Assignment 1: Quadratics and Tangents of Products

by

Tom Cooper

Let's take advantage of the power of a graphing device to explore different quadratics. Let's begin with f(x) = x

That does not solve the problem, so we can try a
vertical shift of
one of these functions. Let's try g(x) = x^{2}
+ 1.

^{2} - a^{2} = 1
and a^{2}
- b^{2} = 1.

It looks promising
as we now
appear to have g(x) and h(x) tangent at two points. Perhaps a
shift in
f(x) will create two tangents there. Let's
try f(x) = x^{2}-1.

That
is not any better. Experiment with the vertical shifts, and you should discover f(x) = x^{2}-1 and g(x)
= x^{2}+2.

It looks like success! Let's Prove it.

f '(x) = 2x, g'(x) = -2x, and h'(x) = (fg)'(x)
=2x(-x^{2}+2)-2x(x^{2}-1) = -4x^{3}+6x. We'll need these to find the slopes of the
tangent lines.

Here is a
picture of f(x), g(x), h(x), and the tangent lines.

Inspired by this success and some intuition gained from the graphs, lets see if we can algebraically find a family of solutions.

If we let f(x) = (x-a)(x+a) and g(x) = -(x-b)(x+b), then h(x) = -(x-a)(x+a)(x-b)(x+b), and we are guaranteed that f and g will always intersect h at (a,0), (-a,0) and (b,0), (-b,0) respectively. Thus we need only find a and b such that f '(a) = h'(a) and g'(b) = h'(b). This is actually easier than it seems.

Since f(x) = x^{2 }- a^{2}
, g(x) = x^{2} - b^{2}, and h(x) = -x^{4}
+ (a^{2}+b^{2})x^{2}
+ a^{2}b^{2},

f '(x) = 2x, g'(x) = -2x, and h'(x) = -4x^{3}
+ 2xa^{2}
+2b^{2}x.

A little algebra with f '(a) = h'(a)
gives, ab^{2}-a^{3}-a=0
and, ba^{2}-b^{3}-b=0 from
g'(b)=h'(b).

Here is a neat
animation with graphing calculator 3.2.