Algebra Know-It-ALL

532 Review Questions and Answers

Question 28-4

When we compare the systems stated in Questions 28-2 and 28-3 and graphed in Figs. 30-3

and 30-4, we can see that the curves have the same shapes in both situations. But in the

second case, the upward-opening parabola has been moved vertically down by 1 unit, while

the downward-opening parabola has been moved vertically up by 1 unit. This has caused the

single intersection point (Fig. 30-3) to “break in two” (Fig. 30-4). What will happen if we

move the upward-opening parabola, shown by the solid curve, further down, and move the

downward-opening parabola, shown by the dashed curve, further up by the same distance?

How will the equations in the system change if we do this?

Answer 28-4

If we move the parabolas this way, the intersection points will move farther from each other.

The negative x-value of one real solution will become more negative, and the positive x-value

Table 30-2. Selected values for graphing the functions

y=x^2 − 1 and y=−x^2 + 1.

Bold entries indicate real solutions.

xx^2 − 1 −x^2 + 1

−2 3 − 3

−10 0

0 −1 1

10 0

2 3 − 3

x

y

(–1,0) (1,0)

Figure 30-4 Illustration for Answer 28-3. The first function

is graphed as a solid curve; the second function is

graphed as a dashed curve. Real-number solutions

appear as points where the curves intersect. On both

axes, each increment is 1 unit.