Some Different Ways to Examine


James W. Wilson and Carolyn Johnson

University of Georgia


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 two
real, positive roots at the point of tangency (the two roots are the same value, therefore, it is usually listed as one root). 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



For the graph , you notice that it is turned upside-down in comparision to all the other graphs. What caused the graph to be turned downward? the negative coefficient of . What effect does the +1 have on the graph? Since it is a positive 1, it causes the graph to be shifted upward one unit. So the negative coefficient caused the downward direction, and the positive 1 caused the upward shift of the graph. This graph contains as points all of the vertices of the other graphs. Notice that the lowest point of each graph is located on the graph. Therefore, is the locus of the vertices, the "path" that contains all the vertices of the graphs. Interesting!

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 = 5, 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




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 (x= -1/2, -5), one negative real root when b = 2 (x = -1), no real roots for -2 < b < 2, one positive real root when b = -2 (x = +1), and two positive real roots when b < -2 (x = +1, +5). All roots are approximations by viewing the above graph.

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



Now we have two hyperbolas instead of just one, both having the same assymtopes. The graph with the +1 seems to be squeezed within the boundaries of the assymtopes. The hyperbola with the -1 seems to be expanded along the outside of the assymtopes. The equations of the assymtopes are x = 0 and y = -x.


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. That occurs when 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.