Assignment 1: Quadratics and Tangents of Products
by
Tom Cooper

Problem: Find two quadratic functions f(x) and g(x) such that h(x) = f(x)g(x) is tangent to f and g each at two different points.
Let's take advantage of the power of a graphing device to explore different quadratics. Let's begin with f(x) = x2 and g(x) = -x2

That does not solve the problem, so we can try a vertical shift of one of these functions. Let's try g(x) = x2 + 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) = x2-1.

That is not any better. Experiment with the vertical shifts, and you should discover f(x) = x2-1 and g(x) = x2+2.

It looks like success!  Let's Prove it.

f '(x) = 2x, g'(x) = -2x, and h'(x) = (fg)'(x) =2x(-x2+2)-2x(x2-1) = -4x3+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) = x2 - a2 , g(x) = x2 - b2, and h(x) = -x4 + (a2+b2)x2 + a2b2,

f '(x) = 2x, g'(x) = -2x, and h'(x) = -4x3 + 2xa2 +2b2x.

A little algebra with f '(a) = h'(a) gives, ab2-a3-a=0 and, ba2-b3-b=0 from g'(b)=h'(b).

If we assume that a and b are nonzero, we get the hyperbolas b2 - a2 = 1 and a2 - b2 = 1.

Here is a neat animation with graphing calculator 3.2.