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 d which yields the following graphs:

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 x = d.  And for y' = 2(x-d) , its zero is x = d.  Furthermore, the derivative, y', always has positive slope for all values of x.