Assignment 3: Quadratics

By Krista Floer

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 they-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 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

Without calculus, show that the locus is a parabola. One approach is suggested by Item 6 in Assignment 2. If we complete the square with first two terms on the right hand side of the equation we get

Factoring, and replacing b by n for purposes of animating a graph we have

A quadratic in the form has coordinates of the vertices at (p, q). So the locus of our original equation will follow the points (p,q) where and . We then use these equations to solve as a system of equations. First we solve for b in the first equation. This gives us b = -2p. Substituting this into the equation for q, we get . From this we get that the locus of is . This can be seen below.

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