Assignment 3



Some Different Ways to Examine

by

James W. Wilson and Angel R. Abney
University of Georgia


It has now become a rather standard exercise, with availble 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 intersets 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

.

To show that the locus of vertices is a parabola, we'll use the vertex formula. Let , where a = 1.

Then, =

=

=

Therefore, we can see that the locus of vertices is the parabola:

, since

See graph.

 

Graphs in the xb plane.


Consider again the equation

Now graph this relation in the xb plane. We get the following graph.

b

x

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.


b

x

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 < -2.

Consider the case when c = - 1 rather than + 1. So we have the equation

See the red curve on the graph:


b

x

Notice that for each value of b, which corresponds to a horizontal on the graph, we get two real roots of the equation above. Note also, for all values of b, we get one positive and one negative x-intercept for the function

.

This is clear to see graphically, but it is also easy to show analitically. The discriminant,

is always greater than zero, which implies that the function has two distinct, real roots for all values of b. In fact, we can say that f will have two distinct, real roots for all values of b if a>0, and c<0.


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 below, the horizontal line at c = 1 is shown.

Thus, the equation

will have two negative roots -- approximately -0.2 and -4.8.

Notice, there is one value of c where the equation will have only 1 real root of multiplicity two -- 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 distinct, real roots: both negative for 0 < c < 6.25, one negative and one equal to zero when c = 0, and one negative and one positive when c < 0.


Graphs in the xa plane.

In the following example the equation

is considered. If the equation is graphed in the xa plane, it is easy to see that the curve is a rational function with a horizontal asymptote at y=0 and a vertical asymptote at x=0. For each value of a 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 a. In the graph below, the horizontal line at a= 2 is shown.

 


Thus, the equation

has two negative roots -- approximately x = -2.28 and x = -.22.

Notice, there are two values of a where the equation will have only 1 real root -- at a = 6.25 (root of multiplicity 2) and a = 0 (root of multiplicity 1). Note, if a = 0, the equation is no longer a parabola, but a line. For a > 6.25, the equation will have no real roots and for 0 < a < 6.25 the equation will have two distinct, real roots, both negative. For a < 0, the equation will have two distinct, real roots, one negative and one positive.

Again, this is easy to show analytically. The discriminant,

is only positive when a < 6.25.



Send e-mail to jwilson@coe.uga.edu
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