## Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions

**Page 126 Essential Question Answer**

We have to take some steps to describe the function that is represented by more than one equation

Identify the intervals for which different rules apply.

Determine formulas that describe how to calculate an output from input in each interval.

Use braces and if-statements to write the function.

In this way, we can describe the function that is represented by more than one equation.

We have given an explanation of, How we can describe the function that is represented by more than one equation.

**Page 127 Exercise 2 Answer**

** Given:** The given graph is

As we can see in the above graph

For the value of “x” is less than 0and greater than or equal to −3, the value of “y” is 0

For the range of value of x, −6≤x<−3, the value of “y” is −2

For the range of value of x , 0≤x<3, the value of “y” is 2

For the range of value of x, 3≤x<6, the value of “y” is 4.

Yes, the given graph represents y as a function of x.

**Given:** The given graph is

To find We have to find the value of the function for the given intervals.

\(f(x)=\left\{\begin{array}{c}, \text { if }-6 \leq x<3 \\, \text { if }-3 \leq x<0 \\, \text { if } 0 \leq x<3 \\, \text { if } 3 \leq x<6\end{array}\right.\)

Use the relation that we have got in from the above of this question.

From the above Equation, we can write the values of the function for the given interval

\(f(x)=\left\{\begin{aligned}-2, \text { if }-6 & \leq x<-3 \\0, \text { if }-3 & \leq x<0 \\2, \text { if } 0 & \leq x<3 \\4, \text { if } 3 & \leq x<6\end{aligned}\right.\)

**Page 127 Exercise 3 Answer**

** Given:** The given graph is

As we can see in the above graph, shows the modulus function.

A modulus function is a function that gives the absolute value of a number or variable.

It produces the magnitude of the number of variables.

It is also termed as an absolute value function.

So we can describe a function that is represented by more than one equation as

f(x)=∣x∣

f(x)=x ,if x is positive ——–(1)

f(x)=−x, if xis negative ——–(2)

**Page 127 Exercise 4 Answer**

**Given:** The given graph is

As we can see in the above graph, shows the modulus function.

we can use two equations to describe the function represented by the graph, as\(f(x)=\left\{\begin{array}{r}x, \text { if } x>0 \\-x, \text { if } x<0\end{array}\right.\)

We can use two equations to describe the function represented by the graph, as

\(f(x)=\left\{\begin{array}{r}x, \text { if } x>0 \\-x, \text { if } x<0\end{array}\right.\)

**Page 129 Exercise 2 Answer**

**Given:** The given function is

\(f(x)= \begin{cases}3 x-1, & \text { if } x \leq 1 \\ 1-2 x, & \text { if } x>1\end{cases}\) and

\(g(x)=\left\{\begin{array}{lc}3 x-1, & \text { if } x \leq-3 \\2, & \text { if }-3<x<1 \\-3 x, & \text { if } x \geq 1\end{array}\right.\)

To find We have to find the value of f(1).

Find the appropriate function for the required value of the function.

After finding the appropriate function, put the value of the domain of the required function in that function.

The domain of the required function is x=1.

We have to find the value of the function f(1), so the appropriate function is f(x)=3x−1,x≤1 because the value of “x” is one.

Now put x=1 in the above function, and we get

f(x)=3x−1

f(1)=3(1)−1

f(1)=3−1

f(1)=2

The value of the function is f(1)=2.

**Page 129 Exercise 3 Answer**

**Given:** The given function is

\(f(x)= \begin{cases}3 x-1, & \text { if } x \leq 1 \\ 1-2 x, & \text { if } x>1\end{cases}\) and

\(g(x)=\left\{\begin{array}{lc}3 x-1, & \text { if } x \leq-3 \\2, & \text { if }-3<x<1 \\-3 x, & \text { if } x \geq 1\end{array}\right.\)

To find We have to find the value of the function f(5).

Find the suitable function according to the domain of the required function.Then put the value of the domain in that function to get the answer.

The domain of the required function is x=5.

So the appropriate function is f(x)=1−2x,x>1 because the domain satisfies the condition of the equation.

We have to find the value of the function f(5).

Now put x=5 in the above function, we get

f(x)=1−2x

f(5)=1−2(5)

f(5)=1−10

f(5)=−9

The value of the function is f(5)=−9.

**Page 129 Exercise 4 Answer**

**Given:** The given functions are

\(f(x)= \begin{cases}3 x-1, & \text { if } x \leq 1 \\ 1-2 x, & \text { if } x>1\end{cases}\) and

\(g(x)=\left\{\begin{array}{lc}3 x-1, & \text { if } x \leq-3 \\2, & \text { if }-3<x<1 \\-3 x, & \text { if } x \geq 1\end{array}\right.\)

To findWe have to find the value of the function f(−4).

Find the suitable function according to the domain value of the required function.

Then put the value of the domain in that function to get the answer.

The domain of the required function is x=−4.

So the appropriate function is f(x)=3x−1,ifx≤1, because the domain of the required function satisfies this function.

We have to find the value of the function f(−4).

Now put x=−4in the above function, and we get

f(x)=3x−1

f(−4)=3(−4)−1

f(−4)=−12−1

f(−4)=−13

The value of the function is f(−4)=−13.

**Page 129 Exercise 5 Answer**

**Given:** The given functions are

\(f(x)= \begin{cases}3 x-1, & \text { if } x \leq 1 \\ 1-2 x, & \text { if } x>1\end{cases}\)nd

\(g(x)=\left\{\begin{array}{lc}3 x-1, & \text { if } x \leq-3 \\2, & \text { if }-3<x<1 \\-3 x, & \text { if } x \geq 1\end{array}\right.\)

To findWe have to find the value of the function g(0).

Find the suitable function according to the domain value of the required function.

Then put the value of the domain in that function to get the answer.

The domain of the required function is x=0.

So the appropriate function is g(x)=2,if−3<x<1, because the domain of the required function satisfies this function.

Now we have to find the value of g(0).

The above function is a constant function, so we can say that for the given value, x=0, the value of the function remains the same. g(0)=2

The value of the function is g(0)=2.

**Page 129 Exercise 6 Answer**

**Given:** The given functions are

\(f(x)= \begin{cases}3 x-1, & \text { if } x \leq 1 \\ 1-2 x, & \text { if } x>1\end{cases}\) and

\(g(x)=\left\{\begin{array}{lc}3 x-1, & \text { if } x \leq-3 \\2, & \text { if }-3<x<1 \\-3 x, & \text { if } x \geq 1\end{array}\right.\)

To find We have to find the value of the function g(−3).

Find the suitable function according to the domain value of the required function.

Then put the value of the domain in that function to get the answer.

The domain of the required function is x=−3.

So the appropriate function is g(x)=3x−1,ifx≤−3 because the domain of the required function satisfies this function.

Now we have to find the value of g(−3)

Now put x=−3 in the above function, and we get

g(x)=3x−1

g(−3)=3(−3)−1

g(−3)=−9−1

g(−3)=−10

The value of the function is g(−3)=−10.

**Page 129 Exercise 7 Answer**

**Given:** The given functions are

\(g(x)=\left\{\begin{array}{lc}3 x-1, & \text { if } x \leq-3 \\2, & \text { if }-3<x<1 \\-3 x, & \text { if } x \geq 1\end{array}\right.\)

To find We have to find the value of the function g(1).

Find the suitable function according to the domain value of the required function.

Then put the value of the domain in that function to get the answer.

The domain of the required function is x=1.

So the appropriate function is g(x)=−3x because the domain of the required function satisfies this function.

Now put x=1in the above function, and we get

g(x)=−3x

g(1)=−3(1)

g(1)=−3

The value of the function is g(1)=−3.

**Page 129 Exercise 9 Answer**

**Given:** The given functions are

\(f(x)= \begin{cases}3 x-1, & \text { if } x \leq 1 \\ 1-2 x, & \text { if } x>1\end{cases}\)

\(g(x)=\left\{\begin{array}{lc}3 x-1, & \text { if } x \leq-3 \\2, & \text { if }-3<x<1 \\-3 x, & \text { if } x \geq 1\end{array}\right.\)

To findWe have to find the value of the function g(−5).

Find the suitable function according to the domain value of the required function.

Then put the value of the domain in that function to get the answer.

The domain of the required function is x=−5.

So the appropriate function is g(x)=3x−1 if x≤−3 because the domain of the required function satisfies this function.

Now put x=−5 in the above function, and we get

g(x) = 3x−1

g(−5) = 3(−5)−1

g(−5) = −15−1

g(−5) = −16

The value of the function is g(−5)=−16.

**Page 129 Exercise 11 Answer**

**Given:** A function \(y=\left\{\begin{array}{l}4-x, \text { ifx }<2 \\x+3, \text { ifx } \geq 2\end{array}\right.\)

To graph the function and describe the domain and range.

Make a table of values and plot the points and sketch the graph from the graph, we can identify the domain which is the set of input values shown on the x-axis, and the range which is the set of output values shown on the y-axis.

The given piecewise function is \(y=\left\{\begin{array}{l}4-x, \text { ifx }<2 \\x+3, \text { ifx } \geq 2\end{array}\right.\)

For x<2, make a table of values usingy=4−x

Sincex<2, the point(2,2) is not considered and it is represented by an open dot.

For x≥2, make a table of values using y=x+3

Since x≥2, the point(2,5) is considered and it is represented by a closed dot.

Plot the points and sketch the graphs on the same coordinates axis.

From the graph, we see that the domain is the set of all real numbers, and the range is y>2

The graph of the function y={4−x,ifx<2x+3,ifx≥2 is given below and the domain and range is described as a set of all real numbers andy≥0 respectively.

**Page 130 Exercise 12 Answer**

**Given:** A function\(y=\left\{\begin{array}{r}2 x, \imath f x<-2 \\2, i f-2 \leq x<2 \\-2 x, \text { ifx }>2\end{array}\right.\)

To graph the function and describe the domain and range.

Make a table of values and plot the points and sketch the graph from the graph, we can identify the domain which is the set of input values shown on the x-axis, and the range which is the set of output values shown on the y-axis.

The given piecewise function is \(y=\left\{\begin{array}{r}2 x, \imath f x<-2 \\2, \text { if }-2 \leq x<2 \\-2 x, \text { ifx } \geq 2\end{array}\right.\)

For x<−2, make a table of values using y=2x

Since x<−2, the point(−2,−4) is not considered and it is represented by an open dot.

For x≥2, make a table of values using y=−2x

Sincex≥2, the point(2,−4) is considered and it is represented by a closed dot.

The line y=2 is a horizontal line drawn between −2≤x<2

Plot the points and sketch the graphs on the same coordinates axis.

From the graph, we see that the domain is the set of all real numbers, and the range is y=2,y≤−4

The graph of the function \(y=\left\{\begin{array}{r}2 x, \text { if } x<-2 \\2, \text { if }-2 \leq x<2 \\-2 x, \text { ifx } \geq 2\end{array}\right.\) is given below and the domain and range are described as a set of all real numbers and y=2,y≤−4

respectively.

**Page 130 Exercise 13 Answer**

**Given:** A function y\(=\left\{\begin{array}{r}

-1, \imath f x \leq-1 \\

0, i f-1<x<2 \\

1, \text { ifx } \geq 2

\end{array}\right.\)

To graph the function and describe the domain and range.

Make a table of values and plot the points and sketch the graph from the graph, we can identify the domain which is the set of input values shown on the x-axis, and the range which is the set of output values shown on the y-axis.

The given piecewise function \(y=\left\{\begin{array}{r}

-1, \text { if } x \leq-1 \\

0, \text { if }-1<x<2 \\

1, \text { ifx } \geq 2

\end{array}\right.\)

The graph of y=−1 is a horizontal line.

The y-coordinate of all of the points on this line are−1

Since x≤−1, the point(−1,−1) is considered and it is represented by a closed dot.

The graph of y=0 is a horizontal line.

The y-coordinate of all of the points on this line are− 0

Since−1<x<2, the points(−1,0),(2,0) are not considered and it is represented by an open dot.

The graph of y=1 is a horizontal line.

The y-coordinate of all of the points on this line are −1

Sincex≥2, the point(2,1)is considered and it is represented by a closed dot.

Plot the points and sketch the graphs on the same coordinates axis.

From the graph, we see that the domain is the set of all real numbers and the range is y=−1,0,1

The graph of the function \(y=\left\{\begin{array}{r}

-1, \imath f x \leq-1 \\

0, \text { if }-1<x<2 \\

1, \text { ifx } \geq 2

\end{array}\right.\) is given below and the domain and range are described as a set of all real numbers andy=−1,0,1

respectively.

**Page 130 Exercise 14 Answer
**

**Given:**a graph

To write a piecewise function for the graph.

From the graph calculate the change in y coordinate with respect to the change in x coordinate of that line and using the general equation of a line the piecewise function for the given graph is written.

The given graph is

The two lines of this graph have values x<0,x≥0

For x<0, the run is−5, and rise is 7 for the graph so the slope of this line is\(\frac{-7}{5}\) and the line does not intersect the y-axis.

The general equation of a line is y=mx+b where m=slope,b= y-intercept.

The equation of this line is y=\(\frac{-7}{5}\)x+0

For all values of x<0, the function is y=\(\frac{-7}{5}\)x

For all values of x≥0,

The line is a horizontal line.

All the points in this line have the same y-coordinate 3

The equation of this line is y=3

The piecewise function for this graph is \(y=\left\{\begin{array}{c}

y=\frac{-7}{5} x, \text { if } x<0 \\

y=3, \text { ify }=x \geq 0

\end{array}\right.\)

The piecewise function for this graph is \(y=\left\{\begin{array}{c}

y=\frac{-7}{5} x, \text { if } x<0 \\

y=3, \text { ify }=x \geq 0

\end{array}\right.\)

**Page 130 Exercise 15 Answer**

**Given:** A graph

To write a piecewise function for the graph.

From the graph calculate the change in y coordinate with respect to the change in x coordinate of that line and using the general equation of a line the piecewise function for the given graph is written.

The given graph is

The three lines of this graph have values −3≤x<0 and 0≤x≤1 and 1<x≤5

For 1<x≤5, the rise and run is the same so the slope is 1 and the line does not intersect the y-axis.

The general equation of a line is y = mx+b where m = slope,b = y-intercept.

The equation of this line is y=x

For all values of −3≤x<0, the line is a horizontal line.

All the points in this line have the same y-coordinate 3

The equation of this line is y=3

For all values of 0≤x≤1, the line is a horizontal line.

All the points in this line have the same y-coordinate 4

The equation of this line is y=4

The piecewise function for this graph is\(y=\left\{\begin{array}{r}

3, i f-3 \leq x<0 \\

4, i f 0 \leq x \leq 1 \\

x, 1<x \leq 5

\end{array}\right.\)

The piecewise function for this graph is \(y=\left\{\begin{array}{r}

3, i f-3 \leq x<0 \\

4, i f 0 \leq x \leq 1 \\

x, 1<x \leq 5

\end{array}\right.\)

**Page 130 Exercise 16 Answer**

Given a postal service charges $4 for shipping any package weighing up to but not including one pound and $1 for each additional pound or portion of a pound up to but not including five pounds.

To write and graph a step function that shows the relationship between the number of pounds a package weighs and the total cost for postage.

Using a table we organize the given information and write a step function based on this and then graph the function that shows the relationship between the number of pounds a package weighs and the total cost for postage

It is given that a postal service charges $4 for shipping any package weighing up to but not including one pound.

It costs an additional $1 for each additional pound or portion of a pound up to but not including five pounds.

If the package weighs one or more than one pound then $1 is added for each additional pound.

The given information is organized in a table.

The above information can be written as a step function that shows the relationship between the number x of pounds a package weighs and the total cost y for postage as

\(y=\left\{\begin{array}{l}4, i f 0<x<1 \\

5, \text { if } 1 \leq x<2 \\

6, \text { if } 2 \leq x<3 \\

7, \text { if } 3 \leq x<4 \\

8, i f 4 \leq x<5

\end{array}\right.\)

The graph of this step function that shows the relationship between the number x of pounds a package weighs and the total cost y for postage is drawn.

A postal service charges $4 for shipping any package weighing up to but not including one pound and $1 for each additional pound or portion of a pound up to but not including five pounds then the step function is

\(y=\left\{\begin{array}{l}

4, \text { if } 0<x<1 \\

5, \text { if } 1 \leq x<2 \\

6, \text { if } 2 \leq x<3 \\

7, \text { if } 3 \leq x<4 \\

8, \text { if } 4 \leq x<5

\end{array}\right.\)

And the graph is a step function that shows the relationship between the number x of pounds a package weighs and the total cost y for postage is