Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 3 Graphing Linear Functions Exercise 3.3

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 3 Graphing Linear Functions

Page 6  Essential Question 9  Answer

To explain how do we use function notation to represent a function.

Using the method of function.

The notation y=f(x) defines a function named f. This is read as “y is a function of x.” The letter x represents the input value or independent variable.

The letter y, or f(x), represents the output value, or dependent variable.

The function notation represents a function that is y=f(x).

Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions

Page 69  Exercise 1  Answer

Given: Function is f(x)=2x−3 To graph the function in the graph

The given expression is f(x) = 2x-3 Now to graph the given expression

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 1

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 3 Graphing Linear Functions Exercise 3.3


The graph that represents the expression f(x)=2x−3

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 5

 

The given expression is g(x)=x​+2 Now to graph the given expression

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 10

 

The solution for the given expression g(x)=x​+2 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 10

 So the question has no solution.

 

The given expression is h(x)=x2−1 Now to graph the equation

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 11

 

 

The solution for the given expression h(x)=x2−1 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 15

 

The given expression is j(x)=2x2−3 Now to graph the given expression

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 16

 

The solution for the given expression j(x)=2x2−3 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 20

 

Page 70  Exercise 2  Answer

Given: Function is f(x)=x+3
To graph the given function.
Using the method of function.

The given function is f(x)=x+3
Now substitute x=−1, we get,
f(−1)=−1+3
f(−1)=2
Then the point is(−1,2)

Now to draw a graph with the given function and mark the point(−1,2)

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 21

 

The solution for the given function f(x)=x+3 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 22

 

Given: Function is f(x)=x+3
To graph the given function.
Using the method of function.

The given function is f(x)=x+3
Now substitute x=0,
f(0)=0+3
f(0)=3
Then the point is(0,3)

To graph the function with the point (0,3)

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 23

 

The solution for the given function f(x)=x+3 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 24

 

Given: Function is
f(x)=x+3
To graph the given function.
Using the method of function.

The given function is f(x)=x+3
Now substitute x=1
f(1)=1+3
f(1)=4
Then the point is(1,4)

To draw a graph for the given function and with the point

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 25

 

The solution for the given function f(x)=x+3 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 26

 

Given: Function is f(x)=x+3
To graph the given function.
Using the method of function.

The given function is f(x)=x+3
Now substitute x=2 in the given function,
f(2)=2+3
f(2)=5
Then the point is(2,5)

 

To draw a graph for the given function and the point is(2,5)

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 27

 

The solution for the given function f(x)=x+3 is

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 28

 

Page 72  Exercise 1   Answer

Given: function is f(x)=x+4

To find its range.
Using the method of function.

The given function is f(x)=x+4
substitute x=4
f(4)=4+4
f(4)=8

To find the function of
f(0)=0+4
f(0)=4

substitute x=2
f(2)=2+4
f(2)=6

The solution for the function f(x)=x+4 is
f(4)=8
f(0)=4
f(2)=6

 

Page 72  Exercise 2  Answer

Given: A function of “x” is given to us.

To find  We have to find the function at x=−4,0,2
We will put the values of “x” in the given function and get the answer.

The given function is g(x)=5x
Putting the values of “x” in the given function, we get:
​x=−4
⇒ g(−4)=5×−4
⇒ g(−4)=−20
x=0
⇒ g((0)=0
x=2
⇒ g(2)=5×2
⇒ g(2)=10

The values of the function at x=−4,0,2 are−20,0,10

 

Page 72  Exercise 3  Answer

Given: A function of “x” is given to us.

To find We have to find the function at =−4,0,2
We will put the values of “x” in the given function and get the answer.

The given function is h(x)=7−2x
Putting the values of “x” in the given function, we get

​x=−4
⇒ h(−4)=7−2×−4
⇒ h(−4)=15

x=0
⇒ h(x)=7

x=2
⇒ h(2)=7−2×2
⇒ h(2)=3

The values of the function at x=−4,0,2 are 15,7,3

 

Page 72  Exercise 4  Answer

Given: A function of “x” is given to us.

To find We have to find the function at x=−4,0,2
We will put the values of “x” in the given function and get the answer.

The given function is s(x)=12−0.25x
Putting the values of “x” in the given function, we get:
​x=−4
⇒ s(−4)=12−0.25×−4
⇒ s(−4)=13
s=0
⇒ s(0)=12
s=2
⇒ s(2)=12−0.25×2
⇒ s(2)=11.5

The values of the function at x=−4,0,2 are 13,12,11.5

 

Page 72  Exercise  5 Answer

Given: A function of “x” is given to us.

To find We have to find the function at =−4,0,2
We will put the values of “x” in the given function and get the answer.

The given function is ​t(x)=6+3x−2
t(x)=4+3x
​Putting the values of “x” in the given function, we get
​x=−4​
⇒ t(x)=4+3×−4
⇒ t(x)=−8
x=0
⇒ t(x)=4
x=2
⇒ t(x)=4+3×2
⇒ t(x)=10

The values of the function at x=−4,0,2 are −8,4,10

 

Page 72  Exercise 6  Answer

Given: A function of “x” is given to us.

To find  We have to find the function at x=−4,0,2
We will put the values of “x” in the given function and get the answer.

The given function is ​u(x)=−2−2x+7​
⇒u(x)=−2x+5
​Putting the values of “x” in the given function, we get
​x=−4
⇒ u(x)=−2×−4+5
⇒ u(x)=13
x=0
⇒ u(x)=5
x=2
⇒ u(x)=−2×2+5
⇒ u(x)=1

The values of the function at x=−4,0,2 are 13,5,1

 

Page 72  Exercise 8  Answer

Given: A function of “x” is given to us. The value of the function is given at a certain “x”.

To find We have to find the value of x so that the function has the given value.

We will put the given value on the left-hand side of the given function and then solve the equation for “x”.

The given function is b(x)=−3x+1
Putting the given value b(x)=−20 in the above function and simplifying the equation for the value of “x”, we get
​−20=−3x+1
−21=−3x
x=7

The value of x so that the function has the given value is 7.

 

Page 72  Exercise 9   Answer

Given: A function of “x” is given to us. The value of the function is given at a certain “x”.

To find  We have to find the value of x so that the function has the given value.

We will put the given value on the left-hand side of the given function and then solve the equation for “x”.

The given function is r(x)=4x−3
Putting the given value r(x)=33 in the above function and simplifying the equation for the value of “x”, we get
​33=4x−3
36=4x
x=9

The value of x so that the function has the given value is 9.

 

Page 72  Exercise 10  Answer

Given: A function of “x” is given to us. The value of the function is given at a certain “x”.

To find We have to find the value of x so that the function has the given value.

We will put the given value on the left-hand side of the given function and then solve the equation for “x”.

The given function is \(m(x)=\frac{-3}{5} x-4\)
Putting the given value m(x)=2 in the above function and simplifying the equation for the value of “x”, we get

​2\(=\frac{-3}{5} x-4\)

6\(=\frac{-3}{5} x\)

x=\(=\frac{-30}{3}\)

x=−10

The value of x so that the function has the given value is−10.

 

Page 72  Exercise 11  Answer

Given: A function of “x” is given to us. The value of the function is given at a certain “x”.

To find We have to find the value of x so that the function has the given value.

We will put the given value on the left-hand side of the given function and then solve the equation for “x”.

The given function is \(=\frac{5}{6} x-3\)
Putting the given valuew(x)=−18 in the above function and simplifying the equation for the value of “x”, we get

​−18\(=\frac{5}{6} x-3\)

−15\(=\frac{5}{6} x\)

x=\(-15 \times \frac{6}{5}\)

x=−3×6

x=−18

The value of x so that the function has the given value is−18.

 

Page 73  Exercise 13  Answer

Given: A linear function of “x” is given to us.

To find We have to complete the table and plot the points obtained on a graph.

We will put the values of “x” in the given function and get values of t(x)
then we will complete the table and plot the graph.

The given function is t(x)=1−2x.
Putting x=−4,−2,0, we get

​x=−4
t(−4)=1−2×−4
t(−4)=9

x=−2
t(−2)=1−2×−2
t(−2)=5

x=0
t(0)=1−2×0
t(0)=1

Putting the values x=2,4, we get

​x=2
t(2)=1−2×2
t(2)=−3

x=4
t(4)=1−2×4
t(4)=−7

The table that shows the value of the function at different values of x is given below

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation table 1

 

The graph of the linear function is given below

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 29

 

The table that shows the value of the function at different values of x is given below

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation table 2

 

The graph of the linear function is given below

Big Ideas MathAlgebra 1Student Journal 1st Edition Chapter 3.3 Function Notation graph 30

 

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