**Sarah Hofmann**

**EMAT
6680 Summer 2006**

**Assignment
1**

Exploration 3: Our
assignment is to find two linear 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 distinct points. We are to
discuss and illustrat our method and results

To just get started,
we try two random linear equations. Who knows, we may get lucky!
Let f(x) = 3x + 1 and g(x) = 5x - 4. The the graph of f(x),
g(x), and h(x) in red, green and yellow respectively are

Well this wasn't
very productive. There are too many different numbers flying
around in these equations to make much sense of what is happening.
Let's try making our equations simpler. What if we make the
slope the same in both equations. What would happen? Lets
try a slope of 1. Now if we make the y-intercepts the same, we will
have two of the same line. That won't help. So lets try to
make one y-intercept +1 and one -1. So we are trying the
equations f(x) = 1x + 1 and g(x) = 1x - 1. Graphing f,g, and h in
red, green, and yellow respectively we find

This is definately
better. The green line is getting close, it just needs to move
down a little. But all we have to do with the red line is move it
down too, but then both lines would be tangent at the same point.
So we nee to change the red line. What if we switched it so
that it had negative slope instead of positive slope? Then we
would just need to move the red line down on the other side of the
parabola. So lets try f(x) = -1x + 1 and g(x) = 1x -1. Then
we get the following graph

Wow! This is a lot
better. Maybe we can make this even easier if we made the center
of the parabola (and the point where the lines cross) are centered on
the y-axis. So we need to make sure that f(0) = g(0). That
is that -0+b = 0+c or b=c. So we need to have an equation like
f(x) = -x + 1 and g(x) = x + 1.

Now it looks like all we have to do is slide the
graphs up and down until we find the right y-intercepts. Let's
look at an animation that has different values for the y-intercepts and
see if that helps.
*Click on the graph above to
find the animation*
.

If we look at this
animation it looks like the correct answer happens when the y-intercept
is between 0 and 1, so lets try a y-intercept of 1/2. So we will
graph the equations for f(x) = -x + 0.5 and g(x) = x + 0.5 below.

This looks right!
Lets double check. We can see that f(x) = h(x) = f(x)g(x)
implies -x+0.5 = (-x+0.5)(x+0.5) implies 0 = x + 0.5 implies x = -0.5.
The equation of the tangent line of h(x) = (-x +0.5)(x + 0.5) at
the point x = -0.5 is y = -x + 0.5 = f(x). So f(x) is tangent to
h(x) at x = -0.5. Similarly g(x) is tangent to h(x) at x = 0.5.
Therefore, we have found the two needed points!

Just for fun, lets see if we can find more sets
of lines based on our first set. Lets try an animation so that
both the y - intercepts are 0.5 and the slopes of f(x) and g(x) are n
and -n respectively.
*Click on the above graph for the animation*
.

We can see from the
animation that in fact, the lines f(x) = nx + 0.5 and g(x) = -nx + 0.5
give the desired result for any nonzero value of n.