Algebra Know-It-ALL

10. We need more points for reference. We can find two more points on the graph by
    plugging in values of x smaller and larger than xmax. Let’s try x=−1. Then

y=− 2 x^2 + 2 x− 5

=− 2 × (−1)^2 + 2 × (−1)− 5

=− 2 − 2 − 5

=− 4 − 5

=− 9

Now we know that (−1,−9) is on the parabola. The x-value that we chose to find that

point,−1, happens to be 3/2 units smaller than xmax. Let’s choose a value of x that’s 3/2

units larger than xmax. That would be x= 2. Then

y=− 2 x^2 + 2 x− 5

=− 2 × 22 + 2 × 2 − 5

=− 8 + 4 − 5

=− 4 − 5

=− 9

This gives us (2, −9) as a third point on the graph. Now that we know three points on the

curve, we can draw an approximation of the graph. Figure C-3 illustrates this parabola.

On both axes, each increment represents 1 unit.

x

y

Each axis

increment

is 1 unit

(1/2,–9/2)

(–1,–9) (2,–9)

Figure C-3 Illustration for the solution to Prob. 10

in Chap. 24.

Chapter 24 685