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. Let’s 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 “a” 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 graphs’ point of intersection and the vertex of the parabolic product will be collinear.

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