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

Generalize.

**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.

Consider the cases when c=5, 3, 1,-1, 3, 5.