,

Some Different Ways to Explore

By: Dr. James W. Wilson

and

Elizabeth Gore

It has now become a rather standard excercise, with available technology, to consturct 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, 2, 3, and overlay the graphs, the following picture is obtained.

We can discuss the "movement " of a parabola as b changes.

Notice that the parabola always passes through the same oint on the y-axis (the point (0,1) using the above equation).

For b < -2, the parabola will intersect the x-axis in two points with negative x values (i.e. the original equation has two positive real roots).

For b = -2, the parabola is tangents to the positive x-axis, and so the original equation has one positive real root. (at the opint of tangency).

For -2 < b < 2, the parabola does not intersect the x-axis; Therefore, the original equation has no real roots. (i.e. the roots are imaginary).

Similarly, for b = 2, the parabola is tangent to the negative x-axis (so the original equation has one negative real root).

And for b > 2, the parabola intersects the x-axis twice ( so the original equation has 2 negative real roots).

Now consider the locus of the vertices of the set of parabolas graphed from

We can show that the locus of the vertices is the parabola which has the equation

How do we do that?

We must use the vertex of the parabola and a point on the parabola.

We know that since the a value of the parabola is negative that the parabola opens downward.

Since our parabola must sit at the vertex of (0,1), we will look at the equation

Lets take the point on the parabola (-1,0) and find out whether a=-1.

We substitute in -1 for x and 0 for y and get the following equation.

Solving for a, we see that indeed

So we can say with confidence that the locus os the vertices is the parabola which has the equaiton

The figure shown below shows the original graph, plus the graph of the locus of the vertices.

What can you generalize?

Generalization:

The locus of the vertices contains the point (1,0) and has a vertexz of (0,1). This gives us a new parabola with the equation

If we substitute our point (1,0) into the equation, we will get

yeilding

so

so we get the equation of the parabola to be

or simply

In general, we get ( 0, c ) as the vertex of our parabola.

Graphs In the xa Plane

There are other ways to examine the roots of a quadratic equation by examining the graph.

One way is to graph in the xa plane instead of the xy plane.

This simply means that we will substitute y for a in the a quadratic equation.

Let us consider the equation

We can graph this relation in the xa plane by graphing the equaiton

Here is what we get:

If we take any particular value of a, and overlay the equation y = a on the graph, we add a line parrallel to the x-axis.

If the line intersects the curve in the curve in the xa plane, the intersection point correspoonds to the roots of the original equation for the value of a.

We can see on this single coordinate plane, that we get no real roots of the original equation when a > 0.25.

When a = 0.25, we get one negative real root.

When a < 0.25 we get two real roots.

Graphs in the xb Plane

Now, instead of graphing in the xy plane, we will graph in the xb plane.

This simply means we will substitute y for b in to a quadratic equation and then graph it.

Consider again the equation

.

Now we will graph this relation into the xb plane. To do this, we will need to graph the equation

This is the graph we get.

If we take a particular value of b and overlay the equation y = b on the graph,

we add a parallel line to the x-axis.

If the line intersects the curve in the xb plane,

the intersection points correspond to the roots of the original equation for the value of b.

Take this graph into consideration.

It is pretty clear that on this plane we get:

When b > 2, we get two real roots.

When b = 2, we get one real root.

When -2 < b < 2, there exist no real roots.

When b < -2, we get two real roots.

What if c = -1 rather than 1? Our equation would look like

What can you conclude about the roots of the new equation for different values of b by examining the above graph?

Graphs in the xc Plane

Finally, instead of graphing in the xy plane, we will graph in the xc plane.

Once again, this just means that we will substitute y for c in the quadratic equation.

Lets look at the equation

To graph this into the xc plane we need to graph

This is what we get

When the equaiton is graphed into the xc plane, it is easy to see that the curve will be a parabola.

If we take any particular value of c, and overlay the equation y = c, we add a line parrallel to the x-axis.

The intersections of these lines with the parabola yeilds the roots of the original equation for c.

It does not take a rocket scientist to see that..

When c > 2.25, you have no real roots

When c = 2.25, there exists one real root.

When c < 2.25 you get two real roots.

Once you get over the hill of actually understanding about changing coordinate planes, you have found a quick way to find roots of a quadratic equation.