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

**Page 58 Exercise 2 Answer**

**Given:- **(2,2)

We have to plot the point on the graph.

We also have to describe the location of the point.

(2,2)on the graph is

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

**Location:**

The required point is two units right from the origin and two units above the x−axis.

The required point on the graph is

**Page 58 Exercise 4 Answer**

**Given:-** (5,2)

We have to plot the point on the graph.

We also have to describe the location of the point.

(5,2) on the graph is

**Location:**

The required point is five units right from the origin and two units above the x−axis.

The required point on the graph is

**Page 58 Exercise 5 Answer**

**Given:-** 5 units down and on the y−axis.

We have to plot the point on the graph.

We also have to describe the location of the point.

The graph for the point that is on the y-axis and 5 units down from the origin is

The required point on the graph is (0,−5)

**Page 58 Exercise 6 Answer**

**Given:-** 2x+1 , x=3

We have to find the value of the expression for the given value of x.

We just have to substitute the given value of the variable in the expression.

Substituting the value of x in the expression

2x+1

x=3

=2(3)+1

=6+1

=7

The value of 2x+1 at x=3 is 7.

**Page 58 Exercise 7 Answer**

**Given:- **16−4x , x=−4

We have to find the value of the expression for the given value of x.

We just have to substitute the given value of the variable in the expression.

Substituting the value of x in the expression

16−4x

x=−4

=16−4(−4)

=16+16

=32

The value of 16−4x at x=−4 is 32.

**Page 58 Exercise 9 Answer**

**Given:-** −9−3x ,x=5

We have to find the value of the expression for the given value of x.

We just have to substitute the given value of the variable in the expression.

Substituting the value of x in the expression

−9−3x

x=5

=−9−3(5)

=−9−15

=−24

The value of −9−3x at x=5 is −24.

**Page 58 Exercise 10 Answer**

**Given:-** Length =24−3x

We just have to substitute the given value of the variable in the expression.

The length of a side of a square =24−3x

Length at x=6

=24−3(6)

=24−18

=6ft

The length of the side of the square when x=6 is 6 ft.