Expert: Josh - 4/9/2009

Question
In the form y=a(x-p)^2+q, would it be possible to reflect along the x-axis(Not at the y=0, but where y=q) and have changed all three values but still have a zero along the y=q line?

Answer
Hi Sean,

I'm not sure that I understand what you are asking.
Anyway, we basically have either an upright or inverted parabola, symmetrical about x=p. Concavity changes depending on the sign of "a". If the constant "a" is positive, it is "U" shape, if "a" is negative, it is "^" shape. In any event, we have a stationary (turning) point, either a global maximum at x=p (if a<0) or a minimum at x=p (if a>0).

The constant q simply provides a vertical offset.

The part that confuses me is "still have a zero along the y=q line". The zero(s) (or solutions to a quadratic equation) are by definition the points where the parabola crosses the x-axis. So, I'm trying to understand what you are referring to.

In the simplest case where q=0, we can flip the parabola vertically (for example, changing it from "U" to "^" if a>0), with the stationary point still located at y=q by changing the sign of "a" from "+a" to "-a". For example: y=3(x-p)^2 is "U" shape with a minimum point (repeated roots in fact) at x=p and y=0. On the other hand, y=-3(x-p)^2 has the same shape but is up-side-down, with a maximum at x=p and again with y=0.

Now, if q is non-zero, say q=1, using the same numerical values, y=3(x-p)^2+1 is simply the same parabola as y=3(x-p)^2 being pushed up by 1 unit along the vertical axis. Note carefully that changing "a" from +3 to -3 reflects the parabola about y=q, NOT the x-axis. If we wish to reflect y=3(x-p)^2+1 about the x-axis (y=0), we can do this in two steps.

Step 1: Reflect y=3(x-p)^2+1 about y=q first by changing "a" to "-a". i.e., consider y=-3(x-p)^2+1. This has a maximum value at (x=p,y=1).

Step 2: Provide a downward shift of 2q units, so that the new function is a mirror image of the original y=3(x-p)^2+1 about the x-axis.

This gives y=-3(x-p)^2+1-2q, where q=1.

So, reflecting y=a(x-p)^2+q about y=q simply requires changing the sign of "a". We don't tough "p" at all because we want the stationary point to stay at x=p. We only change "q" if we want the new function to be reflect about y=k for some k value different from q.