Let f(x) = ax+b and g(x) = cx+d where a, b, c, and d are real numbers with a and c not equal to zero. Then h(x) = (ax+b)*(cx+d). Clearly, h(x) is a polynomial of degree 2. Notice f intersects h at their common root: (-b/a, 0), and g intersects h at their common root: (-d/c, 0).

Consider: f(x) = -x+1/2, g(x) = x+1/2.

Then h(x) = (-x+1/2) (x+1/2) =

Intersection Points of f and h: (1/2, 0)
Intersection Points of g and h: (-1/2, 0)

Notice h'(x) = -2x and h'(1/2) = -1 = slope of f(x). h'(-1/2) = 1 = slope of g(x). Therefore we can see that f and g are tangent to h at the points (-1/2, 0) and (1/2, 0) respectively.

See graph: h(x) = (-x+1/2) (x+1/2) , f(x) = -x+1/2, g(x) = x+1/2.

Can we generalize to find all such functions f and g? We know that the slope of f(x) = ax+b is a, and that f is tangent to h(x) at the point (-b/a, 0). We also know that the derivitive of h(x),

gives us the slope of the tangent line at a point. Therefore, we can set the derivative of h(x) equal to the slope of f at their point of intersection. Using a similar argument for h and g gives us the system of equations:

(1)

(2) .

Reducing this system yields:

(3)

(4) .

This result gives us conditions on a and c.

(5) .

Substituting this result into equaion (3) yields:

(6)

which reduces to

(7) .

Now we have two very important conditions on the parameters: a and c must be additive inverses, while d and b must add up to be one.

Let's consider a = 1, c = -1, b = 1/4, and d = 3/4. Notice that these parameters fit the conditions.

See graph: h(x) = (x+1/4) (-x+3/2) , f(x) = x+1/4, g(x) = -x+3/4

These functions appear, graphically, to work. To be sure, notice h'(x) = -2x +1/2 and h'(-1/4) = 1 = slope of f(x). h'(3/4) = -1 = slope of g(x). Therefore we can see that f and g are tangent to h at the points (-1/4, 0) and (3/4, 0) respectively.

Let's try one more. Let a = 4, c = -4, b = -1/4, and d = 5/4. Notice these parameters also fit the conditions, but will f and g be tangent to h?

See graph: h(x) = (4x-1/4) (-4x+5/4) , f(x) = 4x-1/4,

g(x) = -4x+5/4

Again, these functions appear, graphically, to work. To be sure, notice h'(x) = -32x + 6 and h'(1/16) = 4 = slope of f(x). h'(5/16) = -4 = slope of g(x). Therefore we can see that f and g are tangent to h at the points (1/16, 0) and (5/16, 0) respectively.

I'm convinced!