(shown by the dashed curve) can be modified to produce a system with no real solutions, assuming that
the other quadratic function (shown by the solid curve) stays the same, and also assuming that the contour
of the graph for the modified function stays the same.
Solution
To cause the two solution points to merge, we must move the dashed parabola downward to the same
extent we moved the solid parabola upward in the previous example. That means we must reduce the
constant in the function for the dashed parabola by 16, giving us
y=− 3 x^2 − 5 x− 5
If we reduce the constant to anything smaller than −5, the real solutions of the two-by-two system vanish,
and the parabolas no longer intersect.
Enter the Cubic
The third time around in Chap. 27, we solved this two-by-two system:
x^3 + 6 x^2 + 14 x−y=− 7
and
− 6 x+ 2 y= 2
As with all the other examples, we let x be the independent variable, and then we morphed to
gety as a function of x in both cases. In this situation we got
y=x^3 + 6 x^2 + 14 x+ 7
and
y= 3 x+ 1
We found these solutions:
(x,y)= (−1,−2)
(x,y)= (−2,−5)
(x,y)= (−3,−8)
First, we tabulate some points
Table 28-3 shows several different values of x, along with the results of plugging those values into
the functions and calculating. All three solutions are included, and are written in bold. We also
include two x- values less than −3, and two larger than −1. Because the x-values for the solutions
are consecutive integers, it makes sense to choose consecutive integers on either side of them.
Enter the Cubic 471
