EMAT 6680 Assignment 3

Some Different Ways to Examine

 

by

James W. Wilson and Paula Whitmire

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.

Let's observe the case of a= -3-purple, -2-red, -1 -blue, 0-green, 1- aqua, 2 - yellow, 3 -gray, and b=c=1. The following graph is obtained.

All of the graphs go throught the point (0,1) as the values of "a" changes. For negative values of "a", the parabola points down and therefore has 2 real roots. The parabolas with a positive "a" value all have a vertex above the x-axis so have no real roots. For a=0, the graph is a line. It appears that if the positive"a" parabolas were reflected about y=1 then translated up and shifted to the right , they become the graphs of the downward parabolas. The shift and translation appear to be small as the absolute value of "a" gets larger.

 

 

 

 

 

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

.

Show that the locus is the parabola

 

The locus of the parabolas is

It is the downward parabola is the picture below.

 

 

 

 

 

 

Graphs in the xb plane.

 

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 < -2.

Consider the case when c = - 1 rather than + 1.

 

 

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.\

 

 

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