Damarrio Holloway

Assignment #1

__Tangent Linear Equations__

__Problem #3__

Find two linear
functions f(x) and g(x) such that their product h(x) = f(x).g(x) is a tangent of
both equations f(x) and g(x) at two distinct points.

Below are the graphs of
the equations: f(x) = x + 1

g(x) = -x + 1

h(x) = (x +
1)(-x + 1)

*GRAPH 1*

The reason I
chose these two functions is because of their simplicity. In order to find two linear functions,
where the graph of their product creates a tangent point on each line, the
slopes for the f(x) and g(x) must be the negative reciprocal of each other in
order for the vertex of the parabola and the intersection of the linear graphs
to be colinear. After
examining the three equations in GRAPH 1, the h(x) function crosses each linear
function twice, rather than the required one for a tangent line. By observation, you will notice that in
order for the h(x) function to form tangent lines, the linear functions must be
shifted. At first thought,
changing the y-intercept might change the intersecting points. LetÕs explore increasing the intercepts
to possibly raising the linear functions above the parabola.

f(x) =
x + 2

g(x)
= -x + 2

h(x)
= (x + 2)(-x + 2)

*GRAPH 2*

By increasing
the y-intercepts for the linear functions (Graph 2), the product graph h(x) has
a phase shift upward. Now in this
graph, h(x) still crosses each linear function at two points, but not at the
intersection of the linear functions. Since a lower y-intercept in Graph 1 yields a downward phase
shift than Graph 2, we will explore an even lower y-intercept for f(x) and g(x)
than in Graph 1 using the following functions:

g(x) =
x + 1/2

f(x)
= -x + 1/2

h(x)
= (x + 1/2)(-x + 1/2)

* *

*GRAPH 3*

Using quadratic
functions that are differences of squares helped me manipulate the parabola up
and down the x-axis. Graph 3 shows
the lines of tangency across the parabola. Now lets explore shifts of the graph to the left and right.

Options to the left: f(x) =
-x -1/3 g(x) = x + 4/3 Options to the right: f(x) = -x + 3/2 g(x) = x - 1/2

h(x)
= (-x – 1/3)(x + 4/3) h(x)
= (-x + 3/2)(x – 1/2)

With different
values of ÒaÓ in the quadratic function: f(x) = 3x + 3/2 g(x) = -3x – 1/2 h(x) = (3x + 3/2)(-3x -1/2)

Wider
graph: f(x) = (1/2x + 3/2) g(x) = (-1/2x – 1/2) h(x) = (1/2x
+ 3/2)(-1/2x-1/2)

Once points of
tangency are found for two linear functions along their product, these things
will occur:

1. All of the points of tangency occur on
the x-axis.

2. The linear graphsÕ point of
intersection and the vertex of the parabolic product will be collinear.

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