### Assignment 1.5 Investigation of Linear Functions and their Products 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. Discuss and illustrate the method and the results.

Our problem is stated to begin with two linear functions. Our goal is to find their product such that when they are graphed on one coordinate plane the product [h(x)] is tangent to f(x) and g(x) at two different points. Because we are investigating the relationship between two linear functions and their product it is reasonable to begin with very basic linear functions. We can then use the information learned at each step to reach our ultimate goal.

To start pick the very simplistic of linear functions

f(x) = x
g(x) = x + 1.

Because the slope of each linear function is the same our lines will run parallel to each other. We can assume that the product of f(x) and g(x) therefore will not touch the lines at only two distinct points.

As you can see the function h(x) crosses the linear functions at its roots 0 and -1. It also crosses the linear function f(x) at the point (1,2).

Our next step may be to change the slope of one of the linear functions. Therefore, the lines will not run parallel, but they will intersect each other at one point. Again, I am going to begin at a basic level graphing the functions

f(x) = x
and
g(x) = -x.

Before looking at the graph, we can see that there is going to be a relationship between the linear functions that have been chosen. The opposite slopes will give us two linear functions that are perpendicular to each other when graphed.

What can be predicted then about the behavior of the product of these two linear functions? When we multiply f(x) and g(x) we find that

This will produce a parabola that opens downward from the origin. Will it be tangent to f(x) and g(x) at two distinct points?

As you can see the function h(x) is not tangent to f(x) and g(x) at only two distinct points. We can see that our goal does not seem far from us at this point. If it were possible to move our parabola down below the point of intersection of f(x) and g(x) we will be even closer. To investigate this idea of moving the parabolar "down" on the graph we should take a moment to look at what is happening when we multiply our linear functions.

Let's investigate one more pair of linear functions in order to observe the relationship between their intersection point and h(x). We should continue to use linear functions whose slopes are opposite.

I am going to choose the linear functions f(x) = x + 1 and g(x) = - x + 1 so that they will intersect at a point above the origin on the y-axis. We can determine this point from the given information. If we substitute zero in for the value of x, we can identify the y-intercept of any linear function. In this case substituting zero in f(x) and g(x) gives us a y-intercept of 1.

Before we graph the product of f(x) and g(x), let's observe some characteristics of h(x) algebraically.

The negative coefficient of the first time of the parabola tells us that the parabola will be opening downwards. Substituting zero in for the value of x we will be able to find the y -intercept.

Substituting in zero for the value of y [h(x)] we will be able to find the x-intercept.

Looking at the graph will illustrate what we have determined algebraically.

Our investigation continues as we try to determine a way to "move" the parabola down below the point of intersection of the linear functions. Again, let's visit this algebraically. Let f(x) = ax + c and g(x) = bx + d. We can determine the equation for h(x) by multiplying the two together.

To find the y-intercept of h(x) we can substitute zero in for x. The resulting point would be cd. If we look at where f(x) and g(x) would cross the y-axis we would find their y-intercepts at (0,c) and (0,d) respectively. Therefore, in order for the graph of the parabola to be "lower on the graph" than the linear functions, cd needs to lie on the y-axis below the intercepts c and d.

As we continue to choose functions for f(x) and g(x) we need to keep this in mind. Let's modify the equations that we had used before.

Our last graph had f(x) = x + 1 and g(x) = -x + 1. To create a parabola that is below the intercept, let's replace the constants c and d with the value of 1/2.

Because the range on our graph has the x and y minimums at -10 and 10, our picture is difficult to observe. Patterns may not be easily recognized and usable. Let's change our range for a closer look. I have chosen a range of -2 and 2.

This should be able to give us a closer look and a better understanding of the graphs.

The graph pictured below is a graph of f(x), g(x) and h(x).

Looking at this graph, it is difficult to tell whether or not h(x) is tangent to f(x) and g(x) at the roots of h(x). The picture can begin to give a convincing argument, although a furthur investigation into the graph and the mathematics behind the graph will complete our exploration.

Let's first zoom in our picture to check the alleged points of tangency.

It is difficult to examine the points of tangency due to computer limitations. Another way to determine the points of tangency for any equation is to look at its derviative.

Our equation is

When we take the derivative of the equation we are able to see that the slope is -2x for each value of x. If we plug in the roots of our curve (x = 1/2, -1/2) we can see that our tangent will have the same slope as our original linear equations.
At x = -1/2, the slope of the tangent line will be 1 and the slope of f(x) = x + 1 is also 1.
At x = 1/2, the slope of the tangent line will be -1 and the slope of h(x) = -x + 1 is also -1.

Therefore at the roots of the parabola (x = 1/2 and x = -1/2) the points of tangency are on the original linear equations f(x) and g(x) and on the curve h(x).