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