Philippa M. Rhodes

## Write-up 3 Some Different Ways to Examine

### by Philippa M. Rhodes and Dr. James W. Wilson 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 wquation 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 positive real 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 intersects 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

We see that the vertices are symmetric about the y-axis and that the point (0, 1) is the 'highest' vertex for any of the parabolas graphed from the equation. Thus, we have a parabola concaved downward with vertex (1, 0) and given by

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 horizontal line. If it intersects the curve in the xb plane the itersection points correspond to the roots of the original equation for that value of b. We have the following graph.

This graph reinforces the prior conclusions about the roots of the original equation.

Now consider

By graphing in the xb plane

we see that for each value of b, the equation has two real roots, one negative and one positive.

For the next example, we will set

for c = 7, 6, 5, 1, 0, -1.

( When c = 7, the parabola crosses the y axis at 7. When c = 6, the parabola crosses the y axis at 6, and so on.)

We see that changing c, simply moves the parabola up and down. At some value of c, where 6< c <7, the x-axis is tangent to the parabola. Thus, all points greater than this value will have no real roots and all points below this value will have two real roots.

We can now consider the equation

.

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 the roots of the original equation at that value of c.

There is one value of c where the equation will have only one real root. That is 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 real roots. When 0 < c < 6.25, the two roots are negative. When c = 0, there is one negative root and one 0, and when c < 0, there is one negative and one positive real root of the equation.

Finally, consider the example

for a = -4, -2, 0, 2, 4, and 6.

When a is negative, the parabola is concaved downward, so it has no roots (in this case since c = -8). When a = 0, the eqation becomes linear, y = 2x - 8. Thus, it has one root. When a is positive, the parabola is upward and it has two roots, one positive and one negative.

On the xa plane, graph

We can see that it is true that when a is positive, the equation has one positive and one negative root. Also, there is one root when a = 0, but the other assumption is wrong. There are negative values of a that have one root. By enlarging the graph near the origin,

we see that these negative values are when a is between -0.2 and 0.