by: Tim Lehman

Assignment #3

Some Different Ways to Examine


James W. Wilson and Tim Lehman
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


We will first find the vertices of. Above, we examined b=-3,-2,-1,0,1,2, and 3. If you are having trouble finding the vertices, this will be discussed more in depth later.



Does this match ?














 (-1/2, 3/4)








Thus, it appears the vertices of the graph are along the parabola for any points b.

This graph shows the data from the table above. Of course, this serves only to show the hypothesis probably holds. It remains to prove this.

Brief proof that the locus of vertices of the equation is y=-(x 2) + 1

First, we need to find the vertices at each point. The vertex of a parabola is the point where its derivative is zero. Thus, since, it follows that f '(x)=2x+b.


When f ' (x)=0, x=- (b/2). Using our equation f(x) and plugging in - (b/2) for x, we find that y=- (b2/4)+1. Therefore, the vertices are on the parametric equation


Since - (t/2)2+ 1=-(t2/4) + 1,

we see this parametric equation is equivalent to f(x) = - (x2) + 1.


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.

For one of the authors feeble attempt to show this earlier, click here.

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