Write Up #1

3. 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.

I will start with the two most simplistic linear functions--constant functions--and go from there. In each case the third equation will be the product of the first two.

Of course, this doesn't even produce a parabola.

Well, this is encouraging. However, we definitely do not want the lines to intersect at the vertex of the parabola--they must intersect above the parabola.

Let's try adding some y-intercepts different than the origin.

As you can see, the vertex moved away from the intersection of the lines, but in the wrong direction.

Let's alter one equation so they have different y-intercepts.

Wow! That appeared to worked. This brings rise to two sets of questions:

1) Did it really work? Are the lines tangent to the parabola?

2) Why did it work? What is the pattern?

In response to the first question, a little algebra should do the trick. Each line and the parabola should have exactly one point in common. We can solve them simultaneously by setting them equal to each other:

So, y=0 and the common point is (-2,0). This is the only point where the two graphs intersect, so they are tangent.

Similarly, we can solve the other two equations simultaneously and find that they intersect at the point (-1,0).

Now for the more difficult, and perhaps more interesting question:

Why did this combination of linear equations work? If we take the above example and shift it to the left, it should still work:

Notice that the two examples that have worked have one thing in particular in common: The difference of the absolute values of the y-intercepts is 1. Algebraically this is very important. Let's take a look at the first example and the algebra we used to solve the two equations simultaneously.

Recall, the equation was . An alternate way to solve this equation for x is to assume that x does not equal -2 and divide both sides by x+2.

Notice the results:

This contradictory result is actually very useful. It gives us a strategy for finding other such pairs of linear functions. Let us try a generic case and see what happens.

We now solve the first and third equations simultaneously.

The quadratic formula yields the following two results for x:

Since we want the solution to be unique (such that the two figures will be tangent) we want these two values to be the same:

Thus, this equation represents the relationship between the two linear equations such that the first line will be tangent to the parabola. Now we must further limit our choices of a,b,c,d so that the second line will be tangent to the parabola.

These two relationships, , describe the necessary relationship between the two lines to ensure that they are simultaneously tangent to the given parabola. Solving these two equations for c and d yields the following equations:

We can look at these two equations as formulas for finding all combinations of linear equations and parabolas of the sort we were seeking.

If a=1 and b=2 then c=-1 and d=-1 and we get our first working example. Similarly, a=1 and b=8 determines c=-1 and d=-7 which was our translated example.

Let's try one more example just to show that all works well.

Suppose a=17 and b=-43. Then c=-17 and d=44.

It seems appropriate to make a few observations here.

1) Notice that the points of tangency must always be on the x-axis since we are always setting a parabola equal to one of it's factors.

2) The parabola will always be upside down since the coefficient of the squared term will always be negative.