The Department of Mathematics Education

J. Wilson, EMAT 6680



Some Different Ways to Examine

by

James W. Wilson and Karl Mealor
University of Georgia

It has now become a rather standard exercise, with available technology, to construct graphs to consider the equation

by overlaying several graphs of

for different values of a, b, or c as the other two are held constant. For example, if we set

for , b = -3, -2, -1, 0, 1, 2 and 3 and overlay the graphs, the following picture is obtained.


b = -3, -2, -1, 0, 1, 2, 3

 

Now consider the solutions to the quadratic for b = -3, -2, -1, 0, 1, 2 and 3. The solutions of a quadratic equation are related to the discriminant, . The following table lists the values of the discriminant for the quadratics under consideration:

 Value of b

 Value of discriminant

 Number and type of solutions

 -3

 5

 2 irrational

 -2

 0

 1 rational

 -1

 -3

 2 complex

 0

 -4

 2 complex

 1

 -3

 2 complex

 2

 0

 1 rational

 3

 5

 2 irrational

Reexamine the graphs of in light of the information in the above table. Notice that when b = -2 and 2, the graph is tangent to the x-axis. We would expect this to happen because when y = 0 when b = -2 or 2, the resulting quadratic equation has 1 solution. When -2 < b < 2, the graph does not intersect the x-axis and the related quadratic has no real solutions. When b = -3 and 3, the graph intersects the y-axis in two places. We could use the points of intersection to approximate the irrational solutions to the related quadratic equations as -0.4 and -2.6, and 0.4 and 2.6, respectively.



Now consider the locus of the vertices of the set of parabolas graphed from

.

We wish to demonstrate that the locus of the vertices can be represented by the equation:

as demonstrated by the following graph.

graphed in red

Changing

to work form yields

Thus, the x-coordinates of the vertices of the parabolas determined by

can be represented by

while the y-coordinates can be represented by

.

This last equation can be rewritten as

Replacing -b/2 with x yields

or


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. In the next diagram, we have overlayed the graph of (graphed in red) on the previous graph.

Notice that any horizontal line y = b will intersect the graph in two places, implying that the quadratic will always have two real solutions. This makes sense when the discriminant is considered again. The discriminant of would be or which would always produce a positive value. A positive discriminant implies 2 real solutions.


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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