The Department of Mathematics Education

The Product of Two Linear Functions
each of which is Tangent to the Product Function

by

James W. Wilson
University of Georgia

and

David Barnes
University of Missouri

This may have been an attempt to write a paper with a longer title than the paper itself. In fact, this "paper" is a discussion of our examining a particular problem that having tools like function graphers available might make possible different approaches.

The problem is:
Find two linear functions f(x) and g(x) such that the product
h(x) = f(x).g(x) is tangent to each.

This problem was posed by a group of teachers during a workshop in which the use of function graphers was being explored. Our analysis is presented as a sort of stream of consciousness account of how one might explore the problem with the tools at hand. In fact, we came up with two different streams of consciousness and so we have two senarios that are parallel in that they cover alternative approaches to the problem. The senarios represent a composite of several discussions of the problem with teachers, students, and colleagues. Note, the goal here is using this problem context, not only to solve the problem posed, but to understand the concepts and procedures underlying the problem.

Function graphers are available for almost any computing platform or graphic calculator. Such tools make it possible to look at new topics in the mathematics curriculum or to look at current topics in different ways. This problem has some elements of each. In general it would not be included in the school curriculum, but there is no reason it should not. Further, the use of technological tools to examine visualizations of the functions makes for a different approach to the problem.

Is it possible to find two linear functions, f(x) = mx + b and g(x) = nx + c, such that the function h(x) = f(x).g(x) is tangent to each. A traditional approach would begin with algebraic manipulation. This is useful because it keeps the students occupied, but what do they learn from it? Clearly, h(x) = (mx + b)(nx + c) is a polynomial of degree 2 and h(x) has two roots. The respective roots are when f(x) = 0 and g(x) = 0. This means the graph of h(x) crosses the x-axis at the same two points as f(x) and g(x). Thus, if there are points of tangency then they must occur at these common points on the x-axis. Experienced students, very bright students, and good problem solvers could whittle this information and a lot more information out of this algebraic analysis. Novice students would be more tentative.

Senario One

Lets open up the function grapher and explore with some specific f(x), g(x), and the resulting h(x). Lets try

The graphs on the same set of axes are


For some novices, seeing the graph of the product h(x) = (3x + 2)(2x+1) and the graphs of the two straight lines from the factors on the same coordinate axes provides a new experience. This particular graph has one of the two lines "close" to being tangent to the product curve but the other one is not close. How could the picture be changed?

One idea is to spread the two lines so that one has negative slope. Try

The graphs are

This is better? What can be observed? How can the graph of h(x) be "moved down?" What if the graphs of f(x) and g(x) had smaller y-intercepts? Try

The graphs are

Still not too good but at least the graph of h(x) was "moved down."

Try smaller y-intercepts, such as

and result in graphs of

These graphs seem close, but clearly the line with negative slope is not tangent to the graph of h(x). Looking back over the sequence of graphs (and perhaps generating some others) the graph of h(x) always has a line of symmetry parallel to the y-axis. It seems that the pair of tangent lines will have to have this same symmetry. How? Try making the slopes 3 and -3. The functions are

and the graphs are

A zoom to the right hand side of the graph with give

showing tangency has not been achieved. A zoom to the left hand side shows a similar problem.

One could try adjusting the y-intercepts. In fact, if the y-intercepts were equal, the y-axis would be the line of symmetry. Try

The resulting graphs are

Worse. Try

The graphs are

A zoom to the left shows

and to the right shows

The result seems on target. It remains to confirm f(x) and g(x) each share exactly one point in common with h(x). Again the tradition is to do so algebraically, but it might be instructive to look at some graphs of h(x) - f(x) and h(x) - g(x), such as the following:

Can we immediately generate graphs of other f(x), g(x), and h(x) satisfying the conditions of the problem? Reviewing the graphs and the strategy, its seems that the slopes of the lines can vary, the only condition being that they are m and -m. So, a simpler case might be to let the slopes be 1 and -1, giving

and

It is also of interest to see both of the solutions on the same graph:

Other solutions could be generated by making some other vertical line the axis of symmetry. Indeed substituting

gives the graphs

for which the equations simplify to

Scenario Two

Try

First consider the graphs of f(x) and g(x) and try to sketch in h(x). The graph is

What do these lines tell you about the parabola? What points do you know the curve will go through? Why? What causes it to open the way it does? Now let's add the graph of the parabola and compare it with our sketch.

How is it like our sketch? How is it different? Is there anything we should notice or consider?

It appears that all three graphs seem to intersect at 1 on the y axis. Lets zoom for a closer look.

Maybe changing one of the functions will help with the explanation.
Consider

The three functions no longer intersect at 1 on the y-axis. However, the changed function, f(x), does intersect the curve at its y-intercept.

When g(x) = 1 the parabola intersects f(x). Is the opposite true? Lets graph and see.

It seems that if f(x) = 1 then h(x) = g(x) and if g(x) = 1 then h(x) = f(x). Lets test this by trying to generate h(x) from a new f(x) and g(x). Let

Then add the sketch h(x) and compare with . . .

.

In this process we seem to have also noticed that the lines and the parabola intersect at the points when the lines cross the x-axis. Why would this be true?

Now the goal is to get one line tangent to the parabola. The function g(x) is close to being tangent. If we could just get the two points to slid together then they would become one point -- the point of tangency. (If a line intersects a parabola in exactly one point, what is true about the line?)

Since f(x) takes on a value of 1 when x = 1, then lets try to change g(x) so that g(1) = 0. Lets see we could change the slope or change g(x)'s position up and down.

Lets try them both.

To change slope, g(x) = -2x + 2 and test to see if g(1) = -2(1) + 2 = 0

Or change position, g(x) = -3x + 3 and test g(1) = -3(1) + 3 = 0

This seems to imply that f(x) and h(x) are tangent at the f(x) and h(x)'s common root, if the function g(x) takes on the value 1 at this root. Or in other words if f(a) = 0 and g(a) = 1 then f(x) is tangent to h(x) at a.

What would we have to do to get both f and g tangent to h? That would mean that when f(x) = 0, then g(x) = 1; and when g(x) = 0, then f(x) = 1. Lets first start with an easy function for f and then try to generate a g(x) which satisfies what we want. Lets begin with f(x) = x.

Now when f(x) = 0, then g(x) should take a value of 1. In other words if f(0) = 0, then g(0) = 1. Likewise, when g(x) = 0, f(x) should have a value of 1. Since f(1) = 1 then g(1) = 0. We need a our linear function g(x) to go through (0,1) and (1,0). So our g(x) = -x +1. Lets graph it to check.

That looks right! Now, add the graph of the product and then test it to see if the curves are tangent.

That looks good. (What is the coordinates of the vertex?) Lets zoom in at the roots.

and .

This seems to be a useful direction. Does it work on the previous problem? When we left off, f(x) = x and g(x) = -3x + 3. Can we use our technique to find a different f(x) that works for g(x) to produce h(x) = f(x).g(x) with f(x) and g(x) each tangent to h(x)? We have the following graph.

Since g(1) = 0, then f(1) = 1 and since g(2/3) = 1 then f(2/3) = 0 will be necessary. So f(x) contains the points (1,1) and (2/3, 0). Try g(x) = -3x - 2.

Zoom in for a closer look.

What are the coordinates of the upper vertex of the triangle? What are the coordinates of the vertex of the parabola? Are the lines really tangent to the parabola? How might this be proved? Where else could the function f(x) possibly take on the same value as h(x) if h(x) = f(x).g(x)? And how can we interpret this on the graphs?

Summary

Many issues are hidden in these composite accounts of examination of this problem. We still have the additional problem of writing a concise argument of proof of the demonstration -- that the solution will always have the two lines of slope m and -m crossing on y = 1 and the vertex of the parabola on y = 1/2.

Each senario presents a somewhat different approach. Which would be most helpful in finding two quadratic functions f(x) and g(x) such that their product function h(x) = f(x).g(x) has each tangent? The following graphs show such functions. How can they be generated?

What are some generalizations of the problem (and the solutions)?