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.
BACK TO DNB'S HOMEPAGE BACK TO VRAM'S HOMEPAGE