To begin to solve this problem, the students may start by experimenting with parallel lines. It should be obvious that vertical or horizontal parallel lines will not work since their product does not result in a parabola. Generally, students should notice that parallel lines will not work because, intuitively, a parabola will get increasingly wider as the points get further away from the directrix. Since parallel lines will always be equal distance from each other, it can be seen that at most one can be tangent to the parabola. If the students can not see the reasons for this situation, they may graph several examples like the following ones.

y = -2x + 1

y = -2x- 3

y = (-2x + 1)(-2x - 3)

y = 3x + 2

y = 3x + 5

y = (3x + 2) (3x + 5)

Since parallel lines do not satisfy the desired conditions, the students
must conclude that the lines intersect at some point. They may try several
different graphs using the program, and they should look similar to the
following.

y = -2x + 1

y = 3x - 3

y = (-2x +1)(3x -3)

y = -4x +2

y = -3x +3

y = (-4x + 2)(-3x +3)

Again, the students will see that, generally, these do not satisfy the conditions
of the problem. Hopefully, the next step that the students will take is
to conclude something about the symmetry of the parabolas that they have
just graphed. It can be seen that the parabolas in all of the previous examples
have had a directrix that has been parallel to the to the x-axis. So the
parabola is symmetric around a line perpendicular to the x-axis and which
passes through its focal point. This leads to the idea that two lines that
are symmetrical around a line parallel to the y-axis may be part of the
solution to the problem. Lines that would satisfy this condition would be
lines that have slopes that are opposite of each other. Again, students
can use the computer to graph some of these lines quickly and without error.
Some possible examples may look like the following.

y = -4x + 2

y = 4x + 3

y = (-4x + 2)( 4x + 3)

y = -2x - 2

y = 2x + 5

y = (-2x - 2)(2x + 5)

Again, students should notice that while this is closer to the desired situation
it is still not correct. By looking at the generated graphs, they should
see that the parabola is on the same axis of symmetry as the lines, but
it is either too high or too low. So the students might next try to shift
the lines up or down. Since the parabola is the product of the two lines,
it follows that if one line is shifted up or down, then the resulting parabola
will be adjusted accordingly. Again, several trials might be done on the
computer before one that looks like it satisfies the desired conditions
is found.

y = -2x -2

y = 2x + 4

y = (-2x -2)(2x +4)

y = -2x -1

y = 2x + 3

y = (-2x - 1)(2x +3)

y = -2x - 2

y = 2x + 3

y = (-2x - 2)(2x +3)

While the last graph looks like it has the desired properties, the students
should verify that it actually does have them. It is a simple process to
verify this algebraically, and the following is an outline of this process.
The last graph and its equations will be used as an example. First, the
coordinates of the two potential tangent points must be found, which can
be done using two separate systems of equations.

y = (-2x - 2)(2x +3) = -4x2 - 10x - 6

y = (-2x -2)(2x + 3)

Let us look at the first system of equations. If the graphs intersect, then
they meet at y = -2x -2, for some x. So, -2x - 2 = (-2x - 2)(2x +3), for
some x, which must mean that 2x+3 = 1. Solving this equation for x, we find
x = -1. Thus, for the first set of equations we have (x, y) = (-1, 0) as
the point of intersection for these two equations. In a similar manner,
we find that (x, y) = (-3/2, 0) is the point of intersection for the second
system of equations.

Second, it must be verified that the lines are tangent to the parabola at
their respective points. Calculus is used to solve this part of the problem.
Simply take the derivative of the equation for the parabola. So if f(x)
= (-2x - 2)(2x + 3) = -4x2 - 10x -6, then the derivative is f '(x) = -8x
- 10. Since the derivative is the slope of the tangent line of a curve at
a given point, all that needs to be done is to substitute the two x values
into the derivative and see if the resulting answers match the slopes of
the given lines. So, we take x = -1, and substitute it into the derivative.
f '(-1) = -8(-1) -10 = -2. This matches the slope of the corresponding line,
and so we see that the line is indeed tangent to the parabola at the point
(-1, 0). Similarly, we have that the line y = 2x + 3 is tangent to the parabola
at the point (-3/2, 0). This should prove to the student that these lines
do indeed satisfy the desired criterion.

A harder question for the students is to generalize this situation. In other
words, if given a line that is not horizontal or vertical, can a second
line be found such that the product of the two lines is a parabola and both
lines are tangent to the parabola in two distinct points. The following
is an outline of the algebraic method used to find a generalization.

Let f(x) = ax + b, g(x) = cx + d, and h(x) = f(x)g(x) = (ax + b)(cx + d).
Suppose f(x) and h(x) intersect at some point (x1, y1), then f(x1) = h(x1)
and g(x1) = cx1 +d = 1. Similarly, suppose that g(x) and h(x) intersect
at some point (x2, y2), then g(x2) = h(x2) and so f(x2) = ax2 + b = 1. We
use two different systems of equations, but our demonstration will only
consider the first.

h(x) = f(x) g(x) = acx2 + (ad + bc)x + bd

Now, since we want the line to be tangent at the point (x1, y1), we set
the derivative h '(x1) = a. We also know that x1 = (1 - d)/ c. So using
substitution, we have the equation

Some simplification gives the equation, 0 = a(1 - d) + bc. In a similar
manner, we can manipulate the other system of equations to give us the equation
0 = c(1 - b) + ad. We can then set these two equations equal to each other
and simplify. This results in the following equality, c = -a. Using this
fact, the equation 0 = c(1 - b) + ab, and some more algebra we get the equation
b + d = 1.

So to generalize, if an equation y = ax + b is given, and we want to find
another line such that these two lines are tangent to the parabola resulting
from their product, then the line y = -ax + d, where b + d = 1 is the appropriate
line.

If the students can get to this point in the analysis of the problem, they
should try several systems of equations and visually see that these conditions
on a, b, c, and d satisfy the demands of the problem. This investigation
should work well with advanced high school mathematics students. It should
help the students understand what a derivative of a line really is, and
how it can be used to solve investigations into the nature of mathematics.
Certainly, there are other ways to investigate mathematics using this situation.
Why do the lines and the parabola always intersect at the x-axis? Can we
find equations that will satisfy these conditions, but have the parabola
opening in a direction other than up or down? The teacher should be able
to come up with several other puzzling questions with only a minimum of
investigation. This example of the use of the Algebra Xpressor is just one
in a myriad of possible uses that will help high school students gain a
deeper understanding of mathematics.