Write -Up #2

Some different Ways to Examine

by
James W. Wilson and Michael R. Akes
University of Georgia

It has now become a rather standard exercise in Algebra II, 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 s 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 point with positive x values (ie. 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 -1 < 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 vertices of the set of parabolas created from

We can show that the locus is the parabola

The locus of the vertices is being controlled by the c term at the end and the fact that they all have the same a term. Notice what happens if we change the c=1 to c=2.

Now we have changed the locus of the vertices to

The following picture shows this

We should now consider the a term. Notice what happens if we can the a=1 to a=2 and we leave the a coefficient the same on the locus's graph

It is clear that locus's graph is not the one we are looking for. So we should change

to

The following graph shows that this is now the locus of the vertices.

In general, if we consider the equation

and would like to find the locus of the vertices, we need to consider both the a term and the c term. The b term ,as long as a and c remains the same, will just cause a shift of the parabola with the vertex remaining on the locus of vertices determined by a and c.

Consider 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.

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 , on e negative real root when b=2, no real roots for -2 < b < 2, one positive real root when b=-2, and two positive real root when b=-2, and two positive real roots when b < -2.

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

What we notice is that there is no value for b that will not have two roots, and one of the roots must be positive and one must be negative. This due to the c term at the end being negative. For two number two multiply together and give use a negative one must be positive and the other negative.

In the following example

is considered. 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 at the roots of the original equation at that value of c. In the graph, the graph of c=1 is shown. The equation

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

There is one value of c where the equation will have only 1 real root-- at c=6.25. For c > 6.25 the equation will have no real roots and for c < 6.25 the equation will have two teal 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.

The question at this point is why is there no real roots for c > 6.25. We need to consider the x value at this point and how is corresponds to the b value. The x value for c=6.25 is -2.5. This is exactly 1/2 of the b coefficient of 5. This would lead us to believe that for an equation in the form

that there will not be any real roots for c value that corresponds to an b value that is larger than 1/2 the value of b. Take for example the equation

We will not be able to have a real root for any c that has a corresponding x value that is larger than 4.5.

If we consider the family of parabolas that is graphed for

with the use of several values for b such as b=9, 8, 6, 3, and look at the graph the will generate the locus of all the values for which is the maximum value for which c will have a real root . The equation

will create these maximums.

The use of these two equation together will allow use to know before we ever graph if the c value we are considering has a real root.