EMAT 6680 Assignment 3

Last modified on September 5, 2009
Study the background information. Then examine the investigations and select something from one or more of the investigations for an Assignment 3 write-up.

Background.

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

Without calculus, show that the locus is a parabola. One approach is suggested by Item 6 in Assignment 2. If we complete the square with first two terms on the right hand side of the equation we get

Factoring, and replacing b by n for purposes of animating a graph we have

The point (d,f) is the vertex of each parabola where

That is, the parabolas are in the form . Now, (d, f) is the vertex of a parabola corresponding to any n. Set x = d and y = f and solve each for n. Set these two expressions for n equal to one another and the resulting equation is

Here is a Graphing Calculator animation:

Investigation 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 = 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 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.

Graph other values of c on the same axes.

Investigation 2.

Add the graph of 2x + b = 0 to the picture and discuss its relation to the quadratic formula.

Investigation 3.

Consider graphs in the xc plane.

Investigation 4.

Consider graphs in the xa plane

Investigation 5.

Consider the equation

Explore the pattern of roots in the xb, xc, or xd planes. (Okay, maybe look at the xa plane too!)

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