# EMAT 6680 Assignment 3

## Some Different Ways to Examine

### by James W. Wilson and Kristin Karas 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.

Here are some observations about the parabola:

1. We can discuss the "movement" of a parabola as b is changed.

2. The parabola always passes through the same point on the y-axis ( the point (0,1) with this equation).

3. 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).

4. 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.

5. 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.

We have already seen the graph of this parabola. The next question is what is the locus of the vertices of the set of parabolas that we just lookd at? Take a look at the motion of the parabola below and follow the vertices to see what forms.

.

From here we can determine that the locus is a parabola with vertex (0,1) and x-intercepts at 1 and -1. So, the equation would be represented as y= (x - 1)(x + 1). This product will give us , however, since the graph opens down we will need to change the signs to:

### 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. So the solutions of this equation would be where the blue line intersects the part of the graph opening upward.

Notice the following:

1. For each value of b we select, we get a horizontal line.

2. It is clear on a single graph that we get two negative real roots of the original equation when b > 2.

3. We get one negative real root when b = 2.

4. We get no real roots for -2 < b < 2 because there is not a graph in that section.

5. We get one positive real root when b = -2, and two positive real roots when b < -2.

Consider the case when c = - 1 rather than + 1.

Our graph looks like this: A hyperbola

If we overlap the graphs we can see:

### Let's look at graphs in the xc plane.

In the following example the equation

is considered.

1. If the equation is graphed in the xc plane, it is easy to see that the curve will be a parabola.

2. 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 original equation at that value of c.

3. In the graph, the graph of c = 1 is shown. The equation

will have two negative roots -- approximately -0.2 and -4.8.

Notice:

1. There is one value of c where the equation will have only 1 real root -- at c = 6.25.

2. For c > 6.25 the equation will have no real roots.

3. 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.

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