answer

The given equation is 5x²−6x=8.

The question requires to solve the given equation by completing the square, and then state whether the obtained solutions are real or non-real.

To solve the equation by completing the square, rewrite the equation in the form x²+bx=c. Then, identify the value of b, and calculate (b/2)².

Then, add (b/2)² on both sides of the rewritten equation, factor the left-side expression using algebraic identities, and use the definition of square root.

Finally, solve the obtained equation to determine the required solutions, and observe them to state whether they are real or non-real.

Divide both sides of 5x²−6x=8 by 5.​

5x²−6x/5=8/5

x²−6/5

x=8/5……………(1)​

Comparing the equation (1) to the equation x²+bx=c, it can be determined that the value of b is −6/5.

Therefore, the value of (b/2)² is (−6/5)(2/2)=(−3/5)², that is 9/25.

Add the value of (b/2)², that is 9/25, on both sides of equation (1).

Then, factorise the left-hand side expression using the algebraic identity a²−2ab+b²=(a−b)², and simplify the right-hand side expression.

​x²−6/5

x+9/25=8/5+925

(x−3/5)²=8/5+9/25

(x−3/5)²=49/25

Use the definition of square root to simplify (x−3/5)²=49/25, and then add 3/5 on both sides.​

(x−3/5)²=49/25

x−3/5=±√49/25

x=3/5±√49/25

Apply the quotient property of square roots, and then simplify the square roots.​

x=3/5±√49/√25

x=3/5±7/5

When x is equal to 3/5−7/5, the simplified value of x is 3−7/5, that is −4/5.

When x is equal to 3/5+7/5, the simplified value of x is 3+7/5=10/5, that is 2.

Therefore, the required solutions are x=−4/5 and x=2.

It can be observed that the solutions do not contain the square root of a negative number, or the imaginary unit i.

Therefore, the solutions are real.

The required solutions are x=−4/5 and x=2.

The solutions are real.

Page 147 Problem 23 Answer

The given equation is 7x²+13x=5.

The question requires to solve the given equation by completing the square, and then state whether the obtained solutions are real or non-real.

To solve the equation by completing the square, rewrite the equation in the form x²+bx=c.

Then, identify the value of b, and calculate (b/2)². Then, add (b/2)² on both sides of the rewritten equation, factor the left-side expression using algebraic identities, and use the definition of square root.

Finally, solve the obtained equation to determine the required solutions, and observe them to state whether they are real or non-real.

Divide both sides of 7x²+13x=5 by 7.

​7x²+13x/7=5/7

x²+13/7

x=5/7……………(1)

Comparing the equation (1) to the equation x²+bx=c, it can be determined that the value of b is 13/7.

Therefore, the value of (b/2)² is (13/7)²2=(13/14)², that is 169/196.

Add the value of (b/2)², that is 169/196, on both sides of equation (1).

Then, factorise the left-hand side expression using the algebraic identity a²+2ab+b²=(a+b)², and simplify the right-hand side expression.

​x²+13/7

x+169/196

=5/7+169/196

(x+13/14)²=5/7+169/196

(x+13/14)²=309/196​

Use the definition of square root to simplify (x+13/14)²=309/196, and then subtract 13/14 from both sides.

​(x+13/14)²=309/196

x+13/14=±√309/196

x=−13/14±√309/196

Apply the quotient property of square roots, and then simplify the square roots.

​x=−13/14±√309/√196

x=−13/14±√309/14

x=−13±√309/14

Therefore, the required solutions are x=−13+√309/14 and x=−13−√309/14.

It can be observed that the solutions x=−13±√309/14 do not contain the square root of a negative number, or the imaginary unit i.

Therefore, the solutions are real.

The required solutions are x=−13+√309/14 and x=−13−√309/14.

The solutions are real.

Page 147 Problem 24 Answer

The given equation is −x²−6x−11=0.

The question requires to solve the given equation by completing the square, and then state whether the obtained solutions are real or non-real.

To solve the equation by completing the square, rewrite the equation in the form x²+bx=c. Then, identify the value of b, and calculate (b/2)².

Then, add (b/2)² on both sides of the rewritten equation, factor the left-side expression using algebraic identities, and use the definition of square root.

Finally, solve the obtained equation to determine the required solutions, and observe them to state whether they are real or non-real.

Multiply both sides of −x²−6x−11=0 by −1, and then subtract 11 from both sides.

​−1(−x²−6x−11)=−1(0)

x²+6x+11=0

x²+6x=−11……………(1)

Comparing the equation (1) to the equation x²+bx=c, it can be determined that the value of b is 6.

Therefore, the value of (b/2)² is (6/2)²

=(3)², that is 9.

Add the value of (b/2)², that is 9, on both sides of equation (1).

Then, factorise the left-hand side expression using the algebraic identity a²+2ab+b²=(a+b)², and simplify the right-hand side expression.

​x²+6x+9=−11+9

(x+3)²=−11+9

(x+3)²=−2

Use the definition of square root to simplify (x+3)²=−2.

Then, subtract 3 from both sides, and rewrite the expression using the imaginary unit i.

​(x+3)²=−2

x+3=±√−2

x=−3±√−2

x=−3±i√2

Therefore, the required solutions are x=−3+i√2 and x=−3−i√2.

It can be observed that the solutions x=−3±i√2. contain the imaginary unit i, which is equal to the square root of the negative number −1.

Therefore, the solutions are non-real.The required solutions are x=−3+i√2 and x=−3−i√2.

The solutions are non-real.

Page 147 Problem 25 Answer

The given equation is 7x²−11x+10=0.

The question requires to state the number of solutions without solving the equation, and determine whether the solutions are real or non-real.

To answer the question, compare the equation to ax²+bx+c=0 and identify the values of a,b,c.

Then, calculate the discriminant of the equation, and observe its value to determine the number and type of solutions of the given equation.

It can be observed that the given equation is a quadratic equation of the form ax²+bx+c=0.

Comparing the given equation to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are 7,−11,10 respectively.

Substitute 7 for a, −11 for b, and 10 for c in the formula for discriminant b²−4ac. Then, multiply the terms, and subtract the terms.

​b²−4ac=(−11)²

−4(7)(10)

=121−280

=−159

The discriminant is equal to −159, which is less than 0.

This means that the equation 7x²−11x+10=0 has two non-real solutions.

Thus, the number of solutions of the given equation is two, and the two solutions are non-real.

The given equation has two solutions.

The two solutions of the given equation are non-real.

Page 147 Problem 26 Answer

The given equation is −x²−2/5

x=1.

The question requires to state the number of solutions without solving the equation, and determine whether the solutions are real or non-real.

To answer the question, rewrite the equation in the form ax²+bx+c=0 and identify the values of a,b,c.

Then, calculate the discriminant of the equation, and observe its value to determine the number and type of solutions of the given equation.

Subtract 1 from both sides of −x²−2/5

x=1.​

−x²−2/5

x−1=1−1

−x²−2/5

x−1=0…………(1)

Comparing the equation (1) to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are −1,−2/5,−1 respectively.

Substitute −1 for a, −2/5 for b, and −1 for c in the formula for discriminant b²−4ac. Then, multiply the terms, and subtract the terms.

​b²−4ac=(−2/5)²

−4(−1)(−1)

=4/25−4

=4−100/25

=−96/25

The discriminant is equal to −96/25, which is less than 0.

This means that the equation −x²−2/5 x=1 has two non-real solutions.

Thus, the number of solutions of the given equation is two, and the two solutions are non-real.

The given equation has two solutions.

The two solutions of the given equation are non-real.

Page 147 Problem 27 Answer

The given equation is 4x²+9=12x.

The question requires to state the number of solutions without solving the equation, and determine whether the solutions are real or non-real.

To answer the question, rewrite the equation in the form ax²+bx+c=0 and identify the values of a,b,c.

Then, calculate the discriminant of the equation, and observe its value to determine the number and type of solutions of the given equation.

Subtract 12x from both sides of 4x²+9=12x.

​4×2+9−12x=12x−12x

4x²−12x+9=0…………(1)

Comparing the equation (1) to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are 4,−12,9 respectively.

Substitute 4 for a, −12 for b, and 9 for c in the formula for discriminant b²−4ac. Then, multiply the terms, and subtract the terms.

​b²−4ac=(−12)²

−4(4)(9)

=144−144

=0

The discriminant is equal to 0.

This means that the equation 4x²+9=12x has one real solution.

Thus, the number of solutions of the given equation is one, and the solution is real.

The given equation has one solution.

The one solution of the given equation is real.

Page 150 Problem 28 Answer

The given table shows seven quadratic functions.

The question requires to complete the given table by placing X marks in the column which corresponds to the number and type of solutions of the given equations.

To complete the table, compare each equation to ax²+bx+c=0 and identify the values of a,b,c.

Then, calculate the discriminant of each equation, and observe its value to determine the number and type of solutions of each given equation.

Finally, use the determined number and type of solutions and place the X mark in the corresponding column for each equation to complete the table.

The first given equation is x²−3x+1=0.

Comparing the equation to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are 1,−3,1 respectively.

Calculate the discriminant of the equation using the formula for discriminant b²−4ac. Then, multiply the terms, and combine the terms.

​b²−4ac=(−3)²

−4(1)(1)

=9−4

=5

The discriminant is equal to 5, which is greater than 0.

This means that the equation x²−3x+1=0 has two real solutions.

The second given equation is x²−2x+1=0.

Comparing the equation to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are 1,−2,1 respectively.

Calculate the discriminant of the equation using the formula for discriminant b²−4ac. Then, multiply the terms, and combine the terms.

b²−4ac=(−2)²

−4(1)(1)

=4−4

=0

The discriminant is equal to 0.

This means that the equation x²−2x+1=0 has one real solution.

The third given equation is x²−x+1=0.

Comparing the equation to the equation ax²+bx+c=0, it can be determined that the values of a,b,care 1,−1,1 respectively.

Calculate the discriminant of the equation using the formula for discriminant b²−4ac. Then, multiply the terms, and combine the terms.

​b²−4ac=(−1)²

−4(1)(1)

=1−4

=−3

The discriminant is equal to −3, which is less than 0.

This means that the equation x²−x+1=0 has two non-real solutions.

The fourth given equation is x²+1=0.

Comparing the equation to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are 1,0,1 respectively.

Calculate the discriminant of the equation using the formula for discriminant b²−4ac. Then, multiply the terms, and combine the terms.

​b²−4ac=(0)²

−4(1)(1)

=0−4

=−4

​The discriminant is equal to −4, which is less than 0.

This means that the equation x²+1=0 has two non-real solutions.

The fifth given equation is x²+x+1=0.

Comparing the equation to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are 1,1,1 respectively.

Calculate the discriminant of the equation using the formula for discriminant b²−4ac. Then, multiply the terms, and combine the terms.

​b²−4ac=(1)²

−4(1)(1)

=1−4

=−3

The discriminant is equal to −3, which is less than 0.

This means that the equation x²+x+1=0 has two non-real solutions.

The sixth given equation is x²+2x+1=0.

Comparing the equation to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are 1,2,1 respectively.

Calculate the discriminant of the equation using the formula for discriminant b²−4ac. Then, multiply the terms, and combine the terms.

​b²−4ac=(2)²

−4(1)(1)

=4−4

=0

The discriminant is equal to 0.

This means that the equation x²+2x+1=0 has one real solution.

The seventh given equation is x²+3x+1=0.

Comparing the equation to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are 1,3,1 respectively.

Calculate the discriminant of the equation using the formula for discriminant b²−4ac. Then, multiply the terms, and combine the terms.

​b²−4ac=(3)²−4(1)(1)

=9−4

=5

The discriminant is equal to 5, which is greater than 0.

This means that the equation x²+3x+1=0 has two real solutions.

Complete the given table by placing X mark in the column which represents the number and type of solutions for each equation.

The completed table with the X marks placed in the appropriate columns is:

Page 151 Problem 29 Answer

The given equation is −x²+2x−3=0. Also, a student’s attempt at solving the equation by completing the square is given.

The question requires to determine the error in the student’s solution, and correct it.

To find and correct the error, observe the solution and determine whether the student has correctly written the equation in the form x²+bx=c before factoring the left side of the equation, and observe whether the factoring is correct.

Then, rewrite the equation in the form x²+bx=c. Then, identify the value of b, and calculate (b/2)².

Then, add (b/2)² on both sides of the rewritten equation, factor the left-side expression using algebraic identities, and use the definition of square root.

Finally, solve the obtained equation to determine the correct solutions of the given equation.

When solving a quadratic equation by completing the square, the quadratic equation is written in the form x²+bx=c before factoring the left side of the equation.

It can be observed that in the student’s solution, the quadratic equation is written as −x²+2x=3 before factoring the left side of the equation.

The coefficient of x² in the rewritten equation is −1, instead of 1.

Therefore, the error in the solution is that the student has not written the quadratic equation in the form x²+bx=c before factoring the left side of the equation.

Multiply both sides of −x²+2x−3=0 by −1, and then subtract 3 from both sides.

​−1(−x²+2x−3)=−1(0)

x²−2x+3=0

x²−2x=−3……………(1)

Comparing the equation (1) to the equation x²+bx=c, it can be determined that the value of b is −2.

Therefore, the value of (b/2)² is (−2/2)²

=(−1)², that is 1.

Add the value of (b/2)², that is 1, on both sides of equation (1).

Then, factorise the left-hand side expression using the algebraic identity a²−2ab+b²=(a−b)², and simplify the right-hand side expression.

​x²−2x+1=−3+1

(x−1)²=−3+1

(x−1)²=−2

​Use the definition of square root to simplify (x−1)²=−2.

Then, add 1 on both sides, and rewrite the expression using the imaginary unit i.

​(x−1)²=−2

x−1=±√−2

x=1±√−2

x=1±i√2

Therefore, the correct solutions are x=1+i√2 and x=1−i√2.

The error in the solution is that the student has not written the quadratic equation in the form x²+bx=c before factoring the left side of the equation.

This is because the student has written the quadratic equation in the form −x²+bx=c, where the coefficient of x² is −1, instead of 1.

The correct solutions are x=1+i√2 and x=1−i√2.

Page 151 Problem 30 Answer

 

The given equation is x²+8x+c0.

The question requires to determine the values of c for which the given equation has one real solution, two real solutions, and two non-real solutions.

To determine the values, compare the equation to ax²+bx+c=0 and identify the values of a,b,c. Then, calculate the discriminant of the equation.

Next, take the case where the equation has one real solution, equate the determinant to 0, and solve the equation to determine the value of c.

Similarly, take the cases where the equation has two real or two non-real solutions, form inequalities using the determinant and the number 0, and solve them to determine the values of c for which the equation has two real or two non-real solutions.

It can be observed that the given equation is a quadratic equation of the form ax²+bx+c=0.

Comparing the given equation to the equation ax²+bx+c=0, it can be determined that the values of a,b,c are 1,8,c respectively.

Substitute 1 for a and 8 for b in the formula for discriminant b²−4ac, and multiply the terms.

​b²−4ac=(8)²

−4(1)(c)

=64−4c

The discriminant is equal to 64−4c.

First case: The given equation has one real solution.

If the quadratic equation has one real solution, then the discriminant 64−4c is equal to 0.

Add 4c on both sides of 64−4c=0, and then divide both sides by 4.

​64−4c=0

64=4c

16=c

Therefore, the given equation has one real solution if c=16.

Second case: The given equation has two real solutions.

If the quadratic equation has two real solutions, then the discriminant 64−4c is greater than 0.

Add 4c on both sides of 64−4c>0, and then divide both sides by 4.

​64−4c>0

64>4c

16>c

Therefore, the given equation has two real solutions if c<16.

Third case: The given equation has two non-real solutions. If the quadratic equation has two non-real solutions, then the discriminant 64−4cis less than 0.

Add 4c on both sides of 64−4c<0, and then divide both sides by 4.

​64−4c<0

64<4c

16<c

Therefore, the given equation has two non-real solutions if c>16.

The given equation has one real solution if c=16. two real solutions if c<16, and two non-real solutions if c>16.

Page 152 Exercise 1 Answer

It is given that the projectile motion model h(t)=−16t²+v0t+h0 models the ball’s height in t seconds, where v0 is the initial vertical velocity in feet per second, and h0 is the initial height in feet.

Also, it is given that the ball is hit when it is 4 feet above the ground, and it travelled vertically after being hit at the speed 80 feet per second.

Matt’s claim is that the ball must have reached the height 110 feet.

The question requires to model the ball’s height using the given information, and then use the model to determine whether Matt’s claim is correct or not.

To answer the question, use the initial vertical velocity and initial height of the ball in the projectile equation to determine the model for the ball’s height.

Next, let the height h(t) be equal to 110 feet.

Then, use the model and the height to form a quadratic equation in terms of t.

Next, compare the equation to at²+bt+c=0 and identify the values of a,b, and c.

Finally, determine the discriminant of the equation to determine whether the solutions are real or non-real, and thus, determine whether it is possible for the ball to reach a height of 110 feet.

The initial vertical velocity of the ball is the speed at which it travelled vertically after being hit, that is 80 feet per second.

The initial height of the ball is the height at which it is hit, that is 4 feet.

Substitute 80 for v0 and 4 for h0 in the projectile motion model h(t)=−16t²+v0t+h0.

h(t)=−16t²+80t+4

Thus, the model for the height of the ball in t seconds is h(t)=−16t²+80t+4.

Let the height of the ball be 110 feet. Therefore, h(t)=110.

Substitute −16t²+80t+4 for h(t) in h(t)=110, and subtract 110 from both sides.

​−16t²+80t+4=110

−16t²+80t−106=0……………(1)

The equation based on Matt’s claim is −16t²+80t−106=0.

Comparing the equation (1) to the equation at²+bt+c=0, it can be determined that the values of a,b,c are −16,80,−106 respectively.

Substitute −16 for a, 80 for b, and −106 for c in the formula for discriminant b²−4ac. Then, multiply the terms, and subtract the terms.

​b²−4ac=(80)²−4(−16)(−106)

=6400−6784

=−384

The discriminant is equal to −384, which is less than 0.

This means that the equation −16t²+80t−106=0 has two non-real solutions.

Therefore, there is no real value of t for which the height of the ball is equal to 110 feet.

Thus, the ball cannot reach the height equal to 110 feet, and Matt’s claim is not correct.

The model for the height of the ball in t seconds is h(t)=−16t²+80t+4.

The equation based on Matt’s claim is −16t²+80t−106=0.

Matt’s claim is not correct. The ball cannot reach the height equal to 110 feet.

Page 152 Exercise 2 Answer

It is given that the projectile motion model h(t)=−16t²+v0t+h0 models the ball’s height in t seconds, where v0 is the initial vertical velocity in feet per second, and h0 is the initial height in feet.

Also, it is given that the ball is hit when it is 4 feet above the ground, and it travelled vertically after being hit at the speed 80 feet per second.

The question requires to determine whether the ball reaches a height of 100 feet or not, and explain the reasoning.

Given that the model for the height of the ball in t seconds is h(t)=−16t²+80t+4, as determined in the previous part of this exercise.

To answer the question, let the height h(t) be equal to 100 feet. Then, use the model and the height to form a quadratic equation in terms of t.

Next, compare the equation to at²+bt+c=0 and identify the values of a,b, and c.

Finally, determine the discriminant of the equation to determine whether the solutions are real or non-real, and thus, determine whether it is possible for the ball to reach a height of 100 feet.

Let the height of the ball be 100 feet. Therefore, h(t)=100.

Substitute −16t²+80t+4 for h(t) in h(t)=100, and subtract 100 from both sides.​

−16t²+80t+4=100

−16t²+80t−96=0……………(1)

Comparing the equation (1) to the equation at²+bt+c=0, it can be determined that the values of a,b,c are −16,80,−96 respectively.

Substitute −16 for a, 80 for b, and −96 for c in the formula for discriminant b²−4ac. Then, multiply the terms, and subtract the terms.

​b²−4ac=(80)²

−4(−16)(−96)

=6400−6144

=256

The discriminant is equal to 256, which is greater than 0.

This means that the equation h(t)=100 has two real solutions.

Therefore, there is some real value of t for which the height of the ball is equal to 100 feet.

Thus, the ball does reach a height of 100 feet.

The ball does reach a height of 100 feet. This is because the quadratic equation −16t²+80t−96=0

representing the height of the ball as 100

feet has two real solutions. There exists some real value of t for which the height of the ball is equal to 100 feet.

Page 152 Exercise 3 Answer

It is given that the projectile motion model h(t)=−16t²+v0t+h0 models the ball’s height in t seconds, where v0 is the initial vertical velocity in feet per second, and h0

is the initial height in feet. Also, it is given that the ball is hit when it is 4 feet above the ground, and it travelled vertically after being hit at the speed 80 feet per second.

Also, hmax represents the maximum height of the ball.

The question requires to show how the value of hmax can be found using the discriminant of the quadratic formula.

Given that the model for the height of the ball in t seconds is h(t)=−16t²+80t+4, as determined in part (a) of this exercise.

To determine the maximum height, let the height h(t) be equal to hmax.

Then, use the model and the height to form a quadratic equation in terms of t.

Next, compare the equation to at²+bt+c=0 and identify the values of a,b, and c.

Then, determine the discriminant of the equation and equate it to 0 to form a linear equation in terms of hmax.

Finally, solve the equation for hmax to determine the maximum height, and explain how it is calculated.

Let the height of the ball be maximum. Therefore, h(t)=hmax.

Substitute −16t²+80t+4 for h(t) in h(t)=hmax, and subtract hmax from both sides.​

−16t²+80t+4=hmax

−16t²+80t+(4−hmax)=0……………(1)

Comparing the equation (1) to the equation at²+bt+c=0, it can be determined that the values of a,b,c are −16,80,4−hmax respectively.

Substitute −16 for a, 80 for b, and 4−hmax for c in the formula for discriminant b²−4ac. Then, multiply the terms, and combine the terms.

​b²−4ac=(80)²−4(−16)(4−hmax)

=6400+256−64hmax

=6656−64hmax

The discriminant is equal to 6656−64hmax.

The solutions of the equation −16t²+80t+(4−hmax)=0 represent the time at which the ball reaches the maximum height.

The ball reaches the maximum height only once. This is because after reaching the maximum height, it starts falling down.

Therefore, the equation −16t²+80t+(4−hmax)=0 has only one real solution.

Thus, the discriminant of the equation should be equal to 0.

Equate the discriminant 6656−64hmax to 0, add 64hmax on both sides, and divide both sides by 64.

6656−64hmax=0

6656=64hmax

6656/64=hmax

104=hmax

Therefore, the maximum height of the ball is 104 feet.

Thus, the maximum height is calculated by using the model and the maximum height to form the equation −16t²+80t+(4−hmax)=0, calculating the discriminant and equating it to 0, and finally solving the resulting linear equation for hmax.

The ball does reach a maximum height of 104 feet.

This is calculated by writing the equation −16t² +80t+(4−hmax)=0 using the given model, calculating the discriminant of the equation, equating the discriminant to 0, and solving the resulting equation for hmax.

The discriminant of the equation is equated to 0 because the ball reaches the maximum height only once.

Page 152 Exercise 4 Answer

It is given that the projectile motion model h(t)=−16t²+v0t+h0 models the ball’s height in t seconds, where v0 is the initial vertical velocity in feet per second, and h0

is the initial height in feet. Also, it is given that the ball is hit when it is 4 feet above the ground, and it travelled vertically after being hit at the speed 80 feet per second.

The question requires to determine the time when the ball reaches its maximum height.

Given that the model for the height of the ball in t seconds is h(t)=−16t²+80t+4, as determined in part (a) of this exercise.

Given that the maximum height of the ball is 104 feet, as determined in the previous part of this exercise.

To answer the question, let the height h(t) be equal to the maximum height of 104 feet.

Then, use the model and the height to form a quadratic equation in terms of t, and compare the equation to at²+bt+c=0 to identify the values of a,b,c.

Finally, use the quadratic formula, and simplify the expression to determine the time when the ball reaches its maximum height.

Let the height of the ball be the maximum height of 104 feet. Therefore, h(t)=104.

Substitute −16t²+80t+4 for h(t) in h(t)=104, and subtract 104 from both sides.

​−16t²+80t+4=104

−16t²+80t−100=0……………(1)

Comparing the equation (1) to the equation at²+bt+c=0, it can be determined that the values of a,b,c are −16,80,−100 respectively.

Substitute −16 for a, 80 for b, and −100 for c

in the quadratic formula t=−b±√b²−4ac/2a.

Then, multiply the terms, combine the terms in the square root, and simplify the expression.

t=−(80)±√(80)²−4(−16)(−100)/2(−16)

t=−80±√6400−6400/−32

t=−80±√0/−32

t=−80/−32

t=2.5

Therefore, the time in which the ball reaches its maximum height is 2.5 seconds.

The ball reaches the maximum height in 2.5 seconds.

Page 153 Exercise 5 Answer

A quadratic equation can be used to model a real-world problem.

Then, the equation can be solved for one variable by using completing the square method, factoring, or the quadratic formula, and thus, obtain the required solutions.

For example: There is a rectangular field which has an area equal to 875

square feet, and the ratio of its length and width is 7:5. It is required to determine the dimensions of the field.

The length and width of the field can be written as 7m and 5m respectively.

Since the area is given as 875 square feet, the area can be modelled using the quadratic equation 7m⋅5m=875.

This equation can be solved to obtain the values of m, and thus, obtain the dimensions of the rectangular field.

Thus, the real-world problems can be modelled using quadratic equations, and then solved to determine the required solutions.

Quadratic equations can be used to model real-world problems such as the area of a rectangular field when the ratio of the dimensions is given.

Then, the given area can be substituted in the equation, and the quadratic equation can be solved using completing the square method, factoring, or the quadratic formula to obtain the solutions.

The accepted solution can then be used to determine the dimensions of the rectangular field. Similarly, other real-world problems can be solved using quadratic equations.

Page 154 Exercise 6 Answer

The given equation is x²−16=0.

The question requires to solve the given equation by taking square roots.

To solve the equation, isolate the square x² on one side of the equation using addition, and use the definition of square root to simplify the expression.

Add 16 on both sides of x²−16=0, use the definition of square root, and simplify the square root to solve the equation.

​x²−16+16=0+16

x²=16

x=±√16

x=±4

Thus, the required solutions are x=±4.

The required solutions are x=±4.

Page 154 Exercise 7 Answer

The given equation is 2x²−10=0.

The question requires to solve the given equation by taking square roots.

To solve the equation, isolate the square x² on one side of the equation using addition and division, and use the definition of square root to simplify the expression.

Add 10 on both sides of 2x −10=0, and then divide both sides by 2.

​2x²−10+10=0+10

2x²=10

2x²/2=10/2

x²=5

​Use the definition of square root to solve the equation x²=5.

​x²=5

x=±√5

Thus, the required solutions are x=±√5.

The required solutions are x=±√5.

Page 154 Exercise 8 Answer

The given equation is x²+6x+10=0.

The question requires to solve the given equation by completing the square.

To solve the equation by completing the square, rewrite the equation in the form x²+bx=c. Then, identify the value of b, and calculate (b/2)².

Then, add (b/2)² on both sides of the rewritten equation, factor the left-side expression using algebraic identities, and use the definition of square root.

Finally, solve the obtained equation to determine the required solutions.

Subtract 10 from both sides of x²+6x+10=0.

​x²+6x+10−10=0−10

x²+6x=−10……………(1)

Comparing the equation (1) to the equation x²+bx=c, it can be determined that the value of b is 6.

Therefore, the value of (b/2)² is (6/2)²=(3)², that is 9.

Add the value of (b/2)², that is 9, on both sides of equation (1).

Then, factorise the left-hand side expression using the algebraic identity a²+2ab+b²=(a+b)², and simplify the right-hand side expression.

x²+6x+9=−10+9

(x+3)²=−10+9

(x+3)²=−1

​Use the definition of square root to simplify (x+3)²=−1. Then, subtract 3 from both sides, and rewrite the expression using the imaginary unit i.

​(x+3)²=−1

x+3=±√−1

x=−3±√−1

x=−3±i

Therefore, the required solutions are x=−3+i and x=−3−i.

The required solutions are x=−3+i and x=−3−i.

Page 154 Exercise 9 Answer

The given equation is x²−4x+4=0.

The question requires to solve the given equation by factoring it.

To solve the equation, write the pair of factors of the constant, and identify the pair of factors whose sum is equal to the coefficient of the middle term.

Then, factorise the left-hand side of the given quadratic equation, solve it to obtain the values of x, and thus, obtain the required solution.

The constant in the given quadratic equation is 4.

Write the pair of factors of the constant 4.

1 and 4

2 and 2

−1 and −4

−2 and −2

It can be observed that the sum of the pair of factors −2 and −2 is equal to the coefficient of the middle term in the given quadratic equation, that is −4.

Therefore, the left-hand side of the given quadratic equation can be factorised as (x−2)(x−2).

Thus, the equation becomes (x−2)(x−2)=0.

The equation (x−2)(x−2)=0 is true only when x−2 is equal to 0.

When x−2 is equal to 0, the value of x is equal to 2.

Thus, the required solution is x=2.

The required solution is x=2.

Page 154 Exercise 10 Answer

The given equation is x²−x−30=0.

The question requires to solve the given equation by factoring it.

To solve the equation, write the pair of factors of the constant, and identify the pair of factors whose sum is equal to the coefficient of the middle term.

Then, factorise the left-hand side of the given quadratic equation, solve it to obtain the values of x, and thus, obtain the required solutions.

The constant in the given quadratic equation is −30.

Write the pair of factors of the constant −30.

1 and −30

2 and −15

3 and −10

5 and −6

6 and −5

10 and −3

15 and −2

30 and −1

It can be observed that the sum of the pair of factors 5 and −6 is equal to the coefficient of the middle term in the given quadratic equation, that is −1.

Therefore, the left-hand side of the given quadratic equation can be factorised as (x+5)(x−6).

Thus, the equation becomes (x+5)(x−6)=0.

The equation (x+5)(x−6)=0 is true only when either x+5 is equal to 0, or x−6 is equal to 0.

When x+5 is equal to 0, the value of x is equal to −5.

When x−6 is equal to 0, the value of x is equal to 6.

Thus, the required solutions are x=−5 and x=6.

The required solutions are x=−5 and x=6.

Page 154 Exercise 11 Answer

Assume the quadratic equation ax²+bx+c=0. First, the quadratic equation is rewritten in the form x²+b/a

x+c/a=0 by division, such that the coefficient of x² is 1.Then, the constant in the quadratic equation is identified, that is the value of c/a.

Then, the pair of factors of the constant c/a are written. Next, the pair of factors whose sum is equal to coefficient of the middle term is chosen.

In the rewritten quadratic equation, the coefficient of the middle term is b/a.

Therefore, if there is a pair of factors p and q of the constant c/a, that is pq=c/a, then the sum of the factors p and q is b/a.

This means that if there is no pair of factors of the constant c/a whose sum is not equal to the coefficient of the middle term is b/a, then the quadratic equation cannot be solved by factoring.

Thus, a quadratic equation of the form x²+bx+c=0 can be solved by factoring only if there exists a pair of factors of the constant whose sum is equal to the coefficient of the middle term.

It is possible to solve a quadratic equation by factoring if there exists a pair of factors of the constant whose sum is equal to the coefficient of the middle term.

The constant and coefficient of middle term are taken only after writing the equation such that the coefficient of x² is 1.

Page 154 Exercise 12 Answer

The complex solutions of a quadratic equation can be calculated either by completing the square, or by using the quadratic formula.

To complete the square and calculate the complex solutions of a quadratic equation, first the equation is rewritten in the form x²+bx=c. Next, the value of b is identified, and (b/2)² is calculated.

The value of (b/2)² is added on both sides of the rewritten equation, and the left-side of the equation is factorised.

Finally, the definition of square root is used, and the equation is solved for the required values of x.

Each quadratic equation is in the form ax²+bx+c=0.This equation can be written in the form x²+b/a

x=−c/a using subtraction and division.

Therefore, it is possible to solve the equation by completing the square. Thus, every quadratic equation can be solved by completing the square.

Every quadratic equation can be solved by completing the square because each quadratic equation can be written in the form x²+bx=c using operations of addition, subtraction, multiplication, and/or division.

Page 155 Exercise 13 Answer

The given equation is 2x²−16=0.

The question requires to solve the given equation.

To solve the equation, isolate the square x² on one side of the equation using addition and division.

Then, use the definition of square root and the product property of square roots to simplify the expression.

Add 16 on both sides of 2x²−16=0, and then divide both sides by 2.

​2x²−16+16=0+16

2x²=16

2x²/2

=16/2

x²=8

​Use the definition of square root to solve the equation x²=8, use the product property of square roots, and simplify the expression.

​x²=8

x=±√8

x=±√4⋅2

x=±√4⋅√2

x=±2√2

Thus, the required solutions are x=±2√2.

The required solutions are x=±2√2.

Page 155 Exercise 14 Answer

The given equation is 2x²−6x−20=0.

The question requires to solve the given equation.

To solve the equation, first remove the common factor from the equation.

Then, write the pair of factors of the constant, and identify the pair of factors whose sum is equal to the coefficient of the middle term.

Then, factorise the left-hand side of the quadratic equation, solve it to obtain the values of x, and thus, obtain the required solutions.

Divide both sides of 2x²−6x−20=0 by 2.

​2x²−6x−20/2=0/2

x²−3x−10=0

Thus, the given quadratic equation becomes x²−3x−10=0.

The constant in the quadratic equation x²−3x−10=0 is −10.

Write the pair of factors of the constant −10.

1 and −10

2 and −5

5 and −2

10 and −1

It can be observed that the sum of the pair of factors 2 and −5 is equal to the coefficient of the middle term in the rewritten quadratic equation x²−3x−10=0, that is −3.

Therefore, the left-hand side of the rewritten quadratic equation can be factorised as (x+2)(x−5).

Thus, the equation becomes (x+2)(x−5)=0.

The equation (x+2)(x−5)=0 is true only when either x+2 is equal to 0, or x−5 is equal to 0.

When x+2 is equal to 0, the value of x is equal to −2.

When x−5 is equal to 0, the value of x is equal to 5.

Thus, the required solutions are x=−2 and x=5.

The required solutions are x=−2 and x=5.

Page 155 Exercise 15 Answer

The given equation is 2x²+2x−2=0.

The question requires to solve the given equation.

To solve the equation, rewrite the equation in the form x²+bx=c.

Then, identify the value of b, and calculate (b/2)². Then, add (b/2)² on both sides of the rewritten equation, factor the left-side expression using algebraic identities, and use the definition of square root.

Finally, solve the obtained equation and use the quotient property of square roots to determine the required solutions.

Add 2 on both sides of 2x²+2x−2=0, and then divide both sides by 2.

​2x²+2x−2+2=0+2

2x²+2x=2

2x²+2x/2=2/2

x²+x=1……………(1)

Comparing the equation (1) to the equation x²+bx=c, it can be determined that the value of b is 1.

Therefore, the value of (b/2)² is (1/2)²=1/4.

Add the value of (b/2)², that is 1/4, on both sides of equation (1).

Then, factorise the left-hand side expression using the algebraic identity a²+2ab+b²=(a+b)², and simplify the right-hand side expression.

​x²+x+1/4=1+1/4

(x+1/2)²=1+1/4

(x+1/2)²=5/4

Use the definition of square root to simplify (x+1/2)²=5/4, and then subtract 1/2 from both sides.

​(x+1/2)²=5/4

x+1/2=±√5/4

x=−1/2±√5/4

Apply the quotient property of square roots, and simplify the square root in the denominator.

​x=−1/2±√5/√4

x=−1/2±√5/2

x=−1±√5/2

Therefore, the required solutions are x=−1+√5/2 and x=−1−√5/2.

The required solutions are x=−1+√5/2 and x=−1−√5/2.

Page 155 Exercise 16 Answer

The given equation is x²+x=30.

The question requires to solve the given equation by factoring it.

To solve the equation, first rewrite the given equation in standard form using subtraction.

Then, write the pair of factors of the constant, and identify the pair of factors whose sum is equal to the coefficient of the middle term.

Then, factorise the left-hand side of the quadratic equation, solve it to obtain the values of x, and thus, obtain the required solutions.

Subtract 30 from both sides of x²+x=30.​

x²+x−30=30−30

x²+x−30=0

Thus, the given quadratic equation becomes x²+x−30=0.

The constant in the quadratic equation x²+x−30=0 is −30.

Write the pair of factors of the constant −30.

1 and −30

2 and −15

3 and −10

5 and −6

6 and −5

10 and −3

15 and −2

30 and −1

It can be observed that the sum of the pair of factors 6 and −5 is equal to the coefficient of the middle term in the rewritten quadratic equation x²+x−30=0, that is 1.

Therefore, the left-hand side of the rewritten quadratic equation can be factorised as (x+6)(x−5).

Thus, the equation becomes (x+6)(x−5)=0.

The equation (x+6)(x−5)=0 is true only when either x+6 is equal to 0, or x−5 is equal to 0.

When x+6 is equal to 0, the value of x is equal to −6.

When x−5 is equal to 0, the value of x is equal to 5.

Thus, the required solutions are x=−6 and x=5.

The required solutions are x=−6 and x=5.

Page 155 Exercise 17 Answer

The given equation is x²−5x=24.

The question requires to solve the given equation by factoring it.

To solve the equation, first rewrite the given equation in standard form using subtraction.

Then, write the pair of factors of the constant, and identify the pair of factors whose sum is equal to the coefficient of the middle term.

Then, factorise the left-hand side of the quadratic equation, solve it to obtain the values of x, and thus, obtain the required solutions.

Subtract 24 from both sides of x²−5x=24.

​x²−5x−24=24−24

x²−5x−24=0

Thus, the given quadratic equation becomes x²−5x−24=0

The constant in the quadratic equation x²−5x−24=0 is −24.

Write the pair of factors of the constant −24.

1 and −24

2 and −12

3 and −8

4 and −6

6 and −4

8 and −3

12 and −2

24 and −1

It can be observed that the sum of the pair of factors 3 and −8 is equal to the coefficient of the middle term in the rewritten quadratic equation x²−5x−24=0, that is −5.

Therefore, the left-hand side of the rewritten quadratic equation can be factorised as (x+3)(x−8).

Thus, the equation becomes (x+3)(x−8)=0.

The equation (x+3)(x−8)=0 is true only when either x+3 is equal to 0, or x−8 is equal to 0.

When x+3 is equal to 0, the value of x is equal to −3.

When x−8 is equal to 0, the value of x is equal to 8.

Thus, the required solutions are x=−3 and x=8.

The required solutions are x=−3 and x=8.

Page 155 Exercise 18 Answer

The given equation is −4x²+8=24.

The question requires to solve the given equation.To solve the equation, isolate the square x² on one side of the equation using subtraction and division.

Then, use the definition of square root, the product property of square roots, and the imaginary unit i to simplify and rewrite the expression.

Subtract 8 from both sides of −4x²+8=24, and then divide both sides by −4.

​−4x²+8−8=24−8

−4x²=16

−4x²−4=16−4

x²=−4

​Use the definition of square root to solve the equation x²=−4, use the product property of square roots, and rewrite the expression using the imaginary unit i.

​x²=−4

x=±√−4

x=±√(4)(−1)

x=±√4⋅√−1

x=±2i

Thus, the required solutions are x=±2i.

The required solutions are x=±2i.

Page 155 Exercise 19 Answer

The given equation is x²+30=24.

The question requires to solve the given equation.

To solve the equation, isolate the square x² on one side of the equation using subtraction.

Then, use the definition of square root, the product property of square roots, and the imaginary unit i

to simplify and rewrite the expression.

Subtract 30 from both sides of x²+30=24 and then use the definition of square root to solve the equation.

​x²+30−30=24−30

x²=−6

x=±√−6

Use the product property of square roots to rewrite the square root in x=±√−6, and rewrite the expression using the imaginary unit i.

​x=±√−6

x=±√(−1)(6)

x=±√−1⋅√6

x=±i√6

Thus, the required solutions are x=±i√6.

The required solutions are x=±i√6.

Page 155 Exercise 20 Answer

The given equation is x²+4x+3=0.

The question requires to solve the given equation.

To solve the equation, write the pair of factors of the constant, and identify the pair of factors whose sum is equal to the coefficient of the middle term.

Then, factorise the left-hand side of the given quadratic equation, solve it to obtain the values of x, and thus, obtain the required solutions.

The constant in the given quadratic equation is 3.

Write the pair of factors of the constant 3.

1 and 3

−1 and −3

It can be observed that the sum of the pair of factors 1 and 3 is equal to the coefficient of the middle term in the given quadratic equation, that is 4.

Therefore, the left-hand side of the given quadratic equation can be factorised as (x+1)(x+3).

Thus, the equation becomes (x+1)(x+3)=0.

The equation (x+1)(x+3)=0 is true only when either x+1 is equal to 0, or x+3 is equal to 0.

When x+1 is equal to 0, the value of x is equal to −1.

When x+3 is equal to 0, the value of x is equal to −3.

Thus, the required solutions are x=−3 and x=−1.

The required solutions are x=−3 and x=−1.

Page 155 Exercise 21 Answer

The given equation is 7m⋅5m=875.Assume that there is a rectangular field which has an area equal to 875 square feet, and the ratio of its length and width is 7:5.

The length and width of the field can be written as 7m and 5m respectively.

Therefore, the area 875 square feet of the rectangular field would be the product of 7m and 5m, that is 7m⋅5m=875.

Thus, the given equation can be used to model the situation where the area of a rectangular field is 875 square feet, and the ratio of its length and width is 7:5.

The equation 7m⋅5m=875 can be used to model the situation where the ratio of the length and width of a rectangular field is 7:5, and its area is 875 square feet.

Page 156 Exercise 22 Answer

The given quadratic function is f(x)=ax²+bx+c. Assume that a>0.

When the value of x moves away from the x-coordinate of the vertex of the graph of the function f(x), the value of f(x) increases.

This means that the graph of the function f(x)=ax²+bx+c forms a parabola opening upward when the leading coefficient a>0, and forms a U-shaped graph.

Assume that a<0. When the value of x

moves away from the x-coordinate of the vertex of the graph of the function f(x), the value of f(x) decreases.

This means that the graph of the function f(x)=ax²+bx+c forms a parabola opening downward when the leading coefficient a<0, and forms an inverted U-shaped graph.

Assume that a=0. When a is equal to 0, the given function becomes f(x)=bx+c.

This means that when a=0, the function is linear and not quadratic.

Therefore, the graph of the function is a straight line when a=0.

The graph of the quadratic function opens upward when a>0 because if the leading coefficient a is positive, then the value of f(x) increases as the value of x moves away from the x-coordinate of the vertex of f(x), and forms a U-shaped graph.

The graph of the quadratic function opens downward when a<0 because if the leading coefficient a is negative, then the value of f(x) decreases as the value of x moves away from the x

-coordinate of the vertex of f(x), and forms an inverted U-shaped graph.

When a=0, the given function is a linear function, and not a quadratic function. The graph of the function would be a straight line.

Algebra 2 Volume 1 1st Edition Module 3 Quadratic Equations

Page 127 Problem 1 Answer

A complex number is a number that is of the form a+bi, where a and b are real numbers, and i is the imaginary unit equal to √−1.

Consider the two arbitrary complex numbers a+bi and c+di.The sum of the two complex numbers can be written as (a+bi)+(c+di)

. In the sum, the terms a and c, and the terms bi and di are like terms.By grouping the like terms, the sum can be written as (a+bi)+(c+di)=(a+c)+(bi+di).

The right-hand side of the sum can be simplified by combining the like terms in the parentheses to find the simplified sum of the two complex numbers.

The difference of the two complex numbers can be written as (a+bi)−(c+di). In the difference, the terms a and c, and the terms bi and di are like terms.

By grouping the like terms, the difference can be written as (a+bi)−(c+di)=(a−c)+(bi−di).

The right-hand side of the difference can be simplified by combining the like terms in the parentheses to find the simplified difference of the two complex numbers.

Thus, the sum and difference of two complex numbers can be determined by grouping the like terms, that is the real parts and the imaginary parts, and then combining the like terms.

When a+bi and c+di are multiplied using the distributive property of multiplication, the product is (a+bi)(c+di)=ac+adi+bci+bdi².Using the value of i² and simplifying the expression, the resulting expression is (a+bi)(c+di)=ac+(ad+bc)i−bd.

The right-hand side of the equation can be grouped as (ac−bd)+(ad+bc)i.

Thus, the product of two complex numbers can be determined by using the distributive property of multiplication, combining the like terms, and then using the value of i² to simplify the product.

A number is called a complex number if it can be written in the form a+bi, where a and b are real numbers, and i is the imaginary unit equal to √−1.

Two complex numbers can be added or subtracted by writing the expression for the sum or difference, grouping the like terms, that is the real parts and the imaginary parts, and then combining the like terms.

Two complex numbers can be multiplied by using the distributive property of multiplication, combining the like terms, and then using the value of i² to simplify the product.

Page 127 Problem 2 Answer

The given binomials are 3+4x and 2−x.

The question requires to add the binomials 3+4x and 2−x.

To add the binomials, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Write the sum of 3+4x and 2−x, group the like terms in parentheses, and combine the like terms in the parentheses.​

(3+4x)+(2−x)=(3+2)+(4x+(−x))

=(3+2)+(4x−x)

=5+3x

Thus, the required sum is 5+3x.

The required sum is 5+3x.

Page 127 Problem 3 Answer

The given binomials are 3+4x and 2−x.

The question requires to subtract 2−x from 3+4x.

To subtract the binomials, rewrite the difference as a sum, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Write the difference of 2−x from 3+4x, and rewrite the subtraction as addition.

Then, group the like terms in parentheses, and combine the like terms in the parentheses.

(3+4x)−(2−x)=(3+4x)+(−2+x)

=(3+(−2))+(4x+x)

=(3−2)+(4x+x)

=1+5x

Thus, the required difference is 1+5x.

The required difference is 1+5x.

Page 127 Problem 4 Answer

The given binomials are 3+4x and 2−x.

The question requires to multiply the binomials 3+4x and 2−x.

To multiply the binomials, use the FOIL method, and then combine the like terms.

Write the product of 3+4x and 2−x, use the FOIL method, and combine the like terms.​

(3+4x)(2−x)=6+(−3x)+8x+(−4x²)

=6−3x+8x−4x²

=6+5x−4x²

Thus, the required product is 6+5x−4x².

The required product is 6+5x−4x².

Page 127 Problem 5 Answer

The given equation is (3+4x)+(2−x)=5+3x.

The question requires to write the equation obtained if x is equal to the imaginary unit i.

To obtain the required equation, replace the variable x by the imaginary unit i on both sides of the given equation.

Substitute i for x on both sides of (3+4x)+(2−x)=5+3x.

(3+4i)+(2−i)=5+3i

Thus, the required equation is (3+4i)+(2−i)=5+3i.

The required equation is (3+4i)+(2−i)=5+3i.

Page 127 Problem 6 Answer

The given equation is (3+4x)(2−x)=6+5x−4x².

The question requires to write the equation obtained if x is equal to the imaginary unit i.

Also, it is required to explain how the right side of the obtained equation can be simplified further.

To obtain the required equation, replace the variable x by the imaginary unit i on both sides of the given equation.

Then, use the value of i² to simplify the right side of the equation further.

Substitute i for x on both sides of (3+4x)(2−x)=6+5x−4x².

(3+4i)(2−i)=6+5i−4i²

Thus, the required equation is (3+4i)(2−i)=6+5i−4i².

The right side of the equation contains the square i², which is equal to −1.

Substitute −1 for i² on the right side of (3+4i)(2−i)=6+5i−4i², and simplify the equation.​

(3+4i)(2−i)=6+5i−4(−1)

(3+4i)(2−i)=6+5i+4

(3+4i)(2−i)=10+5i

Thus, the right side of the equation can be simplified as 10+5i.

The required equation is (3+4i)(2−i)=6+5i−4i².

The right side of the obtained equation is 6+5i−4i², which contains the square i².

The right side can be further simplified by substituting −1 for i², and then combining the like terms to simplify the right side as 10+5i.

Page 128 Problem 7 Answer

The given number is −7i, which can be written as the complex number 0−7i.

Therefore, the real part of −7i is 0 and the imaginary part of −7i is −7.

Since the given number can be written as the complex number 0−7i having the real part 0, it belongs to the set of imaginary numbers and the set of complex numbers.

The real part and imaginary part of −7i are 0 and −7 respectively.

The given number −7i belongs to the set of imaginary numbers and the set of complex numbers.

Page 129 Problem 8 Answer

The given number is −1+i, which can be written as the complex number −1+1⋅i.

Therefore, the real part of −1+i is −1 and the imaginary part of −1+i is 1.

Since the given number can be written as the complex number −1+1⋅i having nonzero real and imaginary parts, it belongs to only the set of complex numbers.

The real part and imaginary part of −1+i are −1 and 1 respectively.

The given number −1+i belongs to only the set of complex numbers.

Page 129 Problem 9 Answer

The given sum is (a+bi)+(a−bi). The numbers a and b are real.

The question requires to determine and explain whether the sum (a+bi)+(a−bi) is a real number or an imaginary number.

To answer the question, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Then, observe the real and imaginary parts of the sum, and determine whether the number is real or imaginary.

Group the real and imaginary parts in the sum (a+bi)+(a−bi), and then combine the like terms.​

(a+bi)+(a−bi)=(a+a)+(bi−bi)

=2a+0

=2a

Thus, the required sum is 2a.

The sum 2a can be written as the complex number 2a+0i.

Therefore, the real and imaginary part of 2a are 2a and 0 respectively.

Since the imaginary part of the sum 2a is 0, the sum 2a is a real number.

The sum (a+bi)+(a−bi) is a real number because the sum is equal to 2a, and the imaginary part of 2a is equal to 0.

Page 129 Problem 10 Answer

The given difference is (17−6i)−(9+10i).

The question requires to subtract the complex numbers in the given difference.

To subtract the complex numbers, rewrite the difference as a sum, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Rewrite the subtraction of the complex numbers in (17−6i)−(9+10i) as addition.

Then, group the real and imaginary parts, and combine the like terms.​

(17−6i)−(9+10i)=(17−6i)+(−9−10i)

=(17−9)+(−6i−10i)

=8−16i

Thus, the required difference is 8−16i.

The required difference is 8−16i.

Page 129 Problem 11 Answer

The given difference is (18+27i)−(2+3i).

The question requires to subtract the complex numbers in the given difference.

To subtract the complex numbers, rewrite the difference as a sum, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Rewrite the subtraction of the complex numbers in (18+27i)−(2+3i) as addition. Then, group the real and imaginary parts, and combine the like terms.

​(18+27i)−(2+3i)=(18+27i)+(−2−3i)

=(18−2)+(27i−3i)

=16+24i

Thus, the required difference is 16+24i.

The required difference is 16+24i.

Page 130 Problem 12 Answer

The given product is (6−5i)(3−10i).

The question requires to multiply the complex numbers in the given product.

To multiply the complex numbers, use the distributive property of multiplication and the value of i², and simplify the expression.

Apply the distributive property of multiplication in the given product (6−5i)(3−10i), and then combine the like terms.

​(6−5i)(3−10i)=18−60i−15i+50i²

=18−75i+50i²

Substitute −1 for i², and then simplify the expression.

(6−5i)(3−10i)=18−75i+50(−1)

=18−75i−50

=−32−75i

Thus, the required product is −32−75i.

The required product is −32−75i.

Page 130 Problem 13 Answer

The given product is (8+15i)(11+i).

The question requires to multiply the complex numbers in the given product.

To multiply the complex numbers, use the distributive property of multiplication and the value of i², and simplify the expression.

Apply the distributive property of multiplication in the given product (8+15i)(11+i), and then combine the like terms.

​(8+15i)(11+i)=88+8i+165i+15i²

=88+173i+15i²

Substitute −1 for i², and then simplify the expression.

(8+15i)(11+i)=88+173i+15(−1)

=88+173i−15

=73+173i

Thus, the required product is 73+173i.

The required product is 73+173i.

Page 130 Problem 14 Answer

The given product is (−3+12i)(7+4i).

The question requires to multiply the complex numbers in the given product.

To multiply the complex numbers, use the distributive property of multiplication and the value of i², and simplify the expression.

Apply the distributive property of multiplication in the given product (−3+12i)(7+4i), and then combine the like terms.​

(−3+12i)(7+4i)=−21−12i+84i+48i²

=−21+72i+48i²

Substitute −1 for i², and then simplify the expression.​

(−3+12i)(7+4i)=−21+72i+48(−1)

=−21+72i−48

=−69+72i​

Thus, the required product is −69+72i.he required product is −69+72i.

Page 131 Problem 15 Answer

The given diagram shows the components in a circuit. The given current I is 24+12i amps.

The question requires to determine the voltage of each component of the circuit.

To determine the voltage of each component, use the diagram to write the impedance of each component.

Then, use the impedance and current in the Ohm’s law to determine the required voltage of each component.

From the circuit diagram, it can be observed that the resistor is labelled 4 ohms.

Therefore, the impedance of the resistor can be represented by the complex number 4.

The inductor is labelled 3 ohms. Therefore, the impedance of the inductor can be represented by the complex number 3i.

The capacitor is labelled 5 ohms. Therefore, the impedance of the capacitor can be represented by the complex number −5i.

Substitute 24+12i for I and 4 for Z in the Ohm’s law V=I⋅Z, and multiply the terms.​

V=(24+12i)(4)

=96+48i

Thus, the voltage of the resistor is 96+48i volts.

Substitute 24+12i for I and 3i for Z in the Ohm’s law V=I⋅Z, and multiply the terms, and substitute −1 for i².

V=(24+12i)(3i)

=72i+36i²

=72i+36(−1)

=−36+72i

Thus, the voltage of the inductor is −36+72i volts.

Substitute 24+12i for I and −5i for Z in the Ohm’s law V=I⋅Z, and multiply the terms, and substitute 1 for i².​

V=(24+12i)(−5i)

=−120i−60i²

=−120i−60(−1)

=60−120i

Thus, the voltage of the capacitor is 60−120i volts.

The voltage of the resistor, inductor, and capacitor are 96+48i volts, −36+72i volts, and 60−120i volts respectively.

Page 132 Problem 16 Answer

The given diagram shows the components in a circuit. The given current I is 24+12i amps.

The question requires to determine the sum of the voltages for the three components, and write what is noticed by the result.

Given that the voltage of the resistor, inductor, and capacitor are 96+48i volts, −36+72i volts, and 60−120i volts respectively, as determined in part B of example 4.

To determine the sum of voltages, add the voltages by grouping the like terms and combining them.

Then, compare the sum to the voltage across the power source and write what is noticed.

Write the sum of the voltages of the three components, group the real and imaginary parts in the sum, and then combine the like terms.

(96+48i)+(−36+72i)+(60−120i)=(96−36+60)+(48i+72i−120i)

=120+0i

=120​

Thus, the sum of the voltages for the three components is 120 volts.

From the diagram, it can be observed that the power source is labelled 120 V.

This means that the voltage across the power source is 120 volts.

Thus, the sum of the voltages for the three components is equal to the voltage across the power source, that is 120 volts.

The sum of the voltages for the three components is 120 volts.

The sum of the voltages for the three components is equal to the voltage across the power source, that is 120 volts.

Page 132 Problem 17 Answer

It is given that a second resistor with impedance equal to 2 ohms is added to the circuit diagram given in example 4. The given current I is 18+6i amps.

The question requires to determine the total impedance, and then determine the voltage of each component of the circuit.

To determine the total impedance and voltage of each component, use the diagram to write the impedance of each component as a complex number.

Then, add the complex numbers to determine the total impedance.

Next, use the impedance and current in the Ohm’s law to determine the required voltage of each component.

From the circuit diagram, it can be observed that the first resistor has the impedance 4 ohms.

Therefore, the impedance of the first resistor can be represented by the complex number 4.

The inductor has the impedance 3 ohms. Therefore, the impedance of the inductor can be represented by the complex number 3i.

The capacitor has the impedance 5 ohms. Therefore, the impedance of the capacitor can be represented by the complex number −5i.

The second resistor has the impedance 2 ohms. Therefore, the impedance of the second resistor can be represented by the complex number 2.

The total impedance of the circuit is equal to the sum of the impedances of the four components.

Write the sum of the impedances of the four components, group the real and imaginary parts in the sum, and then combine the like terms.​

4+3i+(−5i)+2=(4+2)+(3i−5i)

=6−2i

Thus, the total impedance is 6−2i ohms.

Substitute 18+6i for I and 4 for Z in the Ohm’s law V=I⋅Z, and multiply the terms.​

V=(18+6i)(4)

=72+24i

Thus, the voltage of the first resistor is 72+24i volts.

Substitute 18+6i for I and 2 for Z in the Ohm’s law V=I⋅Z, and multiply the terms.​

V=(18+6i)(2)

=36+12i

Thus, the voltage of the second resistor is 36+12i volts.

Substitute 18+6i for I and 3i for Z in the Ohm’s law V=I⋅Z, and multiply the terms, and substitute −1 for i².

V=(18+6i)(3i)

=54i+18i²

=54i+18(−1)

=−18+54i

Thus, the voltage of the inductor is −18+54i volts.

Substitute 18+6i for I and −5i for Z in the Ohm’s law V=I⋅Z, and multiply the terms, and substitute −1 for i².

V=(18+6i)(−5i)

=−90i−30i²

=−90i−30(−1)

=30−90i​

Thus, the voltage of the capacitor is 30−90i volts.

The required total impedance is 6−2i ohms.

The voltages of the first resistor, the second resistor, the inductor, and the capacitor are 72+24i volts, 36+12i volts, −18+54i volts, and 30−90i volts respectively.

Page 132 Problem 18 Answer

Take the two arbitrary complex numbers a+bi and c+di.When a+bi and c+di are added, the sum is (a+bi)+(c+di)=(a+c)+(b+d)i, which is a complex number.

The sum (a+c)+(b+d)i is a real number only if the imaginary part (b+d) is equal to 0.

The imaginary part (b+d) is the sum of the imaginary parts of the complex numbers a+bi and c+di.

Therefore, the sum of two complex numbers is a real number only when the sum of their imaginary parts is equal to 0.

The sum (a+c)+(b+d)i is an imaginary number only if the real part (a+c) is equal to 0.

The real part (a+c) is the sum of the real parts of the complex numbers a+bi and c+di.

Therefore, the sum of two complex numbers is an imaginary number only when the sum of their real parts is equal to 0.

The sum of two complex numbers is a real number only when the sum of the imaginary parts of the two complex numbers is equal to 0.

The sum of two complex numbers is an imaginary number only when the sum of the real parts of the two complex numbers is equal to 0.

Page 133 Problem 19 Answer

Consider the two arbitrary binomial linear expressions a+bx and c+dx.

When a+bx and c+dx are multiplied using the distributive property of multiplication, the product is(a+bx)(c+dx)=ac+adx+bcx+bdx².

Grouping the like terms, the resulting product is (a+bx)(c+dx)=ac+(ad+bc)x+bdx².

Consider the two arbitrary complex numbers a+bi and c+di.When a+bi and c+di are multiplied using the distributive property of multiplication, the product is (a+bi)(c+di)=ac+adi+bci+bdi².

Using the value of i² and simplifying the expression, the resulting expression is (a+bi)(c+di)=ac+(ad+bc)i−bd.

The right-hand side of the equation can be grouped as (ac−bd)+(ad+bc)i, which is a complex number.

It can be observed that in the multiplication of two binomial linear expressions in the same variable, and in the multiplication of two complex numbers, first the distributive property of multiplication is used, and then the like terms are grouped and simplified.

Also, it can be observed that in the multiplication of two complex numbers, the result obtained after using the distributive property of multiplication and grouping the terms can be simplified further using the value i²=−1.

The product of two binomial linear expressions lacks the square of the imaginary unit, that is i², and thus, it cannot be simplified further.

The similarities between multiplying two complex numbers and multiplying two binomial linear expressions in the same variable are:

Both involve the application of the distributive property of multiplication (a+b)(c+d)=ac+ad+bc+bd

.Both involve grouping and combining the like terms after using the distributive property of multiplication.

The difference between multiplying two complex numbers and multiplying two binomial linear expressions in the same variable is that the result obtained after using the distributive property of multiplication and grouping the terms in the multiplication of two complex numbers can be simplified further using the value i²=−1.

This is not possible when multiplying two binomial linear expressions because their product does not contain the square i².

Page 133 Exercise 1  Answer

The given binomials are 3+2x and 4−5x.

The question requires to determine the sum of the given binomials, and then explain how the sum can be used to find the sum of the complex numbers 3+2i and 4−5i.

To add the binomials, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Then, replace the variable x by the imaginary unit i on both sides of the given equation to find the sum of the complex numbers 3+2i and 4−5i.

Write the sum of 3+2x and 4−5x, group the like terms in parentheses, and combine the like terms in the parentheses.

(3+2x)+(4−5x)=(3+4)+(2x+(−5x))

(3+2x)+(4−5x)=(3+4)+(2x−5x)

(3+2x)+(4−5x)=7−3x

Thus, the required sum of the given binomials is (3+2x)+(4−5x)=7−3x.

The sum of the complex numbers 3+2i and 4−5i can be obtained by substituting the variable x by the imaginary unit i in the equation (3+2x)+(4−5x)=7−3x.

Substitute i for x on both sides of (3+2x)+(4−5x)=7−3x.

(3+2i)+(4−5i)=7−3i

Thus, the sum of the complex numbers is (3+2i)+(4−5i)=7−3i.

The required sum of the given binomials is (3+2x)+(4−5x)=7−3x.

The sum of the complex numbers 3+2i and 4−5i can be determined by substituting i

for x on both sides of the equation for the sum of the given binomials, that is in the equation (3+2x)+(4−5x)=7−3x.

The resulting sum of the complex numbers 3+2i and 4−5i is (3+2i)+(4−5i)=7−3i.

Page 133 Exercise 2  Answer

The given number is 5+i, which can be written as the complex number 5+1⋅i.Therefore, the real part of 5+i is 5 and the imaginary part of 5+i is 1.

Since the given number can be written as the complex number 5+1⋅i having nonzero real and imaginary parts, it belongs only to the set of complex numbers.

The real part and imaginary part of 5+i are 5 and 1 respectively.

The given number 5+i belongs to only the set of complex numbers.

Page 133 Exercise 3  Answer

The given number is 7−6i, which can be written as the complex number 7+(−6)i.

Therefore, the real part of 7−6i is 7 and the imaginary part of 7−6i is −6.

Since the given number can be written as the complex number 7+(−6)i having nonzero real and imaginary parts, it belongs only to the set of complex numbers.

The real part and imaginary part of 7−6i are 7 and −6 respectively.

The given number 7−6i belongs to only the set of complex numbers.

Page 134 Exercise 4  Answer

The given number is 25, which can be written as the complex number 25+0i.Therefore, the real part of 25 is 25 and the imaginary part of 25 is 0.

Since the given number can be written as the complex number 25+0i having the imaginary part 0, it belongs to the set of real numbers and the set of complex numbers.

The real part and imaginary part of 25 are 25 and 0 respectively.

The given number 25 belongs to the set of real numbers and the set of complex numbers.

Page 134 Exercise 5  Answer

The given number is i√21, which can be written as the complex number 0+(√21)i

.Therefore, the real part of i√21 is 0 and the imaginary part of i√21 is √21

.Since the given number can be written as the complex number 0+(√21)i

having the real part 0, it belongs to the set of imaginary numbers and the set of complex numbers.

The real part and imaginary part of i√21 are 0 and √21 respectively.

The given number i√21 belongs to the set of imaginary numbers and the set of complex numbers.

Page 134 Exercise 6 Answer

The given sum is (3+4i)+(7+11i).

The question requires to add the complex numbers in the given sum.

To add the complex numbers, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Group the real and imaginary parts in the sum (3+4i)+(7+11i), and then combine the like terms.​

(3+4i)+(7+11i)=(3+7)+(4i+11i)

=10+15i

Thus, the required sum is 10+15i.

The required sum is 10+15i.

Page 134 Exercise 7 Answer

The given sum is (−1−i)+(−10+3i).

The question requires to add the complex numbers in the given sum.

To add the complex numbers, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Group the real and imaginary parts in the sum (−1−i)+(−10+3i), and then combine the like terms.​

(−1−i)+(−10+3i)=(−1+(−10))+(−i+3i)

=(−1−10)+(−i+3i)

=−11+2i

Thus, the required sum is −11+2i.

The required sum is −11+2i.

Page 133 Exercise 8 Answer

The given sum is (−9−7i)+(6+5i).

The question requires to add the complex numbers in the given sum.

To add the complex numbers, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Group the real and imaginary parts in the sum (−9−7i)+(6+5i), and then combine the like terms.​

(−9−7i)+(6+5i)=(−9+6)+(−7i+5i)

=−3−2i

Thus, the required sum is −3−2i.

The required sum is −3−2i.

Page 134 Exercise 9 Answer

The given difference is (2+3i)−(7+6i).

The question requires to subtract the complex numbers in the given difference.

To subtract the complex numbers, rewrite the difference as a sum, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Rewrite the subtraction of the complex numbers in (2+3i)−(7+6i) as addition.

Then, group the real and imaginary parts, and combine the like terms.​

(2+3i)−(7+6i)=(2+3i)+(−7−6i)

=(2−7)+(3i−6i)

=−5−3i

Thus, the required difference is −5−3i.

The required difference is −5−3i.

Page 134 Exercise 10 Answer

The given difference is (4+5i)−(14−i).

The question requires to subtract the complex numbers in the given difference.

To subtract the complex numbers, rewrite the difference as a sum, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Rewrite the subtraction of the complex numbers in (4+5i)−(14−i) as addition.

Then, group the real and imaginary parts, and combine the like terms.​

(4+5i)−(14−i)=(4+5i)+(−14+i)

=(4−14)+(5i+i)

=−10+6i

Thus, the required difference is −10+6i.

The required difference is −10+6i.

Page 134 Exercise 11 Answer

The given difference is (−8−3i)−(−9−5i).The question requires to subtract the complex numbers in the given difference.

To subtract the complex numbers, rewrite the difference as a sum, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Rewrite the subtraction of the complex numbers in (−8−3i)−(−9−5i)

as addition. Then, group the real and imaginary parts, and combine the like terms.​

(−8−3i)−(−9−5i)=(−8−3i)+(9+5i)

=(−8+9)+(−3i+5i)

=1+2i

Thus, the required difference is 1+2i.

The required difference is 1+2i.

Page 134 Exercise 12 Answer

The given difference is (5+2i)−(5−2i).

The question requires to subtract the complex numbers in the given difference.

To subtract the complex numbers, rewrite the difference as a sum, group the like terms in the sum using parentheses, and then combine the like terms in the expression.

Rewrite the subtraction of the complex numbers in (5+2i)−(5−2i) as addition.

Then, group the real and imaginary parts, and combine the like terms.

(5+2i)−(5−2i)=(5+2i)+(−5+2i)

=(5−5)+(2i+2i)

=0+4i

=4i

Thus, the required difference is 4i.

The required difference is 4i.

Page 134 Exercise 13 Answer

The given product is (2+3i)(3+5i).

The question requires to multiply the complex numbers in the given product.

To multiply the complex numbers, use the distributive property of multiplication and the value of i², and simplify the expression.

Apply the distributive property of multiplication in the given product (2+3i)(3+5i), and then combine the like terms.

​(2+3i)(3+5i)=6+10i+9i+15i²

=6+19i+15i²

Substitute −1 for i², and then simplify the expression.

(2+3i)(3+5i)=6+19i+15(−1)

=6+19i−15

=−9+19i

Thus, the required product is −9+19i.

The required product is −9+19i.

Page 134 Exercise 14 Answer

The given product is (7+i)(6−9i).

The question requires to multiply the complex numbers in the given product.

To multiply the complex numbers, use the distributive property of multiplication and the value of i², and simplify the expression.

Apply the distributive property of multiplication in the given product (7+i)(6−9i), and then combine the like terms.

​(7+i)(6−9i)=42−63i+6i−9i²

=42−57i−9i²

Substitute −1 for i², and then simplify the expression.

​(7+i)(6−9i)=42−57i−9(−1)

=42−57i+9

=51−57i

Thus, the required product is 51−57i.

The required product is 51−57i.

Page 134 Exercise 15 Answer

The given product is (4−i)(4+i).

The question requires to multiply the complex numbers in the given product.

To multiply the complex numbers, use the distributive property of multiplication and the value of i², and simplify the expression.

Apply the distributive property of multiplication in the given product (4−i)(4+i), and then combine the like terms.

​(4−i)(4+i)=16+4i−4i−i²=16−i2

Substitute −1 for i², and then simplify the expression.

​(4−i)(4+i)=16−(−1)

=16+1

=17

Thus, the required product is 17.

The required product is 17.

Page 135 Exercise 16 Answer

The given diagram shows the components in a circuit. The given current is 12+36i amps.

The question requires to determine the total impedance, and then determine the voltage for each component of the circuit.

To determine the total impedance and voltage of each component, use the diagram to write the impedance for each component as a complex number.

Then, add the complex numbers to determine the total impedance.

Next, use the impedance and current in the Ohm’s law to determine the required voltage for each component.

From the circuit diagram, it can be observed that there is one resistor and one capacitor.

The resistor has the impedance 1 ohm. Therefore, the impedance for the resistor can be represented by the complex number 1.

The capacitor has the impedance 3 ohms. Therefore, the impedance for the capacitor can be represented by the complex number −3i.

The total impedance for the circuit is equal to the sum of the impedances for the two components.

Calculate the sum of the impedances for the two components.

1+(−3i)=1−3i

Thus, the total impedance is 1−3i ohms.

Substitute 12+36i for I and 1 for Z in the Ohm’s law V=I⋅Z, and multiply the terms.

​V=(12+36i)(1)

=12+36i

Thus, the voltage for the resistor is 12+36i volts.

Substitute 12+36i for I and −3i for Z in the Ohm’s law V=I⋅Z, and multiply the terms, and substitute −1 for i².

​V=(12+36i)(−3i)

=−36i−108i²

=−36i−108(−1)

=108−36i

​Thus, the voltage for the capacitor is 108−36i volts.

The required total impedance is 1−3i ohms.

The voltages for the resistor and the capacitor are 12+36i volts and 108−36i volts respectively.

Page 135 Exercise 17 Answer

The given diagram shows the components in a circuit. The given current is 19.2−14.4i amps.

The question requires to determine the total impedance, and then determine the voltage for each component of the circuit.

To determine the total impedance and voltage of each component, use the diagram to write the impedance for each component as a complex number.

Then, add the complex numbers to determine the total impedance.

Next, use the impedance and current in the Ohm’s law to determine the required voltage for each component.

From the circuit diagram, it can be observed that there is one resistor and one inductor.

The resistor has the impedance 4 ohms. Therefore, the impedance for the resistor can be represented by the complex number 4.

The inductor has the impedance 3 ohms. Therefore, the impedance for the inductor can be represented by the complex number 3i.

The total impedance for the circuit is equal to the sum of the impedances for the two components.

Write the sum of the impedances for the two components.

(4)+(3i)=4+3i

Thus, the total impedance is 4+3i ohms.

Substitute 19.2−14.4i for I and 4 for Z in the Ohm’s law V=I⋅Z, and multiply the terms.

​V=(19.2−14.4i)(4)

=76.8−57.6i

​Thus, the voltage for the resistor is 76.8−57.6i volts.

Substitute 19.2−14.4i for I and 3i for Z in the Ohm’s law V=I⋅Z, and multiply the terms, and substitute −1 for i².

V=(19.2−14.4i)(3i)

=57.6i−43.2i²

=57.6i−43.2(−1)

=43.2+57.6i

​Thus, the voltage for the inductor is 43.2+57.6i volts.

The required total impedance is 4+3i ohms.

The voltages for the resistor and the inductor are 76.8−57.6i volts and 43.2+57.6i

volts respectively.

Page 135 Exercise 18 Answer

The given diagram shows the components in a circuit. The given current is 7.2+9.6i amps.

The question requires to determine the total impedance, and then determine the voltage for each component of the circuit.

To determine the total impedance and voltage of each component, use the diagram to write the impedance for each component as a complex number.

Then, add the complex numbers to determine the total impedance.

Next, use the impedance and current in the Ohm’s law to determine the required voltage for each component.

From the circuit diagram, it can be observed that there is one resistor, one inductor, and one capacitor.

The resistor has the impedance 6 ohms. Therefore, the impedance for the resistor can be represented by the complex number 6.

The inductor has the impedance 2 ohms. Therefore, the impedance for the inductor can be represented by the complex number 2i.

The capacitor has the impedance 10 ohms. Therefore, the impedance for the capacitor can be represented by the complex number −10i.

The total impedance for the circuit is equal to the sum of the impedances for the three components.

Calculate the sum of the impedances for the three components.

​6+2i+(−10i)=6+(2i−10i)

=6−8i

Thus, the total impedance is 6−8i ohms.

Substitute 7.2+9.6i for I and 6 for Z in the Ohm’s law V=I⋅Z, and multiply the terms.

V=(7.2+9.6i)(6)

=43.2+57.6i

​Thus, the voltage for the resistor is 43.2+57.6i volts.

Substitute 7.2+9.6i for I and 2i for Z in the Ohm’s law V=I⋅Z, and multiply the terms, and substitute −1 for i².

​V=(7.2+9.6i)(2i)

=14.4i+19.2i²

=14.4i+19.2(−1)

=−19.2+14.4i

Thus, the voltage for the inductor is −19.2+14.4i volts.

Substitute 7.2+9.6i for I and −10i for Z in the Ohm’s law V=I⋅Z, and multiply the terms, and substitute −1 for i².

V=(7.2+9.6i)(−10i)

=−72i−96i²

=−72i−96(−1)

=96−72i

​Thus, the voltage for the capacitor is 96−72i volts.

The required total impedance is 6−8i ohms.

The voltages for the resistor, the inductor, and the capacitor are 43.2+57.6i volts, −19.2+14.4i

volts, and 96−72i volts respectively.

Page 135 Exercise 19 Answer

The given diagram shows the components in a circuit. The given current is 16.8+2.4i amps.

The question requires to determine the total impedance, and then determine the voltage for each component of the circuit.

To determine the total impedance and voltage of each component, use the diagram to write the impedance for each component as a complex number.

Then, add the complex numbers to determine the total impedance.

Next, use the impedance and current in the Ohm’s law to determine the required voltage for each component.

From the circuit diagram, it can be observed that there is one resistor, one inductor, and one capacitor.

The resistor has the impedance 7 ohms. Therefore, the impedance for the resistor can be represented by the complex number 7.

The inductor has the impedance 3 ohms. Therefore, the impedance for the inductor can be represented by the complex number 3i.

The capacitor has the impedance 4 ohms. Therefore, the impedance for the capacitor can be represented by the complex number −4i.

The total impedance for the circuit is equal to the sum of the impedances for the three components.

Calculate the sum of the impedances for the three components.

​7+3i+(−4i)=7+(3i−4i)

=7−i

Thus, the total impedance is 7−i ohms.

Substitute 16.8+2.4i for I and 7 for Z in the Ohm’s law V=I⋅Z, and multiply the terms.​

V=(16.8+2.4i)(7)

=117.6+16.8i

Thus, the voltage for the resistor is 117.6+16.8i volts.

Substitute 16.8+2.4i for I and 3i for Z in the Ohm’s law V=I⋅Z, and multiply the terms, and substitute −1 for i².

​V=(16.8+2.4i)(3i)

=50.4i+7.2i²

=50.4i+7.2(−1)

=−7.2+50.4i

​Thus, the voltage for the inductor is −7.2+50.4i volts.

Substitute 16.8+2.4i for I and −4i for Z in the Ohm’s law V=I⋅Z, and multiply the terms, and substitute −1 for i².

​V=(16.8+2.4i)(−4i)

=−67.2i−9.6i2

=−67.2i−9.6(−1)

=9.6−67.2i

Thus, the voltage for the capacitor is 9.6−67.2i volts.

The required total impedance is 7−i ohms.

The voltages for the resistor, the inductor, and the capacitor are 117.6+16.8i volts, −7.2+50.4i volts, and 9.6−67.2i volts respectively.

Page 136 Exercise 20 Answer

The given expressions are (A) (3−5i)(3+5i), (B) (3+5i)(3+5i), (C) (−3−5i)(3+5i), and (D) (3−5i)(−3−5i).

The given products are −16+30i, −34, 34, and 16−30i.

The question requires to match the given products on the right with their corresponding expressions on the left.

To match the products with the expressions, simplify the expressions using the distributive property of multiplication and the value of i².

Then, match the obtained products with the expressions.

Apply the distributive property of multiplication in the product (3−5i)(3+5i), combine the like terms, substitute −1 for i², and then simplify the expression.

(3−5i)(3+5i)=9+15i−15i−25i²

=9−25i²

=9−25(−1)

=9+25

=34

Thus, the correct match for the product 34 is the expression (A) (3−5i)(3+5i).

Apply the distributive property of multiplication in the product (3+5i)(3+5i), combine the like terms, substitute −1 for i², and then simplify the expression.​

(3+5i)(3+5i)=9+15i+15i+25i²

=9+30i+25i²

=9+30i+25(−1)

=9+30i−25

=−16+30i

Thus, the correct match for the product −16+30i is the expression (B) (3+5i)(3+5i).

Apply the distributive property of multiplication in the product (−3−5i)(3+5i), combine the like terms, substitute −1 for i², and then simplify the expression.​

(−3−5i)(3+5i)=−9−15i−15i−25i²

=−9−30i−25i²

=−9−30i−25(−1)

=−9−30i+25

=16−30i

Thus, the correct match for the product 16−30i is the expression (C) (−3−5i)(3+5i).

Apply the distributive property of multiplication in the product (3−5i)(−3−5i), combine the like terms, substitute −1 for i², and then simplify the expression.​

(3−5i)(−3−5i)=−9−15i+15i+25i²

=−9+25i²

=−9+25(−1)

=−9−25

=−34

Thus, the correct match for the product −34 is the expression (D) (3−5i)(−3−5i).

The correct match for the product −16+30i is the expression (B) (3+5i)(3+5i).

The correct match for the product −34 is the expression (D) (3−5i)(−3−5i).

The correct match for the product 34 is the expression (A) (3−5i)(3+5i).

The correct match for the product 16−30i is the expression (C) (−3−5i)(3+5i).

Page 137 Exercise 21 Answer

The given complex numbers are √3+i√3 and −√3−i√3.

The question requires to prove that √3+i√3 and −√3−i√3 are the square roots of 6i.

To prove the equation, calculate the square of √3+i√3 using the distributive property of multiplication and the value of i².

Then, use the definition of square root to prove that √3+i√3

is the square root of 6i. Similarly, calculate the square of −√3−i√3 and use the definition of square root to prove that −√3−i√3 is the square root of 6i.

Calculate the square of √3+i√3 using the distributive property of multiplication.

Then, combine the like terms, substitute −1 for i², and then simplify the expression.

(√3+i√3)²

=(√3+i√3)(√3+i√3)

=3+3i+3i+3i²

=3+6i+3(−1)

=3+6i−3

=6i

According to the definition of square root, if (√3+i√3)²=6i, then the expression √3+i√3 is the square root of 6i.

Hence, it is proved that √3+i√3 is the square root of 6i.

Calculate the square of −√3−i√3 using the distributive property of multiplication.

Then, combine the like terms, substitute −1 for i², and then simplify the expression.

​(−√3−i√3)²

=(−√3−i√3)(−√3−i√3)

=3+3i+3i+3i²

=3+6i+3(−1)

=3+6i−3 =6i

According to the definition of square root, if (−√3−i√3)²=6i, then the expression −√3−i√3 is the square root of 6i.

Hence, it is proved that −√3−i√3 is the square root of 6i.

It is proved that  √3+i√3 and −√3−i√3 are the square roots of 6i.

Page 137 Exercise 22 Answer

Two complex numbers are given which differ only in the sign of their imaginary parts.

The question requires to determine what type of number is the product of two complex numbers that differ only in the sign of their imaginary parts.

Also, it is required to prove the written conjecture.

To answer the question, take two arbitrary complex numbers that differ only in the sign of their imaginary parts.

Then, write their product as a difference of squares, and use the value of i² to determine what type of number the product is.

Next, calculate the product of the two complex numbers using the distributive property of multiplication and the value of i² to prove the conjecture.

Let the two complex numbers be a+bi and a−bi which differ only in the sign of their imaginary parts.

The product of a+bi and a−bi can be written as the difference of squares a²−(bi)², which can be simplified as a²−b²i     .

The expressions a² and b²i² are real because a and b are real, and i² is equal to −1. This means that the difference of squares a²−(bi)² is a real number.

Therefore, the product of a+bi and a−bi is a real number.

Thus, the product of two complex numbers that differ only in the sign of their imaginary parts is a real number.

Write the product of a+bi and a−bi, and apply the distributive property of multiplication. Then, combine the like terms, substitute −1 for i², and then simplify the expression.

​(a+bi)(a−bi)=a²−abi+abi−b²i²

=a²−b²i²

=a²−b²(−1)

=a²+b²

The product a²+b² is real because a and b are real.

Therefore, the product of the complex numbers a+bi and a−bi is a real number.

Hence, it is proved that the product of two complex numbers that differ only in the sign of their imaginary parts is a real number.

The product of two complex numbers that differ only in the sign of their imaginary parts is a real number.

The conjecture ‘the product of two complex numbers that differ only in the sign of their imaginary parts is a real number’ is proved.

Page 138 Exercise 23 Answer

The given table shows a sequence defined by the recursive rule f(n+1)=(f(n))²+i, where f(0) is equal to 1.

The question requires to generate the first few numbers of the sequence and record the results in the given table.

To determine the few numbers and record the results, substitute 0 for n in the recursive rule, and use the value of f(0) to simplify and obtain the number f(1).

Similarly, substitute n as 1, 2, and 3, and solve the recursive rule using the value of f(n) to find the next three numbers of the sequence.

Finally, record the calculations and results in the given table.

The value of f(0) is equal to 1.

Substitute 0 for n in the recursive rule f(n+1)=(f(n))²+i. Then, substitute 1 for f(0), and simplify the expression.​

f(0+1)=(f(0))²+i

f(1)=(1)²+i

f(1)=1+i​

Substitute 1 for n in the recursive rule f(n+1)=(f(n))²+i. Then, substitute 1+i for f(1), −1 for i², and simplify the expression.​

f(1+1)=(f(1))²+i

f(2)=(1+i)²+i

f(2)=1+2i+i²+i

f(2)=1+2i+(−1)+i

f(2)=3i​

Substitute 2 for n in the recursive rule f(n+1)=(f(n))²+i. Then, substitute 3i for f(2), −1 for i², and simplify the expression.​

f(2+1)=(f(2))²+i

f(3)=(3i)²+i

f(3)=9i²+i

f(3)=9(−1)+i

f(3)=−9+i​

Substitute 3 for n in the recursive rule f(n+1)=(f(n))²+i. Then, substitute −9+i for f(3), −1 for i², and simplify the expression.

f(3+1)=(f(3))²+i

f(4)=(−9+i)²+i

f(4)=81−18i+i²+i

f(4)=81−18i+(−1)+i

f(4)=80−17i​

Record the results by completing the given table using the calculations performed and the num

The first few numbers of the sequence are recorded in the given table as shown:

Page 138 Exercise 24 Answer

The given sequence is defined by f(0)=1 and the recursive rule f(n+1)=(f(n))²+i. It is given that if the magnitudes of the numbers increase without bound, the number f(0) equal to 1

does not belong to the “filled-in” Julia set.

The question requires to use the completed table for f(0)=1 to determine whether the number belongs to the “filled-in” Julia set corresponding to c equal to i or not.

Given the completed table for f(0)=1, as calculated in the previous part of this exercise.

To answer the question, write the number f(0)=1 and the next four numbers in the sequence from the table, and calculate their magnitudes.

Then, observe the magnitudes to determine whether the magnitudes increase without bound, and use the observation to determine if the number f(0)=1

belongs to the “filled-in” Julia set corresponding to c equal to i or not.

The value of f(0) is equal to 1, which can be written as the complex number 1+0i.

Calculate the magnitude of f(0) using the formula √a²+b², and simplify the expression.​

√a²+b²=√1²+0²

=√1+0

=√1

=±1

The magnitude is equal to 1 because it cannot be negative.

The value of f(1) is equal to 1+i.

Calculate the magnitude of f(1) using the formula √a²+b², and simplify the expression.

​√a²+b²=√12+12

=√1+1

=√2

The value of f(2) is equal to 3i, which can be written as the complex number 0+3i.

Calculate the magnitude of f(2) using the formula √a²+b², and simplify the expression.​

√a²+b²=√0²+32

=√0+9

=√9

=±3

The magnitude is equal to 3 because it cannot be negative.

The value of f(4) is equal to 80−17i.

Calculate the magnitude of f(4) using the formula √a²+b², and simplify the expression.

√a²+b²=√80²+(−17)²

=√6400+289

=√6689 ≈ ±81.8

The magnitude is equal to 81.8 because it cannot be negative.

From the magnitudes of the first few generated numbers, it can be observed that there is no bound on the value of the magnitudes as n increases.

Since the magnitudes increase without bound, the number f(0)=1 does not belong to the “filled-in” Julia set corresponding to c equal to i.

No, the number f(0)=1 does not belong to the “filled-in” Julia set corresponding to c

equal to i because for the numbers of the sequence determined in the table, the magnitudes increase without bound.

Page 138 Exercise 25 Answer

The given sequence is defined by f(0)=i and the recursive rule f(n+1)=(f(n))²+i.

The question requires to determine whether the number f(0)=i belongs to the “filled-in” Julia set corresponding to c equal to i or not.

To determine the first few numbers of the sequence, substitute 0 for n in the recursive rule, and use the value of f(0) to simplify and obtain the number f(1).

Similarly, substitute n as 1, 2, and 3, and solve the recursive rule using the value of f(n) to find the next three numbers of the sequence.

Finally, record the calculations and results in a table.

To answer the question, calculate the magnitudes of the number f(0)=i and the next four numbers in the sequence from the table.

Then, observe the magnitudes to determine whether the magnitudes increase without bound, and use the observation to determine if the number f(0)=i belongs to the “filled-in” Julia set corresponding to c equal to i or not.

The value of f(0) is equal to i.Substitute 0 for n in the recursive rule f(n+1)=(f(n))²+i.

Then, substitute i for f(0), and −1 for i².​

f(0+1)=(f(0))²+i

f(1)=(i)²+i

f(1)=i²+i

f(1)=−1+i​

Substitute 1 for n in the recursive rule f(n+1)=(f(n))²+i.

Then, substitute −1+i for f(1), −1 for i², and simplify the expression.

f(1+1)=(f(1))²+i

f(2)=(−1+i)²+i

f(2)=1−2i+i²+i

f(2)=1−2i+(−1)+i

f(2)=−i​

Substitute 2 for n in the recursive rule f(n+1)=(f(n))²+i.

Then, substitute −i for f(2), and −1 for i².​

f(2+1)=(f(2))²+i

f(3)=(−i)²+i

f(3)=i²+i

f(3)=−1+i​

Substitute 3 for n in the recursive rule f(n+1)=(f(n))²+i.

Then, substitute −1+i for f(3), −1 for i², and simplify the expression.

f(3+1)=(f(3))²+i

f(4)=(−1+i)²+i

f(4)=1−2i+i²+i

f(4)=1−2i+(−1)+i

f(4)=−i

Record the results by constructing a table with the columns for values of n, f(n), and f(n+1).

The value of f(0) is equal to i, which can be written as the complex number 0+1⋅i.

Calculate the magnitude of f(0) using the formula √a²+b², and simplify the expression.​

√a²+b²=√0²+12

=√0+1

=√1

=±1

The magnitude is equal to 1 because it cannot be negative.

The value of f(1) is equal to −1+i.

Calculate the magnitude of f(1) using the formula √a²+b², and simplify the expression.​

√a²+b²=√(−1)²+12

=√1+1

=√2

The value of f(2) is equal to −i, which can be written as the complex number 0+(−1)i.

Calculate the magnitude of f(2) using the formula √a²+b², and simplify the expression.​

√a²+b²=√0²+(−1)²

=√0+1

=√1

=±1​

The magnitude is equal to 1 because it cannot be negative.

The value of f(3) is equal to −1+i.

Calculate the magnitude of f(3) using the formula √a²+b², and simplify the expression.

​√a²+b²=√(−1)²+12

=√1+1

=√2

The value of f(4) is equal to −i, which can be written as the complex number 0+(−1)i.

Calculate the magnitude of f(4) using the formula √a²+b², and simplify the expression.

√a²+b²=√0²+(−1)²

=√0+1

=√1

=±1

The magnitude is equal to 1 because it cannot be negative.

From the magnitudes of the first few generated numbers, it can be observed that the magnitudes are bounded.

The values of the magnitudes are either 1 or √2. Therefore, the magnitudes are bounded at √2.

Since the magnitudes remain bounded, the number f(0)=i belongs to the “filled-in” Julia set corresponding to c equal to i.

Yes, the number f(0)=i belongs to the “filled-in” Julia set corresponding to c equal to i because the magnitudes for the numbers of the sequence remain bounded at √2.

The values of the magnitudes are either 1 or √2.

Algebra 2 Volume 1 1st Edition Module 3 Quadratic Equations

Page 112 Problem 1 Answer

The given inequality is n−12>9.

The question requires to solve the given inequality.

To solve the inequality, isolate the variable n on one side of the inequality using addition.

Add 12   on both sides of n−12>9.​

n−12+12>9+12n>21

Thus, the solution of n−12>9 is n>21.

The required solution is n>21.

Page 112 Problem 2 Answer

The given inequality is −3p<−27.

The question requires to solve the given inequality.

To solve the inequality, isolate the variable p on one side of the inequality using division.

Divide both sides of −3p<−27 by −3.​

−3p<−27

−3p/−3>−27

−3p>9

Thus, the solution of −3p<−27 is p>9.

The required solution is p>9.

Page 112 Problem 3 Answer

The given inequality is k/4≥−1.

The question requires to solve the given inequality.

To solve the inequality, isolate the variable k on one side of the inequality using multiplication.

Multiply both sides of k/4 ≥−1 by 4.​

4(k/4)≥4(−1)

k≥−4

Thus, the solution of k/4≥−1 is k≥−4.

Page 112 Problem 4 Answer

The given expression is 16p²/2p⁴.

The question requires to simplified the given expression.

To simplify the expression, rewrite the numbers in the expression as powers of 2, and use the property of exponents am/an=am−n to simplify the expression.

Rewrite 16 and 2 as powers of 2, and use the property a^m/aⁿ = a^m−n.​

16p²/2p⁴=24p²

21/p⁴=24−1

p⁴−2

Subtract the exponents and simplify the expression.​

16p²/2p⁴=23

p²=8

p² Thus, the required simplified expression is 8p².

The required simplified form of 16p²

2p⁴ is 8 p².

Page 112 Problem 5 Answer

The given expression is 5vw⁵⋅2v⁴.

The question requires to simplified the given expression.

To simplify the expression, use the property of exponents am/an=am+n and add the exponents to simplify the expression.

Rewrite the expression, apply the property am/an=am+n, and add the exponents.

5vw⁵⋅2v⁴

=5⋅2⋅v¹⋅v⁴⋅w⁵

=10v¹+4/w⁵

=10v⁵/w⁵

Thus, the required simplified expression is 10v⁵/w⁵.

The required simplified form of 5vw⁵⋅2v⁴ is 10v⁵/w⁵.

Page 112 Problem 6 Answer

The given expression is 3x⁷/y

6x⁴/y².

The question requires to simplified the given expression.

To simplify the expression, use the property of exponents am/an=am−n, subtract the exponents, and divide the terms to simplify the expression.

Rewrite the expression, and use the property am/an=am−n.​

3x⁷/y

6x⁴/y²=3x⁷/y1

6x⁴/y²=3x⁷−4

6y²−1 Subtract the exponents and divide the terms.​

3x⁷/y

6x⁴/y²=3x³

6y1=x³/2y

Thus, the required simplified expression is x³/2y.

The required simplified form of 3x⁷/y

6x⁴/y² is x³/2y.

Page 112 Problem 7 Answer

The given equation is x2−7x+6=0.

The question requires to solve the given equation by factoring it.

To solve the equation, write the pair of factors of the constant, and identify the pair of factors whose sum is equal to the coefficient of the middle term.

Then, factorise the left-hand side of the given quadratic equation, and solve it to obtain the values of x, and thus, obtain the required solutions.

The constant in the given quadratic equation is 6.

Write the pair of factors of the constant 6.

1 and 6

2 and 3

−1 and −6

−2 and −3

It can be observed that the sum of the pair of factors −1 and −6 is equal to the coefficient of the middle term in the given quadratic equation, that is −7.

Therefore, the left-hand side of the given quadratic equation can be factorised as (x−1)(x−6).

Thus, the equation becomes (x−1)(x−6)=0.

The equation (x−1)(x−6)=0 is true only when either x−1 is equal to 0, or x−6 is equal to 0.

When x−1 is equal to 0, the value of x is equal to 1.

When x−6 is equal to 0, the value of x is equal to 6.

Thus, the required solutions are 1 and 6.

The required solutions are 1 and 6.

Page 112 Problem 8 Answer

The given equation is x2−18x+81=0.

The question requires to solve the given equation by factoring it.

To solve the equation, write the pair of factors of the constant, and identify the pair of factors whose sum is equal to the coefficient of the middle term.

Then, factorise the left-hand side of the given quadratic equation, and solve it to obtain the values of x, and thus, obtain the required solution.

The constant in the given quadratic equation is 81.

Write the pair of factors of the constant 81.

1 and 81

3 and 27

9 and 9

−1 and −81

−3 and −27

−9 and −9

It can be observed that the sum of the pair of factors −9 and −9

is equal to the coefficient of the middle term in the given quadratic equation, that is −18.

Therefore, the left-hand side of the given quadratic equation can be factorised as (x−9)(x−9).

Thus, the equation becomes (x−9)(x−9)=0.

The equation is true only when is equal to.

When is equal to, the value of is equal to.

Thus, the required solution is.

The required solution is 9.

Page 113 Problem 9 Answer

The square root of any negative number is an imaginary number.

These numbers are not real numbers, and are of the form bi, where b is a real nonzero number, and i is the imaginary unit equal to √−1.

The quadratic equations of the form x2=−a, where a is positive and real, cannot be solved because the square x² cannot be negative for any real value of x.

Such a quadratic equation can be solved using imaginary numbers by using the definition of square root to rewrite the equation as x=±√−a, and then rewriting ±√−a as the imaginary solutions ±i√a.

If x is negative real number, then the square root √x is called an imaginary number. Imaginary numbers are the non-real numbers of the form bi, where b is a real nonzero number, and i  is the imaginary unit equal to √−1.

If a quadratic equation is of the form x²=−a, where a is a real positive number, then imaginary numbers can be used to solve the equation by using the definition of square root to rewrite the equation as x=±√−a, and then rewriting ±√−a using the imaginary unit i to obtain the solutions ±i√a.

Page 113 Problem 10 Answer

The given equation is x²=16.

The question requires to solve the given equation by graphing.

To solve the given equation, let the left-side and right-side of the given equation be the function rules for f(x) and g(x) respectively.

Next, assign both functions to Y₁ and Y₂ in a graphing calculator, graph the functions, and use the intersect feature of the graphing calculator to find the point of intersection of the two functions.

Finally, use the x-coordinate of the intersection points to find the input values where the graphs intersect, and thus obtain the required solutions.

Write two functions for the expressions on either side of the given equation x²=16.​

f(x)=x²

g(x)=16

On a graphing calculator, press the Y= key.

Let Y₁=f(x), and assign x₂ to Y₁.

Let Y₂=g(x), and assign 16 to Y₂.

Next, set the viewing rectangle to [−10,10] by [−2,18], and press the GRAPH key to obtain the graph of the functions.

From the graph, it can be observed that the graphs of the two functions intersect at the points (−4,16) and (4,16).

The x-coordinates of the points of intersection are −4 and 4.

Therefore, x2=16 when the input value x is equal to either −4 or 4.

Thus, the required solutions are x=4 and x=−4.

The required solutions are x=4 and x=−4.

Page 113 Problem 11 Answer

The given equation is x²=16.

The question requires to solve the given equation by factoring.

To solve the equation, rewrite the equation, and factor it using the algebraic identity x²−y²=(x+y)(x−y).

Then, solve the factored equation using the zero-product property to obtain the values of x, and thus, obtain the required solutions.

Subtract 16 from both sides of x²=16.

x²−16=0

Rewrite the left-hand side expression as a difference of squares, and apply the algebraic identity for the difference of squares x²−y²=(x+y)(x−y).​

x²−16=0

x²−4²=0

(x+4)(x−4)=0​

According to the zero-product property, the equation (x+4)(x−4)=0 is true only when either x+4 is equal to 0, or x−4 is equal to 0.

When x+4 is equal to 0, the value of x is equal to −4.

When x−4 is equal to 0, the value of x is equal to 4.

Thus, the required solutions are x=−4 and x=4.

The required solutions are x=−4 and x=4.

Page 113 Problem 12 Answer

The given equation is x²=16.

The question requires to solve the given equation by taking square roots.

To solve the equation, take the square roots of both sides of the given equation, and simplify the expression.

Take the square roots of both sides of x²=16, and then simplify the square roots.

x2=16

x=±√16

x=±4

Thus, the required solutions are x=±4.

The required solutions are x=±4.

Page 114 Problem 13 Answer

The given equation is x²=5.

The question requires to determine which of the three methods should be used to solve the given equation, and then solve the given equation.

To answer the question, explain whether solving the equation by graphing, solving the equation by factoring, or solving the equation by taking square roots is preferred.

Then, using the preferred method, solve the equation and write its solutions.

The method preferred to solve the equation is taking the square roots.

This is because the given equation is a simple quadratic equation of the form x²=a, and thus, taking the square roots is the easiest and fastest method to obtain the required solutions.

Take the square roots of both sides of x²=5.​

x²=5

x=±√5

Thus, the required solutions are x=±√5.

The method preferred to solve the equation is taking the square roots because the given equation is a simple quadratic equation of the form x²=a.

This means that the easiest and fastest method to solve the equation is taking the square roots.

The required solutions are x=±√5.

Page 114 Problem 14 Answer

The given equation is x²=−9.The value of the square root x²

cannot be equal to the negative number −9 for any value of x. This means that the given equation has no solution.

Therefore, the given equation cannot be solved either by graphing, factoring or taking the square roots.

The given equation cannot be solved by any of the three methods because the given equation has no solution.

The left-hand side expression x² cannot be equal to the right-hand side −9 because the square root cannot be negative.

Page 114 Problem 15 Answer

The given equation is −5x²+9=0.

The question requires to solve the given equation by taking square roots.

To solve the equation, isolate the square x² on one side of the equation using subtraction and division.

Then, use the definition of square root, apply the quotient property of square roots, and simplify the expression.

Subtract 9 from both sides of −5x²+9=0, and then divide both sides by −5.​

−5x²+9−9=0−9

−5x²=−9

x²=−9/−5

x²=9/5

Use the definition of square root to simplify x²=9/5, apply the quotient property of square roots, and simplify the square root in the numerator.

x=±√9/5

x=±√9/√5

x=±3/√5

Multiply and divide the right-hand side of the equation by √5 to rationalize the denominator.

x=±3/√5⋅√5/√5

x=±3√5/5

Thus, the required solutions are x=±3√5/5.

The required solutions are x=±3√5/5.

Page 115 Problem 16 Answer

The given equation is x²−24=0.

The question requires to solve the given equation by taking square roots.

To solve the equation, isolate the square x² on one side of the equation using addition.

Then, use the definition of square root, apply the product property of square roots, and simplify the expression.

Add 24 on both sides of x²−24=0, and then use the definition of square root.

​x²−24+24=0+24

x²=24

x=±√24

Rewrite the expression in the square root, apply the product property of square roots, and simplify the square root.​

x=±√4⋅6

x=±√4⋅√6

x=±2√6​

Thus, the required solutions are x=±2√6.

The required solutions are x=±2√6.

Page 115 Problem 17 Answer

The given equation is −4x²+13=0.

The question requires to solve the given equation by taking square roots.

To solve the equation, isolate the square x² on one side of the equation using subtraction and division.

Then, use the definition of square root, apply the quotient property of square roots, and simplify the expression.

Subtract 13 from both sides of −4x²+13=0, and then divide both sides by −4.​

−4x²+13−13=0−13

−4x²=−13

x²=−13/−4

x²=13/4

Use the definition of square root to simplify x²=13/4, apply the quotient property of square roots, and simplify the square root in the denominator.

x=±√13/4

x=±√13/√4

x=±√13/2

Thus, the required solutions are x=±√13/2.

The required solutions are x=±√13/2.

Page 115 Problem 18 Answer

It is given that a water balloon is dropped from the rooftop of a five-story building, which is 50

feet above the ground. Also, it is given that the third-story window is at 24 feet above the ground.

The question requires to determine the time taken by the water balloon to pass by the third-story window.

It is required to write the equation, solve it, and write the exact and if the answer is irrational, also write the decimal approximation of the answer to the nearest tenth of a second.

To determine the time, write an equation modelling the height of the water balloon using the initial height and the equation h(t)=h0−16t².

Then, substitute the required height as 24, and solve the equation for t².

Finally, use the definition of square root, and apply the quotient property to obtain the value of t, and thus, the required time to the nearest tenth.

The given situation can be modelled using the equation h(t)=h0−16t², where h(t) is the height of the water balloon after t seconds, and h0 is the initial height of the water balloon.

Since the water balloon is dropped from the rooftop, the initial height of the water balloon is h0=50 ft.

The required height of the water balloon is h(t)=24 feet above the ground.

Substitute h0−16t² for h(t) and then 50 for h0 in h(t)=24.​

h0−16t²=24

50−16t²=24​

Subtract 50 from both sides of 50−16t²=24, and then divide both sides by −16.​

−16t²=−26

t²=−26/−16

t²=26/16​

Use the definition of square root to simplify t²=26/16, apply the quotient property of square roots, and simplify the square root in the denominator.​

t=±√26/16

t=±√26/√16

t=±√26/4

The time taken by the water balloon to reach the required height cannot be negative.

Therefore, t=√26/4.

Thus, the water balloon passes by the third-story window after √26/4 ≈ 1.3 seconds.

The equation modelling the given situation is 50−16t²=24.

The required time in which water balloon passes by the third-story window is √26/4≈1.3 seconds.

Page 117 Problem 19 Answer

It is given that a water balloon is dropped from the rooftop of a five-story building, which is 50 feet above the ground.

Also, it is given that the third-story window is at 24 feet above the ground.

The question requires to determine the time taken by the water balloon to hit the ground.

It is required to write the equation, solve it, and write the exact and if the answer is irrational, also write the decimal approximation of the answer to the nearest tenth of a second.

To determine the time, write an equation modelling the height of the water balloon using the initial height and the equation h(t)=h0−16t².

Then, substitute the required height as 0 ft, and solve the equation for t².

Finally, use the definition of square root, and apply the product property and quotient property to obtain the value of t, and thus, the required time to the nearest tenth.

The given situation can be modelled using the equation h(t)=h0−16t², where h(t)

is the height of the water balloon after t seconds, and h0 is the initial height of the water balloon.

Since the water balloon is dropped from the rooftop, the initial height of the water balloon is h0=50 ft.

When the water balloon hits the ground, the height of the balloon is 0 ft.

Therefore, the required height of the water balloon is h(t)=0 feet above the ground.Substitute h0−16t² for h(t) and then 50 for h0 in h(t)=0.​

h0−16t²=0

50−16t²=0​

Subtract 50 from both sides of 50−16t²=0, and then divide both sides by −16.​

−16t²=−50

t²=−50/−16

t²=50/16​

Use the definition of square root to simplify t²=50/16, and apply the quotient property of square roots.

Then, apply the product property of square roots in the numerator and simplify the square root in the denominator.​

t=±√50/16

t=±√50/√16

t=±√25⋅√2/4

t=±5√2/4

The time taken by the water balloon to hit the ground cannot be negative.

Therefore, t=5√2/4.

Thus, the water balloon hits the ground in 5√2/4 ≈ 1.8 seconds.

The equation modelling the given situation is 50−16t²=0.

The required time in which water balloon hits the ground is 5√2/4 ≈ 1.8 seconds.

Page 117 Problem 20 Answer

It is given that the function d(t)=8/3t²  models the distance d in feet that an object falls in t seconds on the moon. Also, it is given that a tool is dropped by an astronaut on the moon.

The question requires to determine the time taken by the tool to fall feet.

It is required to write the equation, solve it, and write the exact and if the answer is irrational, also write the decimal approximation of the answer to the nearest tenth of a second.

To determine the time, substitute the distance as 4 ft in the given equation, and solve the equation for t².

Then, use the definition of square root, and apply the product property and quotient property to obtain the value of t, and thus, the required time to the nearest tenth.

The distance that the tool falls is d(t)=4 feet.Substitute 8/3t² for d(t) in d(t)=4, and then multiply both sides by 3/8.​

8/3t²=43/8⋅8/3

t²=3/8⋅4

t²=3/2

Use the definition of square root to simplify t²=3/2, and apply the quotient property of square roots.​

t=±√3/2

t=±√3/√2

Multiply and divide the right-hand side of the equation by √2 to rationalize the denominator.

Then, simplify the square root in the denominator, and use the product property of square roots to simplify the numerator.​

t=±√3/√2⋅√2/√2

t=±√3⋅√2/2

t=±√6/2

The time taken by the tool to fall a certain distance cannot be negative.

Therefore, t=√6/2.

Thus, the tool falls 4 feet in √6/2 ≈ 1.2 seconds.

The equation for the given situation is 8/3t²=4.

The required time taken by the tool to fall 4 feet is √6/2 ≈ 1.2 seconds.

Page 117 Problem 21 Answer

The given imaginary number is −i√2.

The question requires to determine the square of the given imaginary number.

To determine the square, square the given imaginary number and use the product property of exponents.

Then, use the value of i²and simplify the expression.

Take the square of −i√² and use the product property of exponents (ab)^m=a^m⋅b^m.

(−i√²)²=(−√2)²⋅i²

Apply the exponent on the base, substitute −1 for i², and multiply the terms in the expression.​

(−i√2)²

=(2)(−1)

=−2

Thus, the square of −i√2 is −2.

The required square of −i√2 is −2.

Page 118 Problem 22

The square of i is given by the equation −1=i².Rewriting the left-hand side of the equation as a square, the equation becomes (√−1)²=i².

Using the definition of square root, the equation can be simplified to √−1=±√i2.

Simplifying the right-hand side of this equation, the equation becomes √−1=±i.

Therefore, the positive square root of −1 is i, and the negative square root of −1 is −i.

Thus, the other square root of −1 is −i.

The other square root of −1 is −i because the equation (√−1)²=i² can be written as √−1=±√i²

using the definition of square root, which implies that the positive square root of −1 is i, and the negative square root of −1 is −i.

Page 118 Problem 23

The given imaginary number is −2i.

The question requires to determine the square of the given imaginary number.

To determine the square, square the given imaginary number and use the product property of exponents.

Then, use the value of i² and simplify the expression.

Take the square of −2i and use the product property of exponents (ab)^m=a^m⋅b^m.(−2i)²=(−2)²⋅i²

Apply the exponent on the base, substitute −1 for i², and multiply the terms in the expression.​

(−2i)²=(4)(−1)

=−4

Thus, the square of −2i is −4.

The required square of −2i is −4.

Page 118 Problem 24

The given equation is 4x²+11=6.

The question requires to solve the given equation by taking square roots and allow for imaginary solutions.

To solve the equation, isolate the square x² on one side of the equation using subtraction and division.

Then, use the definition of square root, apply the quotient property of square roots, simplify the expression, and rewrite it in the form bi.

Subtract 11 from both sides of 4x²+11=6, and then divide both sides by 4.​

4x²+11−11=6−11

4x²=−5

x²=−5/4

Use the definition of square root to simplify x^2/=−5/4, apply the quotient property of square roots, rewrite the numerator in the form bi, and simplify the square root in the denominator.​

x=±√−5/4

x=±√−5/√4

x=±√(−1)(5)/2

x=±i√5/2

Thus, the required solutions are x=±i√5/2.

The required solutions are x=±i√5/2

Page 119 Problem 25

The given equation is −5x²+3=10.

The question requires to solve the given equation by taking square roots and allow for imaginary solutions.

To solve the equation, isolate the square x² on one side of the equation using subtraction and division.

Then, use the definition of square root, apply the quotient property of square roots, rewrite the numerator in the form bi, and rationalize the denominator.

Subtract 3 from both sides of −5x²+3=10, and then divide both sides by −5.​

−5x²+3−3=10−3

−5x²=7

x²=7/−5

x²=−7/5

Use the definition of square root to simplify x²=−7/5, apply the quotient property of square roots, and rewrite the numerator in the form bi.​

x=±√−7/5

x=±√−7/√5

x=±√(−1)(7)/√5

x=±i√7/√5

Multiply and divide the right-hand side of the equation by √5 to rationalize the denominator.​

x=±i√7/√5⋅√5/√5

x=±i√35/5

Thus, the required solutions are x=±i√35/5.

The required solutions are x=±i√35/5.​

Page 119 Problem 26

The first given equation is 4x²+32=0.The expression x² is always nonnegative for all real values of x.

This means that 4x²≥0 Therefore, 4x²+32≥32.Therefore, the equation 4x²+32=0 is true only if x² is negative.

Thus, the solutions of the first given equation 4x²+32=0 are imaginary. The second given equation is 4x²−32=0.

The expression x² is always nonnegative for all real values of x.

This means that 4x²≥0. Therefore, 4x²−32≥−32. Therefore, the equation 4x²−32=0 is true only if x² is positive.

Thus, the solutions of the second given equation 4x²−32=0 are real.

It can be observed that the solutions of the first quadratic equation are imaginary, and the solutions of the second quadratic equation are real.

The solutions of the first quadratic equation are imaginary, whereas the solutions of the second quadratic equation are real.

This is because the first equation is true only when x² is negative, and the second equation is true only when x² is positive.

Page 119 Problem 27

Assume the imaginary number bi. The square of this number is equal to (bi)², which can be rewritten as b²/i² using the product property of exponents.

The expression b² is always positive because b is a real nonzero number. The expression i² is the square of √−1.

Therefore, i² is always equal to the negative number −1.

The product b²/i² of the positive number b² and the negative number i² is always negative.

Thus, the expression (bi)², that is the square of the imaginary number bi, is always negative.

The square of an imaginary number is always negative.

Page 119 Problem 28    

Assume the negative number −b. Here, b is a real nonzero number.The square of the negative number −b is √−b.

The negative number −b can be written as the product of −1 and the real nonzero number b.

Therefore, the square root √−b can be written as √(−1)(b).

Using the product property of square roots, √(−1)(b) can be written as the product of the real number √b, and the imaginary number √−1.

Therefore, √−b =√−1⋅√b.

Rewriting √−1 as the imaginary unit i, the square root of −b becomes √−b=i√b.

Thus, the square root i√b of the negative number −b is obtained by using the definition of square root, rewriting the negative number as the product of −1 and a real number, and then using the product property of square roots.

To find the square roots of a negative number:

First, use the definition of square root, and rewrite the negative number in the square root as the as the product of −1 and a real number.

Next, use the product property of square roots to rewrite the square root as the product of a real square root, and the square root √−1.

Finally, rewrite the square root √−1 as the imaginary unit i to obtain the required square roots.

Page 120 Exercise 1 Answer

The given equation is x²−2=7.

The question requires to solve the given equation by graphing.

To solve the given equation, let the left-side and right-side of the given equation be the function rules for f(x) and g(x) respectively.

Next, assign both functions to Y₁ and Y₂ in a graphing calculator, graph the functions, and use the intersect feature of the graphing calculator to find the point of intersection of the two functions.

Finally, use the x-coordinate of the intersection points to find the input values where the graphs intersect, and thus obtain the required solutions.

Write two functions for the expressions on either side of the given equation x²−2=7.​

f(x)=x²−2

g(x)=7

On a graphing calculator, press the Y= key.

Let Y₁=f(x), and assign x²−2 to Y₁.

Let Y₂=g(x), and assign 7 to Y₂.

Next, set the viewing rectangle to [−5,5]

by [−2,8], and press the GRAPH key to obtain the graph of the functions.

From the graph, it can be observed that the graphs of the two functions intersect at the points (−3,7) and (3,7).

The x-coordinates of the two points of intersection of the graphs are −3 and 3.

Therefore, x²−2=7 when the input value x is equal to either −3 or 3.

Thus, the required solutions are x=−3 and x=3.

The required solutions are x=−3 and x=3.

Page 120 Exercise 2 Answer

The given equation is x²−2=7.

The question requires to solve the given equation by factoring.

To solve the equation, rewrite the equation and factor it using the difference of squares x²−y²=(x+y)(x−y).

Then, solve the factored equation using the zero-product property to obtain the values of x, and thus, obtain the required solutions.

Subtract 7 from both sides of x²−2=7.​

x²−2−7=7−7

x²−9=0​

Rewrite the left-hand side expression as a difference of squares, and apply the algebraic identity for the difference of squares x²−y²=(x+y)(x−y).​

x²−9=0

x²−3²=0

(x+3)(x−3)=0​

According to the zero-product property, the equation (x+3)(x−3)=0 is true only when either x+3 is equal to 0, or x−3 is equal to 0.

When x+3 is equal to 0, the value of x is equal to −3.

When x−3 is equal to 0, the value of x is equal to 3.

Thus, the required solutions are x=−3 and x=3.

The required solutions are x=−3 and x=3.

Page 120 Exercise 3 Answer

The given equation is x²−2=7.

The question requires to solve the given equation by taking square roots.

To solve the equation, isolate the square x² on one side of the equation using addition.

Then, take the square roots of both sides of the equation, and simplify the expression.

Add 2 on both sides of x²−2=7.​

x²−2+2=7+2

x²=9​

Take the square roots of both sides of x²=9, and then simplify the square roots.​

x²=9

x=±√9

x=±3

Thus, the required solutions are x=±3.

The required solutions are x=±3.

Page 121 Exercise 4 Answer

The given equation is 4x²=24.

The question requires to solve the given equation by taking square roots.

To solve the equation, isolate the square x² on one side of the equation using division, and use the definition of square root to simplify the expression.

Divide both sides of 4x²=24 by 4, and then use the definition of square root.​

4x²/4=24/4

x²=6

x=±√6

Thus, the required solutions are x=±√6.

The required solutions are x=±√6.

Page 121 Exercise 5 Answer

The given equation is -x2/5+15.

The question requires to solve the given equation by taking square roots.

To solve the equation, isolate the square on one side of the equation using subtraction and multiplication.

Then, use the definition of square root, apply the product property of square roots, and simplify the expression.

The question requires to solve the given equation by taking square roots.

To solve the equation, isolate the square on one side of the equation using subtraction and multiplication.

Then, use the definition of square root, apply the product property of square roots, and simplify the expression.

Subtract 15 from both sides of −x²/5+15=0, and then multiply both sides by −5.​

−x²/5+15−15=0−15

−x²/5=−15

(−5)(−x²/5)=(−5)(−15)

x²=75​

Use the definition of square root to simplify the equation x²=75, rewrite the expression in the square root, apply the product property of square roots, and simplify the square root.

​x=±√75

x=±√25⋅3

x=±√25⋅√3

x=±5√3

Thus, the required solutions are x=±5√3.

The required solutions are x=±5√3.

Page 121 Exercise 6 Answer

The given equation is 2(5−5x²)=5.

The question requires to solve the given equation by taking square roots.

To solve the equation, isolate the square x² on one side of the equation using the distributive property of multiplication, and then subtraction and division.

Then, use the definition of square root, use the quotient property of square roots, and simplify the expression.

Apply the distributive property of multiplication in the left-hand side of 2(5−5x²)=5. Then, subtract 10 from both sides, and divide both sides by −10.​

10−10x²=5−10x²

=−5−10x²−10

=−5−10x²

=1/2

Use the definition of square root to simplify x²=1/2, apply the quotient property of square roots, and simplify the square root in the numerator.​

x=±√1/2

x=±√1/√2

x=±1/√2

Multiply and divide the right-hand side of the equation by √2 to rationalize the denominator.​

x=±1/√2⋅√2/√2

x=±√2/2

Thus, the required solutions are x=±√2/2.

The required solutions are x=±√2/2.

Page 121 Exercise 7 Answer

The given equation is 3x²−8=12.

The question requires to solve the given equation by taking square roots.

To solve the equation, isolate the square x² on one side of the equation using the addition and division.

Then, use the definition of square root, use the quotient property and product property of square roots, and simplify the expression.

Add 8 on both sides of 3x²−8=12, and divide both sides by 3.​

3x²−8+8=12+8

3x²=20

3x²/3=20/3

x²=20/3

Use the definition of square root to simplify x²=20/3, and apply the quotient property of square roots.​

x=±√20/3

x=±√20/√3

Rewrite the expression in the numerator, apply the product property of square roots, and simplify the square root.

x=±√4⋅5/√3

x=±√4⋅√5/√3

x=±2√5/√3

Multiply and divide the right-hand side of the equation by √3 to rationalize the denominator.

x=±2√5/√3⋅√3/√3

x=±2√15/3

Thus, the required solutions are x=±2√15/3.

The required solutions are x=±2√15/3.

Page 121 Exercise 8 Answer

It is given that a squirrel drops an acorn from a tree.

The question requires to determine the time taken by the acorn to fall 20 feet.

It is required to write the equation, solve it, and write the exact and if the answer is irrational, also write the decimal approximation of the answer to the nearest tenth of a second.

To determine the time, write an equation modelling the distance that the acorn falls using the equation d(t)=16t².

Then, substitute the required distance fallen as 20, and solve the equation for t².

Finally, use the definition of square root, and apply the quotient property to obtain the value of t, and thus, the required time to the nearest tenth.

The given situation can be modelled using the equation d(t)=16t², where d(t) is the distance that the acorn falls in t seconds.

The distance that the acorn falls is d(t)=20 feet.

Substitute 16t² for d(t) in d(t)=20.

16t²=20 Divide both sides of 16t²=20 by 16.​

16t²/16=20

16t²=20

16t²=5/4

Use the definition of square root to simplify t²=5/4, apply the quotient property of square roots, and simplify the square root in the denominator.

t=±√5/4

t=±√5/√4

t=±√5/2

The time taken by the acorn to fall a certain distance cannot be negative.

Therefore, t=√5/2.

Thus, the acorn falls 20 feet in √5/2 ≈ 1.1 seconds.

The equation modelling the given situation is 16t²=20.

The required time taken by the acorn to fall 20 feet is √5/2 ≈ 1.1 seconds.

Page 121 Exercise 9 Answer

It is given that a window washer drops a squeegee from 60 feet. Also, it is given that another window washer is working squeegee 20 feet above the ground.

The question requires to determine the time taken by the squeegee to pass by the second window washer.

It is required to write the equation, solve it, and write the exact and if the answer is irrational, also write the decimal approximation of the answer to the nearest tenth of a second.

To determine the time, write an equation modelling the height of the squeegee using the initial height and the equation h(t)=h0−16t².

Then, substitute the required height as 20, and solve the equation for t².

Finally, use the definition of square root, and apply the quotient property to obtain the value of t, and thus, the required time to the nearest tenth.

The given situation can be modelled using the equation h(t)=h0−16t², where h(t) is the height of the squeegee after t seconds, and h0 is the initial height of the squee gee.

Since the squeegee is dropped from 60 feet, the initial height of the squeegee is h0=60 ft.

The required height of the squeegee is h(t)=20 feet above the ground.

Substitute h0−16t² for h(t) and then 60 for h0 in h(t)=20.

​h0−16t²=20

60−16t²=20​

Subtract 60 from both sides of 60−16t²=20, and then divide both sides by −16.​

−16t²=−40

t²=−40

−16t²=10/4

Use the definition of square root to simplify t²=10/4, apply the quotient property of square roots, and simplify the square root in the denominator.

t=±√10/4

t=±√10/√4

t=±√10/2​

The time taken by the squeegee to pass by the second window washer cannot be negative.

Therefore, t=√10/2.

Thus, the squeegee passes by the second window washer in √10/2≈1.6 seconds.

The equation modelling the given situation is 60−16t²=20.

The required time taken by the squeegee to pass by the second window washer is √10/2≈1.6 seconds.

Page 122 Exercise 10 Answer

The given figure shows the lengths of the sides of a rectangle, which has an area equal to 45 cm².

The question requires to determine the lengths of the sides of the rectangle, and write the exact answer and the approximate answer to the nearest tenth.

To determine the lengths, use the formula for area, the given dimensions, and the given area to form an equation modelling the given situation.

Then, solve the equation for x², and use the definition of square root to obtain the value of x.

Finally, use the obtained value in the expressions for the lengths of the sides of the rectangle to determine the exact answer and approximate answer to the nearest tenth.

From the figure, it can be observed that the length and width of the rectangle are equal to 3x cm and x cm respectively.

The area of the rectangle is equal to the product of its length 3x and width x, that is A=(3x)(x).

The given area of the rectangle is A=45 square cm.

Substitute (3x)(x) for A in A=45, and multiply the terms.​

(3x)(x)=45

3x²=45

​Divide both sides of 3x²=45 by 3, and use the definition of square root to simplify the equation.​

3x²/3=45/3

x²=15

x=±√15

The length of the width, that is x, cannot be negative.

Therefore, x=√15.

Write the length of the shorter side of the given rectangle, and simplify the expression to the nearest tenth.​

x=√15≈3.9

Calculate the length of the longer side of the given rectangle, and simplify the expression to the nearest tenth.​

3x=3√15 ≈ 11.6

Thus, the lengths of the sides of the rectangle are √15 ≈ 3.9 cm and 3√15 ≈ 11.6 cm.

The length of the shorter side of the given rectangle is √15 ≈ 3.9 cm, and the length of the longer side of the given rectangle is 3√15 ≈ 11.6  cm.

Page 122 Exercise 11 Answer

The given figure shows the lengths of the sides of a rectangle, which has an area equal to 54 cm².

The question requires to determine the lengths of the sides of the rectangle, and write the exact answer and the approximate answer to the nearest tenth.

To determine the lengths, use the formula for area, the given dimensions, and the given area to form an equation modelling the given situation.

Then, solve the equation for x², use the definition of square root, and use the product property of square roots to obtain the value of x.

Finally, use the obtained value in the expressions for the lengths of the sides of the rectangle to determine the exact answer and approximate answer to the nearest tenth.

From the figure, it can be observed that the length and width of the rectangle are equal to 3x cm and x cm respectively.

The area of the rectangle is equal to the product of its length 3x and width x, that is A=(3x)(x).

The given area of the rectangle is A=54 square cm.Substitute (3x)(x) for A in A=54, and multiply the terms.​

(3x)(x)=54

3x²=54​

Divide both sides of 3x²=54 by 3, and use the definition of square root to simplify the equation.​

3x²/3=54/3

x²=18

x=±√18

Rewrite the expression in the square root, apply the product property of square roots, and simplify the square root.​

x=±√9⋅2

x=±√9⋅√2

x=±3√2

The length of the width, that is x, cannot be negative.

Therefore, x=3√2.

Write the length of the shorter side of the given rectangle, and simplify the expression to the nearest tenth.​

x=3√2 ≈ 4.2

Calculate the length of the longer side of the given rectangle, and simplify the expression to the nearest tenth.​

3x=3⋅3√2

=9√2 ≈ 12.7

Thus, the lengths of the sides of the rectangle are 3√2 ≈ 4.2 cm and 9√2 ≈ 12.7 cm.

The length of the shorter side of the given rectangle is 3√2 ≈4.2 cm, and the length of the longer side of the given rectangle is 9√2 ≈ 12.7 cm.

Page 122 Exercise 12 Answer

The given imaginary number is 3i.

The question requires to determine the square of the given imaginary number.

To determine the square, square the given imaginary number and use the product property of exponents.

Then, use the value of i² and simplify the expression.

Take the square of 3i and use the product property of exponents (ab)^m=a^m⋅b^m.

(3i)²=(3)²⋅i²

Apply the exponent on the base, substitute −1 for i², and multiply the terms in the expression.​

(3i)²=(9)(−1)

=−9

Thus, the square of 3i is −9.

The required square of 3i is −9.

Page 122 Exercise 13 Answer

The given imaginary number is i√5.

The question requires to determine the square of the given imaginary number.

To determine the square, square the given imaginary number and use the product property of exponents.

Then, use the value of i² and simplify the expression.

Take the square of i√5 and use the product property of exponents (ab)^m=a^m⋅b^m.

(i√5)²=(√5)²⋅i²

Apply the exponent on the base, substitute −1 for i², and multiply the terms in the expression.​

(i√5)²

=(5)(−1)

=−5

Thus, the square of i√5 is −5.

The required square of i√5 is −5. 

Page 122 Exercise 14 Answer

The given imaginary number is −i√2/2.

The question requires to determine the square of the given imaginary number.

To determine the square, square the given imaginary number and use the product property of exponents.

Then, use the value of i² and simplify the expression.

Take the square of −i√2/2 and use the product property of exponents

(ab)^m=a^m⋅b^m.(−i√2/2)²=(−√2/2)²⋅i²

Apply the exponent on the base, substitute −1 for i², and simplify the expression.

(−i√2/2)²=(2/4)(−1)=−2

4=−1/2

Thus, the square of −i√2/2 is −1/2.

The required square of −i√2/2 is −1/2.

Page 122 Exercise 15 Answer

The given equation is 1/2x²+12=4.

The question requires to determine whether the given equation as real solutions or imaginary solutions by solving it.

To answer the question, isolate the square x² on one side of the equation using subtraction and multiplication.

Then, use the definition of square root, apply the product property of square roots, and rewrite the expression.

Finally, observe the solutions to determine whether they are real or imaginary.

Subtract 12 from both sides of 1/2x²+12=4, and then multiply both sides by 2.​

1/2x²+12−12=4−12

1/2x²=−8x²=−16​

Use the definition of square root to simplify x²=−16, rewrite the expression, apply the product property of square roots, and simplify the expression.​

x=±√−16

x=±√(16)(−1)

x=±√16⋅√−1

x=±4i

Therefore, the solutions are x=±4i.

The solutions include the imaginary unit i.

Thus, the given quadratic equation has imaginary solutions.

The given quadratic equation has imaginary solutions. The solutions are x=±4i.

Page 122 Exercise 16 Answer

The given equation is 5(2x²−3)=4(x²−10).

The question requires to determine whether the given equation as real solutions or imaginary solutions by solving it.

To answer the question, isolate the square x² on one side of the equation using the distributive property of multiplication, and the operations of addition, subtraction, and division.

Then, use the definition of square root, apply the quotient property of square roots, and rewrite the expression.

Finally, observe the solutions to determine whether they are real or imaginary.

Apply the distributive property of multiplication in 5(2x²−3)=4(x²−10). Then, subtract 4x² from both sides, add 15 on both sides, and divide both sides by 6.​

10x²−15=4x²−40

6x²−15=−40

6x²=−25

x²=−25/6

Use the definition of square root to simplify x²=−25/6, apply the quotient property of square roots, and rewrite the numerator in the form bi.

x=±√−25/6

x=±√−25/√6

x=±√(25)(−1)/√6

x=±5i/√6

Multiply and divide the right-hand side of the equation by √6 to rationalize the denominator.​

x=±5i/√6⋅√6/√6

x=±5i√6/6

Therefore, the solutions are x=±5i√6/6.

The solutions include the imaginary unit i.

Thus, the given quadratic equation has imaginary solutions.

The given quadratic equation has imaginary solutions. The solutions are x=±5i√6/6. 

Page 123 Exercise 17 Answer

The given equation is x²=−81.

The question requires to solve the given equation by taking square roots and allow for imaginary solutions.

To solve the equation, use the definition of square root, apply the product property of square roots, rewrite the expression in the form bi.

Use the definition of square root to simplify x²=−81, rewrite the expression, apply the product property of square roots, and simplify the expression.​

x=±√−81

x=±√(81)(−1)

x=±√81⋅√−1

x=±9i​

Thus, the required solutions are x=±9i.

The required solutions are x=±9i.

Page 123 Exercise 18 Answer

The given equation is x²+64=0.

The question requires to solve the given equation by taking square roots and allow for imaginary solutions.

To solve the equation, isolate the square x² on one side of the equation using subtraction.

Then, use the definition of square root, apply the product property of square roots, rewrite the expression in the form bi.

Subtract 64 from both sides of x²+64=0 , and use the definition of square root to simplify the equation.

x²=−64

x=±√−64

Rewrite the expression in the square root, apply the product property of square roots, and simplify the expression.​

x=±√(64)(−1)

x=±√64⋅√−1

x=±8i

Thus, the required solutions are x=±8i.

The required solutions are x=±8i.

Page 123 Exercise 19 Answer

The given equation is 5x²−4=−8.

The question requires to solve the given equation by taking square roots and allow for imaginary solutions.

To solve the equation, isolate the square x² on one side of the equation using addition and division.

Then, use the definition of square root, apply the quotient property of square roots, simplify the expression, and rewrite it using the imaginary unit i.

Add 4 on both sides of 5x²−4=−8, and then divide both sides by 5.​

5x²−4+4=−8+4

5x²=−4

x²=−4/5

Use the definition of square root to simplify x²=−4/5, apply the quotient property of square roots, and rewrite the numerator in the form bi.

x=±√−4/5

x=±√−4/√5

x=±√(4)(−1)/√5

x=±2i/√5

Multiply and divide the right-hand side of the equation by √5 to rationalize the denominator.​

x=±2i/√5⋅√5/√5

x=±2i√5/5

Thus, the required solutions are x=±2i√5/5.

The required solutions are x=±2i√5/5.

Page 123 Exercise 20 Answer

The given equation is 7×2+10=0.

The question requires to solve the given equation by taking square roots and allow for imaginary solutions.

To solve the equation, isolate the square x² on one side of the equation using subtraction and division.

Then, use the definition of square root, apply the quotient property of square roots, simplify the expression, and rewrite it using the imaginary unit i.

Subtract 10 from both sides of 7x²+10=0, and then divide both sides by 7.​

7x²+10−10=0−10

7x²=−10

x²=−10/7

Use the definition of square root to simplify x²=−10/7, apply the quotient property of square roots, and rewrite the numerator in the form bi.

x=±√−10/7

x=±√−10/√7

x=±√(10)(−1)/√7

x=±i√10/√7

Multiply and divide the right-hand side of the equation by √7 to rationalize the denominator.

x=±i√10/√7⋅√7/√7

x=±i√70/7

Thus, the required solutions are x=±i√70/7.

The required solutions are x=±i√70/7.

Page 123 Exercise 21 Answer

The given figures show the lengths of the sides of two squares. It is given that the area of the larger square is 42 cm² greater than the area of the smaller square.

The question requires to determine the lengths of the sides of the squares, and write the exact answer and the approximate answer to the nearest tenth.

To determine the lengths, use the formula for area and the given dimensions to write the areas of the squares.

Then, use the given information and the expression for the areas to form an equation modelling the given situation.

Then, solve the equation for x², and use the definition of square root to obtain the value of x.

Finally, use the obtained value in the expressions for the lengths of the sides of the two squares to determine the exact answer and approximate answer to the nearest tenth.

From the figure, it can be observed that the length of the side of the larger and smaller square are 2x cm and x cm respectively.

Therefore, the areas of the larger square and smaller square are (2x)² sq. cm and (x)² sq. cm respectively.

The area of the larger square, that is (2x)², is 42 cm² more than the area of the smaller square, that is (x)².

Therefore, (2x)²=42+(x)².

Simplify the equation (2x)²=42+(x)², and subtract x² from both sides.

(2x)²=42+(x)²

4x²=42+x²

3x²=42​

Divide both sides of 3x²=42 by 3, and use the definition of square root to simplify the equation.​

3x²/3=42

3x²=14

x=±√14

The length of the side of the smaller square, that is x, cannot be negative.

Therefore, x=√14.

Write the length of the side of the smaller square, and simplify the expression to the nearest tenth.

x=√14 ≈ 3.7

Calculate the length of the side of the larger square, and simplify the expression to the nearest tenth.​

2x=2√14≈7.5

Thus, the lengths of the sides of the smaller square and larger square are √14≈3.7 cm and 2√14≈7.5 cm respectively.

The length of the side of the smaller square is √14≈3.7 cm, and the length of the side of the larger square is 2√14≈7.5 cm.

Page 123 Exercise 22 Answer

The given figures show the lengths of the sides of two squares.

It is given that when the area of the larger square is decreased by 28 cm², the resulting area is equal to half the area of the smaller square.

The question requires to determine the lengths of the sides of the squares, and write the exact answer and the approximate answer to the nearest tenth.

To determine the lengths, use the formula for area and the given dimensions to write the areas of the squares.

Then, use the given information and the expression for the areas to form an equation modelling the given situation.

Then, solve the equation for x², use the definition of square root, and then the product property of square roots to obtain the value of x.

Finally, use the obtained value in the expressions for the lengths of the sides of the two squares to determine the exact answer and approximate answer to the nearest tenth.

From the figure, it can be observed that the length of the side of the larger and smaller square are 2x cm and x cm respectively.

Therefore, the areas of the larger square and smaller square are (2x)² sq. cm and (x)² sq. cm respectively.

When the area of the larger square is decreased by 28 cm², the resulting area is (2x)²−28.

The half of the area of the smaller square is 1/2(x)².

The area (2x)²−28 is equal to half the area of the smaller square, that is 1/2(x)².

Therefore, (2x)²−28=1/2(x)².

Simplify the equation (2x)²−28=1/2(x)², subtract 1/2x² from both sides, and add 1/2x² on both sides.​

4x²−28=1/2x²

4x²−1/2x²−28=0

7/2x²−28=0

7/2x²=28​

Multiply both sides of 7/2x²=28 by 2/7, and use the definition of square root to simplify the equation.​

2/7⋅7/2x²=2/7⋅28x²=8

x=±√8

Rewrite the expression in the square root, use the product property of square roots, and simplify the square root.​

x=±√4⋅2

x=±√4⋅√2

x=±2√2

The length of the side of the smaller square, that is x, cannot be negative.

Therefore, x=2√2.

Write the length of the side of the smaller square, and simplify the expression to the nearest tenth.​

x=2√2≈2.8

Calculate the length of the side of the larger square, and simplify the expression to the nearest tenth.​

2x=2⋅2√2

=4√2 ≈ 5.7

Thus, the lengths of the sides of the smaller square and larger square are 2√2 ≈ 2.8 cm and 4√2 ≈ 5.7 cm respectively.

The length of the side of the smaller square is 2√2 ≈ 2.8  cm, and the length of the side of the larger square is 4√2≈5.7 cm.

Page 124 Exercise 23 Answer

It is given that the height of the ball can be modelled using the equation h(t)=h0−16t², where h is the height of the ball after t seconds as it falls.

Also, it is given that the ball reaches the maximum height 67 feet, and is caught by an outfielder when it is 3 feet above the ground.

The question requires to determine the time in which the outfielder catches the ball after the ball starts descending.

To determine the time, write an equation modelling the height of the ball using the initial height and the equation h(t)=h0−16t² .

Then, substitute the required height as 3, and solve the equation for t².

Finally, use the definition of square root to obtain the value of t, and thus, the required time.

The given situation can be modelled using the equation h(t)=h0−16t², where h(t) is the height of the ball after t seconds, and h0 is the initial height of the ball.

Since the ball reaches the maximum height 67 feet, the initial height of the ball as it falls is h0=67 ft.

The height of the ball when it is caught is h(t)=3 feet above the ground.Substitute h0−16t² for h(t) and then 67 for h0 in h(t)=3.

h0−16t²=3

67−16t²=3

Subtract 67 from both sides of 67−16t²=3, and then divide both sides by −16.​

−16t²=−64

t²=−64

−16t²=4​

Use the definition of square root to simplify t²=4, and then simplify the square root.​

t=±√4

t=±2

The time in which the outfielder catches the ball after the ball starts descending cannot be negative.

Therefore, t=2.

Thus, the outfielder catches the ball 2 seconds after the ball starts to descend.

The outfielder catches the ball 2 seconds after the ball starts to descend.

Page 124 Exercise 24 Answer

The outfielder catches the ball 2 seconds after the ball starts to descend, as calculated in the previous exercise.

This means that the ball descends from the maximum height 67 feet to 3 feet in 2 seconds.

Therefore, the time taken by the ball to hit the ground, after it starts descending, will also be approximately equal to 2 seconds.

Assume that the time taken by the ball to reach the maximum height is equal to the time taken by the ball to hit the ground from its maximum height.

Therefore, the time taken by the ball to reach the maximum height will be approximately equal to 2 seconds.

The total time for which the ball was in the air is equal to the time taken by it to reach the maximum height, and the time taken by it to descend from the maximum height 67 feet to 3

feet.

Thus, the total time is equal to approximately 2+2, that is about 4 seconds.

The total time is equal to approximately 2+2, that is about 4 seconds.

This is because the total time is equal to the time taken by it to reach the maximum height, and the time taken by it to descend from the maximum height 67 feet to 3 feet.

It is assumed that the time taken by the ball to reach the maximum height is equal to the time taken by the ball to hit the ground from its maximum height, which is approximately 2 seconds.

Page 125 Exercise 25 Answer

It is given that the aspect ratio of the images is the ratio of the image width to its height, and the aspect ratio of the images on the given HDTV screen is 16:9.

Also, it is given that the area of the HDTV screen is equal to 864 square inches.

The question requires to determine the dimensions of the screen.

To determine the dimensions, use the given aspect ratio to form expressions for the image width and image height.

Then, use the formula for area, the expressions for the dimensions, and the given area to form an equation modelling the area.

Then, solve the equation for x², and use the definition of square root to obtain the value of x.

Finally, use the obtained value in the expressions for the image width and image height to determine the exact and approximate dimensions of the screen.

The given aspect ratio of the images on the given HDTV screen is 16:9.

This means that the ratio of the image width to the image height on the given HDTV screen is 16:9.

Let the image width and image height be equal to 16x and 9x respectively.

The area of the HDTV screen is equal to the product of the image width 16x and the image height 9x, that is A=(16x)(9x).

The given area of the rectangle is A=864 square inches.

Substitute (16x)(9x) for A in A=864, and multiply the terms.​

(16x)(9x)=864

144x²=864

​Divide both sides of 144x²=864 by 144, and use the definition of square root to simplify the equation.​

144x²

144=864

144x²=6

x=±√6

If the variable x is negative, then the image width 16x and the image height 9x are negative, which is not possible.

Therefore, x=√6.

Calculate the image width of the HDTV screen, and simplify the expression to the nearest hundredth.​

16x=16√6≈39.19

Calculate the image height of the HDTV screen, and simplify the expression to the nearest hundredth.​

9x=9√6≈22.05

Thus, the width and height of the HDTV screen are 16√6≈39.19 inches and 9√6≈22.05 inches respectively.

The width of the HDTV screen is 16√6≈39.19 inches, and the height of the HDTV screen is 9√6 ≈ 22.05 inches.

Page 126 Exercise 26 Answer

It is given that a suspension bridge has two parabolic cables connected between the towers, and vertical cables are suspended from the parabolic cables.

The given table shows the displacements of three vertical cables from the shortest cable, and the height of those cables.

The question requires to determine the quadratic function for the height of the parabolic cable above the road as a function of the displacement from the lowest point of the cable.

Then, it is required to determine the distance between the towers if the maximum height of the parabolic cable is 48 m above the road.

To determine the model, let the equation be h(x)=ax²+bx+c.

Then, use the first row data from the given table to obtain the value of c.

Next, use the data from the second and third rows to form two equations, and solve them to determine the values of the remaining constants.

Finally, use the constants in the assumed equation to determine the required equation.

To determine the distance between the towers, substitute the height as 48

m in the quadratic equation, and use the definition of square root to obtain the value of x.

Then, use the value to find the horizontal displacement of the parabolic cable from the left tower to its lowest position, and the horizontal displacement of the parabolic cable from the right tower to its lowest position.

Finally, a Let the required equation be h(x)=ax²+bx+c, where h(x) is the height of the parabolic cable when the horizontal displacement from the lowest point is x m.

The lowest point of the parabolic cable is the point at which the shortest vertical cable is suspended.

Also, the height of the parabolic cable above the ground is equal to the height of the vertical cable suspended at that point.

When the displacement of a vertical cable from the shortest vertical cable is 0 m, the height of the vertical cable is 3 m.

This means that when the horizontal displacement of the parabolic cable from its lowest point is 0 m, the height of the parabolic cable above the ground is 3 m.

Substitute 0 for x and 3 for h(x) in h(x)=ax²+bx+c, and simplify the expression.​

3=a(0)²+b(0)+c

3=a(0)+b(0)+c

3=0+0+c

3=c​

When the displacement of a vertical cable from the shortest vertical cable is 1 m, the height of the vertical cable is 3.05 m.

This means that when the horizontal displacement of the parabolic cable from its lowest point is 0 m, the height of the parabolic cable above the ground is 3.05 m.

Substitute 1 for x, 3.05 for h(x), and 3 for c in h(x)=ax²+bx+c, simplify the expression, and subtract 3 from both sides.

3.05=a(1)²+b(1)+3

3.05=a(1)+b(1)+3/0.05=a+b…………………(1)​

When the displacement of a vertical cable from the shortest vertical cable is 2 m, the height of the vertical cable is 3.2 m.

This means that when the horizontal displacement of the parabolic cable from its lowest point is 2 m, the height of the parabolic cable above the ground is 3.2 m.

Substitute 2 for x, 3.2 for h(x), and 3 for c in h(x)=ax²+bx+c, simplify the expression, and subtract 3 from both sides.​

3.2=a(2)²

+b(2)+3

3.2=a(4)+b(2)+3

0.2=4a+2b…………………(2)​

Equation (1) can be rewritten as a=0.05−b.

Substitute 0.05−b for a in equation (2), multiply the terms, and solve the equation for b.​

0.2=4(0.05−b)+2b

0.2=0.2−4b+2b

0.2=0.2−2b

0=−2b

0=b​

Substitute 0 for b in equation (1).​

0.05=a+0

0.05=a​

Thus, the equation h(x)=ax²+bx+c becomes:​

h(x)=0.05x²+0x+3

h(x)=0.05x²+3

Thus, the equation modelling the height of the parabolic cable is h(x)=0.05×2+3.

Let the height of the parabolic cable be equal to its maximum height 48. Therefore, h(x)=48.

Substitute 0.05x²+3 for h(x) in h(x)=48, subtract 3 from both sides, and divide both sides by 0.05.

Then, use the definition of square root, and simplify the square root.

0.05x²+3=48

0.05x²=45

x2=900

x=±√900

x=±30

The horizontal displacement of the parabolic cable from its lowest point, that is x, cannot be negative.

Therefore, x=30.

The displacement of the parabolic cable from its lowest point is equal to 30 m when the cable is at the maximum height.

This means that at the point where the parabolic cable connects to the left tower, that is at its maximum height on the left side, the horizontal displacement of the parabolic cable from the shortest vertical cable is 30 m.

Similarly, at the point where the parabolic cable connects to the right tower, that is at its maximum height on the right side, the horizontal displacement of the parabolic cable from the shortest vertical cable is 30 m.

The distance between the towers is equal to the sum of the horizontal displacement of the parabolic cable from the left tower to its lowest position, and the horizontal displacement of the parabolic cable from the right tower to its lowest position.

Thus, the distance between the towers is equal to 30+30=60 m.

The quadratic equation modelling the height of the parabolic cable is h(x)=0.05x²+3, where h(x)  is the height of the parabolic cable when the horizontal displacement from the lowest point is x m.

The distance between the towers is equal to 30+30=60 m.

HMH Algebra 2 Volume 1 1st Edition Module 5 Chapter 5 Polynomial Functions

Page 249 Problem 1 Answer

It is given a polynomial in intercept form. It is required to explain that how do you sketch the graph of a polynomial in intercept form.

To sketch a graph of a polynomial, find the intercept then check the symmetry.

Use the multiplicities of the zeros to determine the behaviour of the polynomial at the $x$ intercepts.Determine the end behaviour by examining the leading term.

Use the behaviour and the behaviour at the intercept to sketch the graph.

The explanation to sketch the graph of a polynomial in intercept form is to find the intercept then check the symmetry.

Use the multiplicities of the zeros to determine the behaviour of the polynomial at the $x$ intercepts.

Determine the end behaviour by examining the leading term . Use the behaviour and the behaviour at the intercept to sketch the graph.

Page 249 Problem 2 Answer

The following functions are given, f(x)=x,

f(x)=x²,

f(x)=x³,

f(x)=x⁴ ,

f(x)=x⁵, and

f(x)=x⁶.

It is required to draw the graph of each function to determine the domain, range and end behavior of the functions.

To find the required answer draw the graph using the graphing calculator and analyze it to determine domain, range and end behavior and make a table.

Graph all the functions using a graphing calculator.

Observe the graph and complete the table filling the domain, rang and end behavior of the given functions.

The graph of all the function is given below.

The complete table is as follows.

Page 249 Problem 3 Answer

The following functions are given, f(x)=−x,

f(x)=−x²,

f(x)=−x³ ,

f(x)=−x⁴ ,

f(x)=−x⁵and

f(x)=−x⁶ .

It is required to draw the graph of each function to determine the domain, range, and end behavior of the functions.

To find the required answer draw the graph using the graphing calculator and analyze it to determine domain, range, and end behavior and make a table.

Graph all the functions using a graphing calculator.

Observe the graph and complete the table filling the domain, rang, and end behavior of the given functions.

The graph of the functions is as follows.

The complete table is as follow.

Page 250 Problem 4 Answer

It is given that two functions,f(x)=xⁿ, f(x)=−xⁿ  where n is positive whole number.

It is required to generalize the result of this explore between these two given functions.

To generalize the result, substitute n with some positive whole number for both the function then analyse the result to get the general relation between them.

For the function f(x)=xⁿ.

Substitute n with some positive whole number.

For n=1,

​f(x)=x¹

​​​​​​​​​​​=x

For n=2,

f(x)=x²

For n=3

f(x)=x³

For the function f(x)=−xn.

Substitute n with some positive whole number.

For n=1,

​f(x)=−x¹

​​​​​​​​​​=−x​

For n=2,

f(x)=−x²

For n=3,

f(x)=−x³

Now, compare the obtained result of both the function’s for n=1,2,3.

The relation between f(x)=xⁿ and

f(x)=−xⁿ is

f(x)=−f(x).

The generalize result of this explore between these two given function f(x)=−xⁿ and f(x)=−f(x) is f(x)=−f(x).

Page 252 Problem 5 Answer

It is required to determine whether the graph of a polynomial function in intercept form crosses the x-axis or is tangent to it at an x−intercept.

The graph of p(x)=a(x−x₁)(x−x₂)…(x−x_n) has x₁,x₂,…, and  x_n as its x-intercepts.

Hence, it is called intercept form.In equation p(x)=a(x−x₁)(x−x₂)…(x−x_n) if x₁,x₂,…, and  x_n has non zero value imply function will cross x- axis, but if variable is 0, then this implies that the point is tangent to it at an x- intercept.

From equation p(x)=a(x−x₁)(x−x₂)…(x−x_n) it can be determined whether the graph of a polynomial function in intercept form crosses the x- axis or is tangent to it at an x- intercept.

Page 252 Problem 6 Answer

It is required to find how the graph of the function changes when a factor of−1 is introduced.Consider the function f(x)=x⁴.

When a factor of−1 is introduced, it becomes f(x)=−x⁴.

Plot the graph f(x)=−x⁴, further observe the graph.

It can be observed that there is one distinct factor which is 0.

Further, there are no x- intercepts, as graph turn at point (0,0).

So, the graph is tangent to x- axis at x- intercepts: (0,0) and has only one turning point.

There is one global maximum value at(0,0) and global minimum none.

Therefore, the graph becomes a reflection of the its positive function.

When a factor of −1 is introduced in any function, the graph of the new function is symmetric to the previous one. Also, it is symmetric. The attributes remain the same.

Page 253 Problem 7 Answer

It is given equation f(x)=−(x−4)(x−1)(x+1)(x+2)

It is required to sketch the graph of the polynomial function.

First identify the end behavior of the graph. Then, find its x-intercepts. Now, identify the nature of the graph and draw the graph.

Identify the end behavior of the function,  f(x)=−(x−4)(x−1)(x+1)(x+2).

As x tends to +∞, then f(x) tends to −∞.

As x tends to −∞, then f(x) tends to −∞.

Identify the x- intercept of graph. Then, use the sign of  f(x) on the intervals determined by the x- intercept to find where the graph is above the x- axis and where It’s below the x- axis.

The x- intercept are,x=4,x=1,x=−1, and x=−2.

Tabulate the data and find the nature of f(x) according to each term.

Plot the graph.

The graph of f(x) is above the x- axis on the intervals −2<x<−1 and 1<x<4.

It’s below the x- axis on the intervals x<−2, −1<x<1, and x>4.

The sketch of the graph of  f(x)=−(x−4)(x−1)(x+1)(x+2) is given below.

Page 255 Problem 8 Answer

It is given that f(x)=−x²(x−4).

It is required to sketch the graph of the polynomial function.

First, identify the end behavior of the graph. Then, find its x-intercepts. Now, identify the nature of the graph and draw the graph.

Identify the end behavior of the function, f(x)=−x²(x−4).

As x tends to +∞, then f(x) tends to−∞.

As x tends to −∞, then f(x) tends to +∞.

Identify the x- intercept of graph. Then, use the sign of  f(x) on the intervals determined by the x- intercept to find where the graph is above the x- axis and where It’s below the x- axis.

The x- intercept are,x=0, andx=4.

Tabulate the data and find the nature of f(x) according to each term.

Plot the graph. The graph of f(x) is above the x- axis on the intervals x<0 and 0<x<4. It’s below the x- axis on the intervalsx>4.

The graph of the function is given below:

Page 255 Problem 9 Answer

It is given an open box is to be made from a sheet of cardboard that is 9 inches long and 5 inches wide and square flap of side x inches at each corner.

It is required to find the value of x that will produce the box with maximum volume to the nearest tenth.

First, write the dimensions of the box and obtain a volume function. Make the graph of the volume function to find the value of the dimensions.

Find the x²– intercept of graph. Identify the graphs x- intercept, and then use the sign of  f(x) on the intervals determined by the x- intercept to find where the graph is above the x- axis and where it’s below the x- axis.

Then, plot the graph and obtain the values.

Find the dimensions of the box.

The length of the box is (9−2x) inches.

The width of the box is (5−2x)inches.

The height of the box is x inches.

Write the volume function. Use the formula,V=Length×width×Height

V=(9−2x)(5−2x)x

Make the graph of the volume function to find the value of the dimensions.

Find the x- intercept of graph. Assume f(x)=(9−2x)(5−2x)x.

Identify the graphs x- intercept. The x- intercept are, x=5/2,x=0, and x=9/2.

Find the intervals of function. So, the graph of f(x)=(9−2x)(5−2x) x is determined by the following three constraints.

9−2x>0, or x<9/2

5−2x>0, or x<5/2

x>0.

Taken together, these constraints give a domain of 0<x<9/2.

The graph for function as follow f(x)=(9−2x)(5−2x)x.

So, f(x)=21.0 when x≈1. Which means that the box has a maximum volume of   cub 21 ic inches when square flaps with side length of 1 inches are made in the corners of the sheet of cardboard.

It is observed that when x is equal to 1 box will have maximum volume.

Page 257 Problem 10 Answer

It is required to discuss, although the volume function has three constraints on its domain, then why domain involves only two of them.

For solving equation function it is required to know point of intercept, for volume as a function, curve will intercept at three points on x- axis, hence involves three constrain.

As each dimension must be greater than zero for real world problem.

One of the constrain is bounded by other two.

Therefore, domains involve two constrains only.

In the volume function, one of the constrain is bounded by other two. So, domain involves only two of them.

Page 257 Problem 11 Answer

It is given an open box is to be made from a sheet of cardboard that is 25 inches long and 13 inches wide and square flap of side x inches at each corner.

It is required to find the value of x that will produce the box with maximum volume to nearest tenth.

First, write the dimensions of the box and obtain a volume function. Make the graph of the volume function to find the value of the dimensions.

Find the x- intercept of graph. Identify the graphs x- intercept, and then use the sign of  f(x) on the intervals determined by the x- intercept to find where the graph is above the x- axis and where it’s below the x- axis.

Find the dimensions of the box.

The length of the box is (25−2x) inch.

The width of the box  is (13−2x) inches.

The height of the box is xinches.

Write the volume function. Use the formula, V=Length×width×Height

V=(25−2x)(13−2x)x

Make the graph of the volume function to find the value of the dimensions.

Find the x- intercept of graph. Assume f(x)=(25−2x)(13−2x)x.

Identify the graphs x- intercept.

x=25/2,x=0, and x=13/2.

Find the intervals of function. So, the graph of f(x)=(13−2x)(25−2x)x is determined by the following three constraints.

13−2x>0, or x<13/2

25−2x>0, or x<25/2 x>0.

Taken together, these constraints give a domain of 0<x<25/2.

Graph for function as follow f(x)=(13−2x)(25−2x)x.

So, f(x)=402.22 when x≈3. Which means that the box has a maximum volume of 402 cubic inches when square flaps with side length of 3 inches are made in the corners of the sheet of cardboard.

It is observed that when x is equal to 3 box will have maximum volume.

Page 257 Problem 12 Answer

It is required to find why is the constant factor important when graphing the function and the function here is a polynomial function with intercept form.

The sign of the function values tells you whether the graph is above or below the x- axis on a particular interval.

Constant factor is determined by sign of the function, further help to describe the end behavior of the curve.

The constant factor changes the sign of the curve and also changes the end behavior it is an important factor.

Page 258 Problem 13 Answer

It is given that f(x)=x⁷.

It is required to find function’s domain, range, and the end behavior.

Use a graphing calculator, plot the graph. Then, using the graph observe the nature and find the domain range and end behaviour,

Using a graphing calculator the graph of the function f(x)=x⁷ is as follows:

From the above graph, it is observed that domain of the function is (−∞,+∞),  range of the function(−∞,+∞) also end behavior can be expressed as,As, x tends to +∞, then f(x) tends to+∞.

As, x tends to−∞, then f(x) tends to −∞.

Domain: (−∞,+∞)

Range: (−∞,+∞)

End behavior: Asx→+∞, f(x)→+∞ and as x→−∞, f(x)→−∞.

Page 258 Problem 14 Answer

It is given thatf(x)=−x⁹.

It is required to find function’s domain, range, and the end behavior.

Use a graphing calculator, plot the graph. Then, using the graph observe the nature and find the domain range and end behavior.

Using a graphing calculator, the graph of the functionf(x)=−x⁹ is as follows:

Assumey=−x⁹.

The function can take all the values of the real line and give all the output values of the real line.

Therefore, the domain and range of the function will be. (−∞,+∞)

From the graph, observe that the function approaches ∞ at x=∞ and it approaches−∞ at x=−∞.

The graph of the function, f(x)=−x⁹ is given below.

The following values are obtained from the graph of the function.

Domain:(−∞,+∞)

Range: (−∞,+∞)

End behavior: Asx→+∞, f(x)→−∞and as x→−∞, f(x)→∞.

Page 258 Problem 15 Answer

It is given that f(x)=x¹⁰.

It is required to find function’s domain, range, and the end behavior.

Use a graphing calculator, plot the graph. Then, using the graph observe the nature and find the domain range and end behavior.

Using a graphing calculator, the graph of the function f(x)=x¹⁰ is as follows:

Assume y=x¹⁰.

The function can take all the values of the real line but it will give only positive output values of the real line because of an even exponent.

Therefore, the domain of the function is R and range of the function will be R+∪{0}.

From the graph, observe that the function approaches ∞at x=∞ and it approaches ∞ at x=−∞.

The graph of the function, f(x)=x¹⁰is given below.

The following values are obtained from the graph of the function.

Domain: R

Range: R+∪{0}

End behavior: Asx→+∞, f(x)→∞ and as x→−∞, f(x)→∞.

Page 258 Problem 16 Answer

It is given that f(x)=−x⁸.

It is required to find function’s domain, range, and the end behavior.

Use a graphing calculator, plot the graph. Then, using the graph observe the nature and find the domain range and end behavior.

Using a graphing calculator, the graph of the function f(x)=−x⁹ is as follows:

Assume y=−x⁸. The function can take all the values of the real line but it will give only negative output values of the real line because of a negative at the beginning.

Therefore, the domain of the function is R and range of the function will be R−∪{0}.

From the graph, observe that the function approaches−∞

at x=∞ and it approaches −∞ at x=−∞.

The graph of the function, f(x)=−x⁸ is given below.

The following values are obtained from the graph of the function.

Domain: R

Range: R−∪{0}

End behavior: As x→+∞, f(x)→−∞ and as x→−∞, f(x)→−∞.

Page 258 Problem 17 Answer

It is given that f(x)=x(x+1)(x+3).

It is required to graph the function using a  graphing calculator and use the graph to determine the turning points and the type of maximum and minimum values.

Use a graphing calculator, plot the graph. Then, using the graph observe the nature and find the required values.

Use a graphing calculator and plot f(x)=x(x+1)(x+3).

Now, from the graph observe that graph changes its concavity at two points. So, there are two turning points.

Those two turning points are (−2.215,2.113) and (−0.45,−0.63).

The function is at peak at the point  (−2.215,2.113) and at it lowest peak at (−0.45,−0.63).

Therefore,(−2.215,2.113) is local maxima and (−0.45,−0.63) is the local minima.

The graph of the function is given below and the function has following attributes.

Turning points: 2

Turning points coordinates: (−2.215,2.113), (−0.45,−0.63).

Local maxima:(−2.215,2.113)

Local minimum: (−0.45,−0.63)

Global maximum or minimum: None

Page 258 Problem 18 Answer

It is given that f(x)=(x+1)²(x−1)(x−2).

It is required to graph the function using a  graphing calculator and use the graph to determine the turning points and the type of maximum and minimum values.

Use a graphing calculator, plot the graph. Then, using the graph observe the nature and find the required values.

Use a graphing calculator and plot f(x)=(x+1)²(x−1)(x−2).

Now, from the graph observe that graph changes its concavity at three points. So, there are three turning points.

Those three turning points are (0.157,2.08), (1.593,−1.623) and(−1,0).

The function is at peak at the point  (0.157,2.08) and at minimum at (−1,0).

Therefore, (0.157,2.08) is local maxima and (−1,0) is the local minima. The global minima is(1.593,−1.623).

The graph of the function is given below and the function has following attributes.

Turning points: 3

Turning points coordinates: (0.157,2.08), (1.593,−1.623) and (−1,0).

Local maxima:(0.157,2.08)

Local minimum: (−1,0)

Global minimum: (1.593,−1.623)

Page 258 Problem 19 Answer

It is given that f(x)=−x(x−1)(x+2)².

It is required to graph the function using a  graphing calculator and use the graph to determine the turning points and the type of maximum and minimum values.

Use a graphing calculator, plot the graph. Then, using the graph observe the nature and find the required values.

Use a graphing calculator and plot f(x)=−x(x−1)(x+2)².

Now, from the graph observe that graph changes its concavity at three points. So, there is only one turning point, (0.25,8.543).

And the function has global minimum atx=0.25.

The graph of the function is given below and the function has following attributes.

Turning points: 1

Turning points coordinates: (0.25,8.543)

Local maxima: None

Local minimum: None

Global minimum:(0.25,8.543)

Page 259 Problem 20 Answer

It is given that f(x)=−x(x−2)².

It is required to graph the function.

Use a graphing calculator, and plot the graph of the given function.

Use a graphing calculator and plot f(x)=−x(x−2)².

The graph of the function is given below.

Page 259 Problem 21 Answer

It is given that f(x)=−x(x+1)(x−2)(x−3).

It is required to graph the function.

Use a graphing calculator, and plot the graph of the given function.

Use a graphing calculator and plot f(x)=−x(x+1)(x−2)(x−3).

The graph of the function is given below.

Page 260 Problem 22 Answer

It is given an open-top box out of cardboard that is 6 inches long and 3 inches wide, you make a square flap of side length x inches in each corner by cutting along one of the flap’s sides and folding along the other.

Once you fold up the four sides of the box, you glue each flap to the side it overlaps.

It is required to find the value of x

that maximize the volume of the box.

To find the required answer, find the dimension of the box once the flaps have been made and the sides have been folded up.

Create a volume function for the box. Maximize the volume for a given x, differentiate the volume with respect to x, and equal to zero, to find the value of x that maximizes the volume.

Write expression for the dimension of the box.

The length of the box is 6−2x units.

The width of box is3−2x units.

The height of the box is x units.

Now write the volume function and determine its domain. Use the formula, V=Length×width×Height.

Substitute length as 6−2x, breadth as 3−2x, and height as x in the above formula and simplify.

​V=(6−2x)(3−2x)x

=(6×3−6×2x−2x×3+2x×3x)×x

=(18−12x−6x+6x²)×x

=4x³−18x²+18x

Thus, the volume function is V=4x³−18x²+18x.

To maximize the volume for a given x, differentiate the volume with respect to x and equal to zero.

Use the power rule of differentiation,d/dx(xⁿ)=nxⁿ−1.

​dV/dx=0

d/dx(4x³−18x²+18x)=0

12x³−36x+18=0

Here, the x has two values 2.366 and 0.634. Since the value of x cannot be more than half the width 3,  choose x=0.634, that maximize the volume.

Find the maximum volume.

Substitute x as 0.634 in V=4x³−18x²+18x and simplify.

​Vmax

=4(0.634)³−18(0.634)²+18(0.634)

=1.019−7.235+11.412

=5.196

Hence, the maximum volume is 5.196 inches 3.

The value of x that maximize the volume of the box is 0.634. Also, the maximum volume is 5.196 inches 3.

Page 261 Problem 23 Answer

It is given a graph of a polynomial that cuts the x axis at 4 points.

It is required to find a quadratic function in intercept form for the given graph whose x-intercepts are integers assuming that the constant factor a is either 1 or−1.

First, identify the x-intercept that means find out the points where the curve cut the x axis, make them in factor form.

Then, write the polynomial as product of all the factors and assume the constant as 1 to get the function.

Analyse the given graph, where it cuts the x axis. Therefore, it cuts the axis at x=−3,x=0,x=4, and x=2.

Let P(x) be the polynomial.

As x=−3 is a zero so, (x+3) is a factor.

As x=0 is a zero so, x is a factor.

As x=4 is a zero so, (x−2) is a factor.

As x=2 is a zero so, (x−4) is a factor.

The polynomial is factored form will beP(x)=ax(x+3)(x−2)(x−4).

Check the end behavior of the polynomial.

The end behavior of the polynomial is x→∞ then, P(x)→∞ and when x→−∞ then, P(x)→∞ Therefore, a=1.

Thus,P(x)=x(x+3)(x−2)(x−4) is the required function.

The quadratic function in intercept form for the given graph isP(x)=x(x+3)(x−2)(x−4).

Page 261 Problem 24 Answer

It is given a graph of a polynomial that touches the x axis at 2 points.

It is required to find a quadratic function in intercept form for the given graph whose x-intercepts are integers assuming that the constant factor a is either 1 or −1.

First, identify the x-intercept that means find out the points where the curve cut the x axis, make them in factor form.

Then, write the polynomial as product of all the factors and assume the constant as 1 to get the function.

Analyse the given graph, where it cuts the x-axis. Therefore, it cuts the axis at x=3, and x=−2.

Let P(x) be the polynomial.

As x=3 is a zero of multiplicity so, (x−3)² is a factor.

Asx=−2 is a zero of multiplicity so,(x+2)² is a factor.

The polynomial is factored form will be P(x)=a(x−3)²(x+2)².

Check the end behaviour of the polynomial.

The end behaviour of the polynomial is x→∞ then, P(x)→−∞ and whenx→−∞ then, P(x)→−∞.

Therefore, a=−1.

Thus, P(x)=−(x−3)²(x+2)² is the required function.

The quadratic function in intercept form for the given graph is P(x)=−(x−3)²(x+2)².

Page 262 Problem 25 Answer

It is given a function f(x)=(x−1)²(x+2).

It is required to find the statement that  the x-intercept are x=1 and x=−2 is applicable or not for the given function.

To find, the statement true or not draw the graph of f(x)=(x−1)²(x+2) using a graphing calculator.

Find out the x intercept. Then check the obtained point are same or not with the given statement to find out is the given statement applicable or not.

Draw the graph of f(x)=(x−1)²(x+2) using a graphing calculator.

From the graph, it is observed that the curve cuts the x axis at x=−2 and the curve touches the x axis at x=1.

It means the curve is tangent to the x-axis at x=1.

Therefore the statement that the x-intercept are x=1 and x=−2 is applicable.

Statement A: The x-intercepts of f(x) are x=1 and x=−2 is true for the function f(x)=(x−1)²(x+2).

Page 262 Problem 26 Answer

It is given a function f(x)=(x−1)²(x+2).

It is required to find the statement that  the x-intercept are x=−1 and x=2 is applicable or not for the given function.

To find, the statement true or not draw the graph of f(x)=(x−1)²(x+2) using a graphing calculator.

Find out the x-intercept. Then check the obtained point are same or not with the given statement to find out is the given statement applicable or not.

Draw the graph of f(x)=(x−1)² (x+2) using a graphing calculator.

From the graph, it is observed that the curve cuts the x axis atx=−2 and the curve touches the x axis at x=1.

It means the curve is tangent to the xaxis atx=1.

Therefore,  the statement that the x intercept arex=−1 and x=2 is not applicable for the given function.

Statement B: The x-intercepts of f(x) are x=−1 and x=2 is not true for the function f(x)=(x−1)²(x+2).

Page 262 Problem 27 Answer

It is given a function f(x)=(x−1)²(x+2).

It is required to find whether the statement that the graph cross the $x$ axis and x=1, and is tangent to the x axis atx=−2 applies to the function or not.

To find, the statement true or not draw the graph of f(x)=(x−1)²(x+2)

using a graphing calculator. Find out the x intercept.

Then, check the obtained point are same or not with the given statement to find out is the given statement applicable or not.

Draw the graph of f(x)=(x−1)²(x+2) using a graphing calculator.

From the graph it is observed that the curve does not cross the x axis at x=1 and it is not tangent to x=−2.

Therefore, this is not applicable for the given function.

Statement C: The graph cross the x axis and x=1, and is tangent to the x axis at x=−2 is not true for the function f(x)=(x−1)²(x+2).

Page 262 Problem 28 Answer

It is given a function f(x)=(x−1)²(x+2).

It is required to find the statement that the graph cross the x-axis atx=−1, and is tangent to the x axis at x=2 is applicable or not for the given function.

To find, the statement true or not draw the graph of f(x)=(x−1)²(x+2) and find out the x-intercept means find the point where the curve cut the x axis.

Then check the obtained point are same or not with the given statement to find out is the given statement applicable or not.

Draw the graph of f(x)=(x−1)²(x+2).

From the graph it is observed that the curve does not cross the x axis at  x=−1, and is tangent to the x axis at x=2 is not applicable for the given function.

Statement D: The graph crosses the x-axis atx=−1, and is tangent to the x axis at x=2 not applicable for the given function

f(x)=(x−1)²(x+2).

Page 262 Problem 29 Answer

It is given a function, f(x)=(x−1)²(x+2).

It is required to tell whether the graph is tangent to the x-axis at x=1 and crosses the x-axis at x=−2.

Plot the lines on the graph off(x)=(x−1)²(x+2) which is inferred from part (A).

Then, observe the graph and get the required answer.

Plot the linex=1 and x=−2 on the graph off(x)=(x−1)²(x+2).

From the graph observe that the graph is tangent to the x-axis atx=1 and crosses the x-axis at x=−2.

Therefore, the statement applies.

The statement E applies to the graph of f(x)=(x−1)²(x+2).

Page 262 Problem 30 Answer

It is given a functionf(x)=(x−1)²(x+2).

It is required to find whether the statement that the graph is tangent to the x-axis at x=−1, and is tangent to the x-axis at x=2 applies to the function or not.

To find, the statement true or not draw the graph of f(x)=(x−1)²(x+2)using a graphing calculator.

Find out the x intercept. Then, check the obtained point are same or not with the given statement to find out is the given statement applicable or not.

Draw the graph of f(x)=(x−1)²(x+2) using a graphing calculator.

From the graph it is observed that the curve is not tangent to the x-axis at x=−1, and is tangent to the x-axis at x=2.

Therefore, the is not applicable for the given function.

Statement C: The graph is tangent to the x-axis at x=−1, and is tangent to the x-axis at x=2 is not true for the function f(x)=(x−1)²(x+2).

Page 262 Problem 31 Answer

It is given a function f(x)=(x−1)²(x+2).

It is required to find whether the statement that the graph is a local, but not global minimum occurs at interval −2<x<1, and a local, but not global, maximum occurs at x=1 applies to the function or not.

The graph inferred from part(a) is as follows.

Observe that the graph is increasing in the interval −2<x<1.

So, it cannot be the global minimum.

The global maximum cannot occur at x=1 because the value of function is zero at that point but actually the value of function increases.

Therefore, the given statement is not valid.

The statement the graph is a local, but not global minimum occurs at interval −2<x<1, and a local, but not global, maximum occurs at x=1

does not apply to the function or not.

Page 262 Problem 32 Answer

It is given a function f(x)=(x−1)²(x+2)

.It is required to find whether the statement that the graph is a local, but not global maximum occurs at interval −2<x<1 , and a local, but not global, minimum occurs at x=1

applies to the function or not.

The graph inferred from part(a) is as follows.

Observe that the graph is increasing in the interval −2<x<1.

So, it is the global minimum.

Therefore, the given statement is valid.

The statement the graph is a local, but not global maximum occurs at interval −2<x<1, and a local, but not global, minimum occurs at x=1 applies to the function.

Page 262 Problem 33 Answer

It is given a function f(x)=(x−1)²(x+2).

It is required to find whether the statement that the graph is a local, but not global minimum occurs at interval −1<x<2, and a local, but not global, maximum occurs at x=2 applies to the function or not.

The graph inferred from part(a) is as follows.

By observing the graph it is deduced that the given statement is invalid.

The statement the graph is a local, but not global maximum occurs at interval −1<x<2, and a local, but not global, minimum occurs at x=2 does not apply to the function.

Page 262 Problem 34 Answer

It is given a function f(x)=(x−1)²(x+2).

It is required to find whether the statement that the graph is a local, but not global maximum occurs at interval −1<x<2, and a local, but not global, minimum occurs at x=2

applies to the function or not.

The graph inferred from part(a) is as follows.

By observing the graph it is deduced that the given statement is invalid.

The statement the graph is a local, but not global maximum occurs at interval−1<x<2, and a local, but not global, minimum occurs at x=2 does not apply to the function.

Page 266 Exercise 1, Answer

It is to require to identify the transformation of the graph f(x)=x³ to produce the graph of the function

g(x)=(−1/4(x+2))³+3.

The given cubic function is of the form h(x)=a(1/b(x−h))³+k.

For g(x), h=−2,k=3, b=−1/4, and a=1.

Thus, according to the rule of the cubic function h(x), first, do a vertical translation of 3 units up.

Then, a horizontal translation of 2 units left followed by a horizontal sketch by a factor of 4.

Then, finally reflect the curve obtained about x-axis.

The required transformation is vertical translation of 3 units up.

Then, a horizontal translation of 2 units left followed by a horizontal sketch by a factor of 4.

Finally, reflect the curve obtained about x-axis.

Page 266 Exercise 2, Answer

It is to require to identify the transformation of the graph f(x)=x³ to produce the graph of the function h(x)=1/3(x−4)³.

The given cubic function is of the form h(x)=a(1/b(x−h))³+k.

For g(x), h=4,k=0, b=1, and a=1/3.

Thus, according to the rule of the cubic function h(x), first, do a horizontal translation of 4 units right.

Then, a vertical compression by a factor of 3.

The required transformation is a horizontal translation of 4  units right followed by a vertical compression by a factor of 3.

Page 266 Exercise 3, Answer

It is given that s(x)=x(x+2)(x+1)².

It is required to determine the number of turning points, global maximums and minimums, and local maximums and minimums.

First, make the graph of the function using a graphing calculator. Then, observe the maximum and minimum, and all other required values.

Graph the function s(x)=x(x+2)(x+1)²using a graphing calculator.

The turning points lie at (−1.70,−0.25), (−1,0), (−0.29,−0.25)

From the graph observe that there are no global minima or global maxima because the concavity of the curve is changing frequently.

The local minima occur at (−1.70,−0.25), and (−0.29,−0.25).

The maxima occur at (−1,0).

According to the graph, there are three turning points, one local maximum, no global minimum.

The turning points are (−1.70,−0.25), (−1,0), (−0.29,−0.25).

The local minima is at (−1.70,−0.25), and (−0.29,−0.25), and the maxima occur at(−1,0).

Page 266 Exercise 4, Answer

It is given that h(x)=x²(x−3)(x+2)(x−2).

It is required to determine the number of turning points, global maximums and minimums, and local maximums and minimums.

First, make the graph of the function using a graphing calculator. Then, observe the maximum and minimum, and all other required values.

Graph the function h(x)=x²(x−3)(x+2)(x−2) using a graphing calculator.

The turning points lie at (0,0),(2.53,−7.24),(1.24,6.66), and (−1.47,17.76).

From the graph observe that there are no global minima or global maxima because the concavity of the curve is changing frequently.

The local minima occur at (0,0), and (2.53,−7.24).

The maxima occur at (1.24,6.66).

According to the graph, there are three turning points, one local maximum, no global minimum.

The turning points are (0,0),(2.53,−7.24),(1.24,6.66), and .

The local minima is at (2.53,−7.24), and (0,0), and the maxima occur at (1.24,6.66).

Page 266 Exercise 5, Answer

It is required to write a real-world situation that could be modeled by the equation V(ω)=ω(5ω)(3ω).

Consider a cuboidal tank of length ω,width 5ω units and height 3ω units.

The volume of a cuboidal tank is the product of its length, breadth, and height.

Let V be volume of cuboid then volume of cuboid isV(ω)=ω(5ω)(3ω).

The equation V(ω)=ω(5ω)(3ω) can be modeled as the volume of a cuboidal tank with side lengths, ω, 5ω, and 3ω.

Page 266 Exercise 6, Answer

It is given function that g(x)=−1/4(x+4)³.

It is to require to identify changes to y when the graph of f(x)=x³ is transformed to get the graph of g(x).

Identify the transformation that is done to the y coordinates and apply them to the coordinates.

On y-axis, compress f(x)= x³ by factor 0.25 along the vertical axis.

The y-coordinates of the reference points are −1, 0, and 1.

Therefore, changes in y are−0.25(−1)=0.25,−0.25(0)=0,and−0.25(1)=−0.25.

The changes toy are −1 becomes 0.25, 0 becomes 0, and 1 becomes 0.25.

The values of y after the changes is 0.25, 0, and 0.25.

Page 267 Exercise 7, Answer

It is given function f(x)=x³and g(x)=−1/4(x+4)³.

It is required to identify the transformation using the changes to x, and y and complete the table.

To do so, use the values of x obtained in question 1 and 2. Substitute it in the function f(x) and g(x) and fill the tables with the required values.

To get the graph of g(x)=−1/4(x+4)³, the graph of f(x)=x³compressed vertically and translated to left.

On substituting values obtained from question 1 and 2.

The complete table as follows:

Page 267 Exercise 8, Answer

It is given that g(x)=x²(x−3).

It is required to graph the function using a graphing calculator and state the number of turning points, and the x- intercepts.

Use a graphing calculator, plot the function. Observe the turning points.

Given, g(x)=x²(x−3).The graph is as follows.

Observe the graph, it has turning points are (0,0) and (2,−4).

The x intercepts are at x=0, and x=3.

The function is at peak at the point  (0,0) and at its lowest peak at (2,−4).

Therefore, (0,0) is local maxima and (2,−4) is the local minima.

The graph of the function is given below.

Number of turning points: 2 Turning points: (0,0) and (2,−4).

x-intercept: x=0,3

Local maxima: (0,0)

Local Minimum:(2,−4)

Page 267 Exercise 9, Answer

It is given thath(x)=(x−4)(x−3)(x+2)².

It is required to graph the function using a graphing calculator and state the number of turning points, and the x- intercepts.

Use a graphing calculator, plot the function. Observe the turning points.

Given,h(x)=(x−4)(x−3)(x+2)².

The graph is a Observe the graph, it has three turning points, that are, (−2,0), (0.705,55.332), and (3.545,−7.624).

The x-intercepts are at x=−2, and x=3, and x=4.

The function is at peak at the point  (0.705,55.332) and at its lowest peak at two points, (−2,0) and (3.545,−7.624).

Therefore, (0.705,55.332) is local maxima and (−2,0), and (3.545,−7.624) is the local minima.

The graph of the function is given below.

Number of turning points: 3

Turning points: (−2,0),(0.705,55.332), and (3.545,−7.624).

x-intercept: x=−2, 4, 3

Local maxima: (0.705,55.332)

Local Minimum: (−2,0) and (3.545,−7.624)

Page 267 Exercise 10, Answer

It is required to give an example of a cubic function.The volume of a cube or cylinder is an example of a cubic function from the real world

Assume a cube with length, x, width x, and height x.

Then, its volume will be the product of all the side lengths.

Therefore, the volume will be x³, which is a cubic function.

Hence, the volume of a cube can be treated as an example of a cubic function from real world.

The volume of a sphere as a function of the radius of the sphere is a cubic function.

Similarly, the volume of a cube as a function of the length of one of its sides is a cubic function.

Page 268 Exercise 11, Answer

Three functions are given.

It is required to tell whether the vertex of the graph translated to the right and up when compared to f(x)=x³.

Check the transformations of y=(x+3)³+2.

Use a graphing calculator, and graph y=(x+3)³+2.

Observe that the vertex is translated left and up.

Check the transformations of y=5x³+7.

Use a graphing calculator, and graph y=5x³+7.

Observe that the vertex is translated up only.

Check the transformations of y=(x−4)³+2.

Use a graphing calculator, and graph y=(x−4)³+2.

Observe that the vertex is translated four units to the right and 2 units up.

The following are the answers for the given polynomials:

y=(x+3)³+2:No y=5x³+7: No y=(x−4)³+2 : Yes

Page 268 Exercise 12, Answer

It is given that h(x)=x(x−1)(x+3)³.

It is required to find whether the given three sentences are true or false.

Make the graph of the given function using the graphing calculator. Now, observe the curve and answer the sentences.

Using a graphing calculator, plot the function.

The graph has three turning points. So, the statement the graph has four turning points is false.

Observe the graph, it crosses the x-axis at −3, 0, and 1.

The statement that the graph crosses the x-axis at −3, −1, 0, and 1 is not true.

A maximum or minimum is said to be global if it is the largest or smallest value of the function, respectively, on the entire domain of a function.

The domain of the given function is all real numbers.

From the graph, observe that it does not have any specific smallest value or any specific largest value with respect to the domain.

Therefore, there are no global maxima and minima.

The statement A and B are false and statement C is true.

Page 268 Exercise 13, Answer

It is given a graph of a polynomial that cuts the x-axis at 4 points.

It is required to find a quadratic function in intercept form for the given graph whose x-intercepts are integers assuming that the constant factor a is either 1 or −1.

First, identify the x intercept that means find out the points where the curve cut thexaxis, make them in factor form.

Then, write the polynomial as product of all the factors and assume the constant as 1 to get the function.

Analyse the given graph, where it cuts the xaxis. Therefore, it cuts the axis at x=−2,x=0,x=3, andx=5.

Let P(x) be the polynomial.

As x=−2 is a zero so, (x+2) is a factor.

As x=0 is a zero so, x is a factor.

As x=3 is a zero so, (x−3) is a factor.

As x=5 is a zero so, (x−4) is a factor.

The polynomial is factored form will be P(x)=ax(x+2)(x−3)(x−5).

Check the end behaviour of the polynomial.

The end behaviour of the polynomial is x→∞ then, P(x)→∞  and when x→−∞ then, P(x)→∞.

Therefore, a=1. Thus, P(x)=x(x+2)(x−3)(x−5) is the required function.

The quartic function in intercept form for the given graph isP(x)=x(x+2)(x−3)(x−5).