*Product of
Two Quadratics*

* *

*By Courtney
Cody*

For Assignment #1, I chose to
explore question #4 for which I wish to find two quadratic functions f(x) and
g(x) such that their product h(x) = f(x) * g(x) is tangent to each of f(x) and
g(x) at two different points.

Let us begin by observing two
simple quadratic functions f(x)=x^2 and g(x)=2x^2, which have a different
leading coefficient, and their product.

**Graph A**

By looking at the graph we
see the product of these two quadratic functions, which is pictured in blue,
does not appear to be tangent to either of f(x) or g(x). The blue curve only appears wider
at its base when compared to f(x) and g(x) and intersects each of the functions
at two points. Hence, we need to
change f(x) and g(x) such that there is more of a difference in the equations
than just leading coefficient.

Suppose we change the two
quadratic functions such that the signs of the leading coefficient are
opposite. For example, let
f(x)=x^2 and g(x)=-2x^2.

**Graph B**

Since the graph does not appear
to be getting closer to satisfying the conditions of the assignment since h(x)
is not tangent to either f(x) and g(x) at any points, more changes need to be
made to the equations of f(x) and g(x).

In order to have f(x) and
g(x) intersect at a couple of points, let us decrease f(x) by a constant and
increase g(x) by a constant. For
example, let f(x)=x^2-1 and g(x)=-x^2+1 and consider the graph of these two
functions and their product, h(x), below:

**Graph
C**

This graph looks much
better! The function h(x) in blue
is already tangent to f(x) at one point.
If we look at the red curve and notice that if we change it such that
the blue curve is completely contained and touching it, then the h(x) will be tangent
to g(x) on each of the sides.
Hence, let us change g(x) such that it has a wider slope in order to
just slightly contain h(x). For example, let the leading coefficient of g(x) be
-1.

**Graph D**

While changing g(x) such that
its slope is wider brought us closer to satisfying the conditions of the
assignment, the blue curve still lies just outside of g(x). What would happen
if we moved the red curve up by a unit?
Let us explore this change.

**Graph E**

Wow! This looks exactly like what we are
looking for to answer the assignment.
Not only does the red curve contain the blue curve such that they are
touching as we desired, but the blue curve also touches the purple curve at
exactly two points as well! Hence,
when f(x)=x^2-1 and g(x)=-x^2+2, the product h(x)=f(x)*g(x) is tangent to each
of f(x) and g(x) at two different points.

Is this the only solution to
the problem? What if f(x) and/or
g(x) contain a linear term in their equations? Let us explore.

Suppose we take f(x) and g(x)
as they are in the previous graph and add a linear term. We now obtain the following graph:

**Graph F**

This does not look
right! By adding a linear term,
the graphs of the functions became skewed and asymmetrical. Additionally, h(x)
is no longer tangent to either f(x) or g(x) at any points. It seems important for f(x) and g(x) to
be centered on the same x-value.
In the previous example, by adding an x term to each of f(x) and g(x),
we actually shifted f(x) to the left one unit and g(x) to the right one
unit. What if we made the
equations such that both f(x) and g(x) were shifted to the left? By changing the coefficient of the
linear term in g(x) to be -1 and keeping f(x) the same, we obtain the following
graph:

**Graph G**

Once again, we have satisfied
the conditions of the assignment since h(x) is tangent to f(x) and g(x) at two
different points.

What would happen if the
coefficients of the quadratic and linear terms were not 1? Is there a way to manipulate the
equations such that the conditions of the assignment still hold? Consider the following graph:

**Graph H**

Let us declare a general
formula for f(x) and g(x).

In general, there are
infinitely many solutions of the form **f(x) = a _{1}x^2 + b_{1}x
-1** and