Presented By:

Dana TeCroney

The challenge in
this assignment is to find two functions f(x) and g(x) such that their product:

h(x) = f(x)*g(x)

is tangent at two
distinct points.

Not only will I
give an example to solve the given question, but this method will be
generalized to show that when give h(x) = ax^{2} +bx +c, there are an
infinite number of lines that could produce such a quadratic.

To begin, consider
the polynomial h(x) = -x^{2} + 6x - 5

Choose a slope for
f(x), say m = 3. Because of the
symmetry of parabolas, the slope of g(x) will be m = -3. Now, values for the constants in f(x)
and g(x) must be determined.

Consider f(x) = 3x
+ d

As 'd' changes,
the line shifts such that sometimes f(x) and h(x) intersect in two places,
sometimes in one place, and at other times not at all. For a graph of this, press HERE.

We are concerned
with where f(x) and h(x) are tangent, i.e. intersect once. To find this value we want to know
where h(x) = f(x) at only one point, or, where h(x) - f(x) = 0 has only one
root. To see the effects of
changing ÔdÕ press HERE. The graph would look like this:

Another way to
look at this is when the discriminate of h(x) – f(x) = 0:

h(x) – f(x)
= -x^{2} + 6x – 5 – (3x + d) =0

= -x^{2} + 3x + (-5 - d) = 0

The discriminate would be equal to zero
where

3^{2 }– 4(-1)(-5 - d) = 0

Or, when d = -11/4

Therefore, f(x) = 3x – 11/4

Here is a graph of f(x) (red) and h(x) (purple):

The same process
can be followed to find g(x).
Remember, the slope of g(x) was -3:

g(x) = -3x + e

What should our
ÔeÕ be?

Again we want h(x)
and g(x) to be tangent, so we want the discriminate of h(x)- g(x) = 0:

-x^{2} + 6x – 5 – (-3x
+ e) = 0

=> -x^{2} + 9x + (-3 – e)
= 0

The discriminate
of this would be equal to zero where:

9^{2} – 4(-1)(-3 – e) =
0

Or where e = 61/4

Therefore g(x) =
-3x + 61/4

Here is the
resulting graph:

In general,
consider a quadratic of the form h(x) = ax^{2} + bx + c. Choose an
arbitrary slope for f(x), say m.
The slope for g(x) must be –m.

h(x) = ax^{2} + bx + c

f(x) = mx + d

g(x) = -mx + e

To find values for
constants d and e, one must find where h(x) = f(x) and where h(x) = g(x)
tangentially.

In other words the
discriminates of the function h(x) – f(x) = 0 is zero and where the
discriminate h(x) – g(x) = 0 is zero.