**Exploration With Second Degree Equations**

By: Denise Natasha Brewley

Consider the following equation of the form, *y = (x-d) ^{2}
- 2. *We will vary the constant

Note that varying *d *just translates each function
along the x-axis. That is, varying *d* is a horizontal translation on the graph. If we wanted to determine the zeros for each
of these graphs, we would simply just solve for *x* in each case by setting *y = 0*. This produces zeros of the form,
*x = d ±Ö2*. It is also important to note
here that varying *d *does not change the shape of each graph. It only
changes its position along the x-axis. Using the animate feature in
Graphing Calculator 3.2, the locus of points of the vertex of each graph would
be the line *y = -2,* which we already knew because of the horizontal
translation.

Now again consider the equation *y = (x-d) ^{2} - 2. *
It follows that the derivative is

*y' = 2(x-d).*

There are a few observations that we can make here. First,
for all values of *d *indicated in the graph above*, *the zeros for
the corresponding derivatives all pass through the line of symmetry.
So we can say that for *y = (x-d) ^{2} - 2, *its line of symmetry is