The Department of Mathematics Education

J. Wilson, EMAT 6680

Some Different Ways to Examine


James W. Wilson and Jonathan Lawson
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


Show that the locus is the parabola


Looking at the equation of y = -x2 + 1, we can see that the x-intercepts are going to be x = -1 and 1 (by letting y equal zero).  

0 = -x2 + 1

x2 = 1

x = -1, 1

If we also look at the standard equation of a quadratic equation, 

y = a(x - h)2 + k,

where (h, k) is the vertex of the equation and a determines if it opens up or down and if it is narrower or wider than the a standard parabola.  So let us put the equation y = -x2 + 1 into standard form by rewriting it into 

y = -(x - 0)2 + 1, 

and it is plain to see that the vertex of this parabola is (0, 1) and that it opens down, just like the locus of the vertices of the parabolas above.  So if we take the last two points, and  , that have yet to be checked by the vertices we see that the also work in the equation y = -x2 + 1.  Plugging these points into the equation we can see they also work.  Since every point works and the x-intercepts and the vertex are all defined correctly by the equation y = -x2 + 1, we can see that the locus of the vertices of the parabola are this equation.

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.

Send e-mail to
Return to EMAT 6680 Home Page