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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 sketched a picture of what I wanted. The graph had to be somehow akin to this:

The problem begins with simple explorations in Algebra Xpresser. Since the lines must obviously have different slopes, I chose a few simple equations and their products to see what happens. I looked at a few such as these:

f(x) = x, g(x) = -x f(x) = x + 1, g(x) = -x + 1

f(x) = 2x + 1, g(x) = -2x + 1


These were not working. I needed to push the parabola down some. I tried different signs on the y-intercept.

f(x) = x + 1, g(x) = -x - 1



I was still not there. I have had only integer y-intercepts, so I started varying them.
f(x) = 2x + 1/2, g(x) = -2x + 1/2 f(x) = 3x + 1/3, g(x) = -3x + 1/3



The first pair was it. Now I tried it for different values of slope.

f(x) = 3x + 1/2, g(x) = -3x + 1/2


These had to be my equations

f(x) = ax + 1/2
g(x) = -ax + 1/2
Now for proof...
The proof begins with a specific case and ALGEBRA! I chose the equations
f(x) = 2x + 1/2
g(x) = -2x + 1/2.

Solve this system to find where they intersect.

y = 2x + 1/2
y = -2x + 1/2

add equations:
2y = 0 + 1

solve:
y = 1/2


Plug this y value into either f(x) or g(x) to find x.

1/2 = 2x + 1/2
0 = 2x
0 = x

The point of intersection for the two lines is (0,1/2).


Now find where the lines intersect the parabolas by solving another system.

y = 2x + 1/2
y = 4x^2 + 1/4

rewrite equation
-y = -4x^2 - 1/4

add
0 = -4x^2+2x+1/4

factor and solve
0 = (2x + 1/2)(2x + 1/2)
x = -1/4

Plug this x value into either equations to find y.

y = 2(-1/4) + 1/2
y = -1/2 + 1/2
y = 0

The point of intersection for these is (-1/4, 0).
Similarly, the point of intersection for the parabola and the other line

y = -2x + 1/2
y = -4x^2 + 1/4


will be (1/4, 0).
A picture of this particular case from Algebra Xpresser follows:

Now for proof of a general case.

f(x) = ax + 1/2
g(x) = -ax + 1/2
h(x) = f(x).g(x) = -ax^2 + 1/4


Add f and g to find the intersection of the two lines :

2y = 0 + 1
y = 1/2

Plug y value into f or g to find x value:

1/2 = ax + 1/2
0 = ax
0 = x

The point of intersection of the two lines is (0, 1/2).


The point of intersection of f and h will be found by solving this stystem:

y = ax + 1/2
y = -a^2x^2 + 1/4

rewrite:
y = ax +1/2
-y = a^2x^2 - 1/4

add:
0 = a^2x^2 +ax +1/4

solve:
0 = (ax + 1/2)^2

-1/2 = ax
-1/2a = x


The point of intersection of f and h is (-1/2a, 0) and, likewise, of g and h is (1/2a, 0).
Naturally, these points are the roots of both the lines and the parabola, as we would expect.


We now see that we cannot translate the lines up and down (at least not in this form) because meeting the criteria of the problem is contingent on the fact that the y-intercept for both lines is 1/2. This all begs the question, "Can we translate the lines right or left?"

Keeping the equations in standard form we can see how to keep the same general form as above.

f(x) = (x - 1) + 1/2, g(x) = -( x - 1) + 1/2

f(x) = (5x +3) + 1/2, g(x) = -(5x + 3) + 1/2

We can see a very interesting graph if we overlay several of these. When we graph the horizontal line y = 1/2, we see where all the pairs of lines intersect. Graphing the vertical line y = 1/4 shows the vertex of each parabola has y value equal to 1/4. Lastly, by considering y = 0 (or the x-axis) we see the roots of each triple of equations (the pair of lines and the tangent parabola).

From this picture, we can see the family of triples
f(x) = ax + 1/2
g(x) = -ax + 1/2
h(x) = f(x) . g(x) = (ax + 1/2) (-ax + 1/2) = -a^2x^2 +1/4
which satisfy the given conditions. My sketch after first reading the problem is incorrect to the extent that I did not consider the fact that one line must have a negative slope, therefore making the parabola open down.
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