Locus
of Vertices

#### by

#### Rachael Brown

The idea behind this investigation is to figure
out what the locus of the vertices of the set of parabolas graphed
from y = x^2 + bx +1. Then the challenge is to extend this to
all parabolas.

I began by graphing a few parabolas on Graphing
Calculator. Then I started to think algebraically. I wondered
how to find the vertex of any parabola given in standard form.
Here is my algebra to find the vertex:

In the example, we are focusing on a = 1, b
= b, and c = 1. So the vertex of these parabolas would be of the
form

In order to get a picture of what these points
look like for a variety of values of b, I decided to use parametric
equations. I set the x = x-coordinate and the y = y-coordinate.
Here is a picture of what the graph looks like:

Are you surprised that its a parabola? To find
the equation of this parabola, let's solve the x equation for
t and plug it into the y equation.

So, now that we've found the equation for the
parabola that is the locus of the vertices of parabolas with the
equation y = x^2 + bx +1, let's see if we can generalize this.
We know that the vertices of parabolas with the equation y = ax^2
+ bx + c is

So we can graph this the same way that we did
earlier. Unfortunately, I can only use one variable at a time
to vary. So let's let a = n, b = t, and c = 2. Here is a picture
of what we get:

I can also create a picture where c = n and
a = 2 or any other number. Click here
to open up a Graphing Calculator file to see what this looks like.

Next, lets see if we can algebraically find
the equation for the locus of vertices. It seems like its always
going to be a parabola. Let's see...

To generalize, we have shown that for any parabola
of the form y = ax^2 + bx + c, the locus of its vertices forms
a parabola with equation y = c - ax^2.

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