Find two linear functions

Let *f*(*x*) = a*x*+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:

Reducing this system yields:

This result gives us conditions on a and c.

Substituting this result into equaion (3) yields:

which reduces to

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*)
= (4*x*-1/4) (-4*x*+5/4) , *f*(*x*)
= 4*x*-1/4,

*g*(x) = -4*x*+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!