Exercise 14 Page 165

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

An x-intercept of 2 means that the graph passes through the x-axis at the point ( 2,0).

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.

8x-4y=16

Substitute

x= 0

8* 0-4y=16

ZeroProdProp

Use the Zero Product Property

- 4y=16

DivEqn

.LHS /(- 4).=.RHS /(- 4).

- 4y/- 4=16/- 4

CalcQuot

Calculate quotient

y=- 4

A y-intercept of -4 means that the graph passes through the y-axis at the point (0, -4).

Graphing the Equation

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