Exercise 123 Page 49

We will graph this equation by finding and plotting its intercepts, then connecting them with a line. To find the x- and y-intercepts, we will need to substitute 0 for one variable, solve, then repeat for the other variable.

Finding the x-intercept

Think of the point where the graph of an equation crosses the x-axis. The y-value of that ( x, y) coordinate pair is equal to 0, and the x-value is the x-intercept. To find the x-intercept of the given equation, we should substitute 0 for y and solve for x.

y=-3/2x+8

Substitute

y= 0

0=-3/2x+8

AddEqn

LHS+3/2x=RHS+3/2x

3/2x=8

MultEqn

LHS * 2=RHS* 2

3x=16

DivEqn

.LHS /3.=.RHS /3.

x=16/3

An x-intercept of 163 means that the graph passes through the x-axis at the point ( 163,0).

Finding the y-intercept

Let's use the same concept to find the y-intercept. Consider the point where the graph of the equation crosses the y-axis. The x-value of the ( x, y) coordinate pair at the y-intercept is 0. Therefore, substituting 0 for x will give us the y-intercept.

y=-3/2x+8

Substitute

x= 0

y=-3/2( 0)+8

ZeroPropMult

Zero Property of Multiplication

y=8

A y-intercept of 8 means that the graph passes through the y-axis at the point (0, 8).

Graphing the equation

We can now graph the equation by plotting the intercepts and connecting them with a line.

Again to graph this equation we will start with finding and plotting its intercepts. After that we will connect them with a line. To find the x- and y-intercepts, we will need to substitute 0 for one variable, solve, then repeat for the other variable.