Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
5. Standard Form
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Exercise 20 Page 326

You will need to substitute 0 and solve for a variable twice before graphing.

Practice makes perfect

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.
-2x+y=8
-2x+ 0=8
-2x=8
x=-4

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

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.
-2x+y=8
-2( 0)+y=8
0+y=8
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.