You're almost completely correct. For the very fact that there are vertical asymptotes (which, by definition, go to infinity), there are neither absolute min's or max's.
However, consider what a relative min/max is compared to a absolute min/max.
Find all relative and absolute extrema in the function, k(x) = (x^2 - 4)^-2, on the interval [-3, 3].
My work:
k'(x) = [(-4x)] / [(x^2 - 4)^3]
Critical Points: x = 2, x = -2
Since there are vertical asymptotes at x = +/- 2, there are no relative extrema.
There is no absolute maximum because the function has no upper bound.
When x = -3, k(-3) = .04
When x = 3, k(3) = .04
Therefore, the function has absolute minimum of k(-3) = k(3) = .04
Basically there are:
No relative extrema
No absolute maximum
Absolute minimum at k(-3) and k(3)
Is this correct?
Any help is appreciated
There are no absolute max points because the peak of the graph occurs at x=2, which is not defined (as you found). There are also no absolute min points because it has a horizontal asymptote, causing the graph to get closer and closer to 0 for infinity. However, since you have the bounds of [-3,3], there is almost certainly either a relative max, min, or both. Here's a hint - you found it already, you just mislabeled it.
Remember, 95% of the time, if there are bounds given, there will be a relative max, min, or both.