Assignment 3: Locus of *ax*^{2} + *bx* + *c* = 0

by Shawn Broderick

For this assignment, I have chosen to do the first question.

To refresh our memory of the directions and explanation:

Investigation 1.

It has now become a rather standard exercise, with availble technology, to construct graphs to consider the equation *ax*^{2} +* bx* + *c* = 0 and to overlay several graphs of *y* = *ax*^{2 }+ *bx* + *c* 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 *ax*^{2 }+ *bx* + *c* = 0
can be followed. For example, if we set *y* = *x*^{2 }+ *bx* + 1 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 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 intersets 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 *y* = *x*^{2 }+ *bx* + 1. Without calculus, show that the locus is a parabola. We shall proceed from here to show how to find the locus of this equation without calculus.

This investigation is going to be carried out in Geometer's Sketchpad.

File: assignment3.gsp