The Department of Mathematics Education
J. Wilson, EMAT 6680
Last modified on August 18, 2002.
The attached 4-page paper is the start of an article that
might appear in a journal such as the Mathematics Teacher -- the
audience being mathematics teachers who might use some of the ideas for
It is a start; incomplete, unclear, maybe in error; maybe glossing over significant points and stressing some obvious or trivial points.
Sign on as a co-author. Rewrite and complete the article. This means you must come to grips with whatever points are to be essential, what to add, what to delete, and what to edit. The "different" approaches to this topic are really in the graphs in the xb, xc, or xa planes. You might want to examine a bunch of these before trying to re-write.
It has now become a rather standard exercise, with available technology, to construct graphs to consider the equation
and to overlay several graphs of
for different values of a, b, or c as the other two are held constant. From these graphs, discussion of the patterns for the roots of
can be followed. For example, if we set
for b = -3, -2, -1, 0, 1, 2, 3, and overlay the graphs, the following picture is obtained.
We can discuss the "movement" of a parabola as b is changed. The
parabola always passes through the same point on the y-axis (the point (0,1)
with this equation). For b < -2 the parabola will intersect the x-axis in
two points with positive x values (i.e. the original equation will have two
real roots, both positive). For b = -2, the parabola is tangent to the x-axis
and so the original equation has one real and positive root at the point of
tangency. For -2 < b < 2, the parabola does not intersect the x-axis --
the original equation has no real roots. Similarly for b = 2 the parabola is
tangent to the x-axis (one real negative root) and for b > 2, the parabola
intersects the x-axis twice to show two negative real roots for each b.
Now consider the locus of the vertices of the set of parabolas graphed from
Show that the locus is the parabola
The vertex of a parabola is defined as the ordered pair (x,y) for which the named parabola reaches its global maximum or minimum. In this case, the global maxima are reached at the values of x for which the value of first order derivative is 0, that is, the value of x where 2x + b = 0 for a given value b. By inspection, the function reaches its global minima at values
at the ordered pair
Since the graph of the locus interpolates these ordered pairs, we may say that for any rational number b
Consider again the equation
Now graph this relation in the xb plane. We get the following graph.
If we take any particular value of b, say b = 3, and overlay this equation on the graph we add a line parallel to the x-axis. If it intersects the curve in the xb plane the intersection points correspond to the roots of the original equation for that value of b. We have the following graph.
For each value of b we select, we get a horizontal line. It is clear on a
single graph that we get two negative real roots of the original equation when
b > 2, one negative real root when b = 2, no real roots for -2 < b <
2, One positive real root when b = -2, and two positive real roots when b <
Consider the case when c = - 1 rather than + 1.
Reciprocity of Roots
We know that in the case where c = +1 that the graph depicts all values of x and b such that
These observations immediately follow:
1. The function is undefined at x = 0 and has a corresponding vertical asymptote.
2. The function has two extrema, one in quadrant II and one in quadrant IV where the first order derivative
that is, 1 and –1 respectively.
3. The value of the function at its extreme is 2 and –2 respectively. The absolute value of the function cannot assume a value less than 2.
4. There is a one-to-one correspondence between values of x greater or equal to 1 and its reciprocal value less than or equal to 1 which will satisfy the equation for any value of b of absolute value greater to or equal to 2.
In contrast, when c=-1, notice that the function has no extrema. While the equation
remains undefined at zero and so has a corresponding vertical asymptote, the first order derivative test for extrema
has no real solutions. In further contrast, if it is true for any value x that
then it is not possible for it to be true for its like-signed reciprocal with the trivial exception of 1 and -1 since
Instead, there is a one-to-one correspondence between all (without exception) values of x and the opposing sign reciprocal.
Uniqueness of Roots
For the equation
to have real roots, there must be a factorization
with n1 and n2 such that
The intersection of the graph of b=5 with the graph of the equation and shows there to be, at most two roots for any value of b. It can be shown that these roots are unique by the following proof. Suppose n1 and n2 are numbers such that . Next suppose there also exist two numbers m1 and m2 each distinct from n1 and n2 such that .
This is a contradiction which established the uniqueness of n1 and n2 as the graph suggest.
Graphs in the xc plane.In the following example the equation
is considered. If the equation is graphed in the xc plane, it is easy to see that the curve will be a parabola. For each value of c considered, its graph will be a line crossing the parabola in 0, 1, or 2 points -- the intersections being at the roots of the orignal equation at that value of c. In the graph, the graph of c = 1 is shown. The equation
will have two negative roots -- approximately -0.2 and -4.8.
There is one value of c where the equation will have only 1 real root -- at c = 6.25. For c > 6.25 the equation will have no real roots and for c < 6.25 the equation will have two roots, both negative for 0 < c < 6.25, one negative and one 0 when c = 0 and one negative and one positive when c < 0. This can be seen as a necessary consequence of the quadratic formula. Each root must satisfy
which only has real solutions for .
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