Let us begin with the simplest case. The pair of functions
will graph as parallel lines through the points (2,0) and (5,0)
with slope of 1.
The product of these two functions y = (x - 2)(x - 5) is a
parabola also passing through the points (2,0) and (5,0). The
graph, showing all three functions, is as follows.
The same function is represented by the equation
Clearly, we could have started with any two lines through (2,0)
and (5,0) respectively, such as
and taken their product
to produce two lines and their corresponding parabola through
the given points.
Going the other direction, we could take any of the family
of parabolas through the points (2,0) and (5,0) with major axis
along x = 3.5, ,and factor the equation into two linear factors
representing lines through the given points.
Consider again the parabola with the equation
Let us look at rewriting it as
and graphing it. The domain over which the function is defined
in the real numbers is restricted to where 7x - 10 > 0. In
this region, however, the graphs of the two equations are the
same.
The spurious vertical line results at the boundary of the defined region -- where 7x-10 approaches zero.
The rewritten equation is in the form of the difference of
two squares and in factored form we have
That is, we have the function as the product of two functions
and the graph of these two functions (on the same axis) is
as follows.
The union of these two graphs appears to be a parabola. Is
it? (Again
A single equation for the union of two curves can be found by
taking the product of the relations producing the curves.
We set
and simplifying gives
The graph is as follows. This is the equation of a parabola
passing through the points (2,0) and (5,0) having an axis of symmetry
with slope of 1.
Let us consider again the equation
For any constant a,
is the same graph. Now we have a slightly different way of
putting this in the form of the difference of two squares:
and the factor functions are
Rewriting as the product of two relations to get the union
of the graphs gives
This simplifies to
Now we can use Graphing Calculator 3.0 to graph a family of
parabolas, all passing through 2 and 5 on the x-axis. The previous
parabola graph was the case for a = 0.
Perhaps some interesting additional graphs would be between
a = 3 and a = 4. Here are the graphs for a = 3, 3.25, 3.5, 3.75,
and 4.
The graph for a = 3.5 deserves a separate look.
As a approaches 3.5 from below, the family of parabolas all
open upward and become more narrow. As A approaches 3.5 from above
the parabolas all open downward and become more narrow. At a =
3.5, however, the parabola reduces to a pair of parallel lines.
This indicates the equation
is factorable. We have
The graph of this relation is a pair of parallel lines. These
are the lines with which we began this investigation and had we
graphed a product of the two relations
we would have obtained the union of the two individual graphs.
There are many parabolas through x = 2 and x = 5 that we have
not yet examined with these equations. For example, what set of
equations would generate the following graphs?
If we graph the product
It is clear that the graph will be the parallel lines defined
by
On the other hand, the product
must also produce a pair of parallel lines when it is graphed. This is a generalized quadratic equation of the form
Examination of the generalized quadratic was once standard fare in courses on analytical geometry. Families of graphs are formed by fixing all but one of the parameters and varying the remailing one.
As long as A = 1, C = -7, and F = 10, the graph of
will cross the x-axis at 2 and 5. We know that for particular
coefficients B, D, and E the resulting graphs could be conics
or degenerate conics, that is, conics, circles, or parallel lines.
For example, when B = -2, D = 1, and E = 0, the graph is a parabola.
When B = 0, D = 1, and E = -5, for example, the graph is a
circle.
Some examples of other figures are
Hyperbolas
Two Lines
Lines and Hyperbolas
The above three graphs on the same axis shows the two hyperbolas and this pair of lines but the pair of lines can not be asymptotes to either of the hyperbolas since the graphs share the points (2,0) and (5,0).
Ellipses can be formed in several ways. The task here is to show sets of ellipses through the points 2 and 5 on the x-axis. The general equation is
What values of B, C, and E will give ellipses? Try C = 1 to get the coefficients of the two squared terms equal. Then try B = 1 and E = 8.
Here we have a rather messy collection of a circle, two ellipses, some parabolas and some hyperbolas, all crossing the x-axis at 2 and 5.
