**Math 6680 –
Jackie Gammaro**

**Assignment #1 - Problem
#3**

**Find two linear
functions f(x) and g(x) such that their
product h(x) = f(x)**

**I. Problem Solve - Trial and Error**

After a few tries, I let *f(*** x)
= 2x + 4** and

The following is a graph of *f*(*x*),* g*(*x*), and* h*(*x*).

**II. What are the equations?**

** **

After this little success I tried to generalize. I noticed the slopes of the two linear
functions were opposites of each other and that their was a relationship
between the y-intercepts of *f(**x) and g(x).*

Thus I let *f(*** x)
= mx +
b, g(x) = -mx
– b + 1, **and

**III. ÒWhat are the points of intersection?Ó**

I noticed the intersections of *f(x) *with *h(x)* and of *g(x) *with *h(x)* lie on the y-axis, thus when the functions equal zero. How can I show this
algebraically?

If *f(*** x)
= mx +
b, g(x) = -mx
– b + 1, **and

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

* mx + b = (mx
+b)(-mx –b+1) -mx-b+1 = (mx+b)(-mx-b+1)*

*
1 = -mx –b+1 1 = mx + b*

*
0 = -mx
-b 1-b
= mx*

* x = x = *

* f = *

* *

Thus, the points
of intersection are
and
.

**IV. ****Do the points of intersection always lie on the x-axis?**

I also wondered, and ÒIf they
do, then WHY?Ó

1. First I examined the graphs of *f(x –k), g(x –k) *and*
h(x – k)* and then determined algebraically, that horizontal
translations of the functions donÕt change the points of intersection off the
x-axis.

*2.
*I then
examined the graphs of *f(x) + k*, *g(x) + k* and *h(x) + k*. Just from viewing the graphs in Graphing Calculator, I
determined the stipulation that h(x) has to be tangent to f(x) and g(x) at the
points of intersection was not visible. However, points of intersection do lie
on the x-axis. In this case m, b,
k =1.

*
*

3. At
this point I wondered, ÒWHY the *x*-axis?

I studied the graphs, I studied my algebra, and it wasnÕt clicking. Then the lightbulb went off!

I had to determine when *f(**x) = h(x)* and *g(x) = h(x)* and I know *h(x) =
f(x)*g(x).*

So I let *f(**x)* = a and *g(x) = b*, and
assuming *x *ëå, substituted into the previous equations which results
to

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

*a** = ab
b = ab*

*a* = 0
or *b* = 1 or a,b=0 * b* = 0 or *a*
= 1 or a,b=0