Algebra Know-It-ALL

Examine the discriminant d for this equation:

d= 22 − 4 × 1 × (−15)

= 4 − (−60)

= 4 + 60

= 64

How can we change the stand-alone constant, which is −15 right now, to bring the discriminant down

to 0? We can find out by substituting 0 for d, and by substituting a letter constant for −15 in the above

equation. If use k as the letter constant, we get

0 = 22 − 4 × 1 ×k

which simplifies to

0 = 4 − 4 k

and further to

4 k= 4

This resolves to k= 1. In the quadratic we got by mixing, let’s change the stand-alone constant k from − 15

to 1. That gives us

x^2 + 2 x+ 1 = 0

If we increase the constant any more, then d becomes negative, and the system has no real solutions.

Now we know that if we increase the stand-alone constant by 16 in the quadratic represented by the

solid parabola in Fig. 28-2, the two solution points will merge in the graph. So let’s increase that constant

by 17! When we do that, the function becomes

y=− 2 x^2 − 3 x+ 13

The graph of this function is a parabola with the same contour as the original one, but displaced by 17 units

upward in the coordinate plane. For extra credit, try solving the system

y=− 2 x^2 − 3 x+ 13

and

y=− 3 x^2 − 5 x+ 11

and see for yourself that it has no real solutions.

Here’s another challenge!

By examining Fig. 28-2, describe how the quadratic function

=− 3 x^2 − 5 x+ 11

470 More Two-by-Two Graphs