Question

How do you identify all asymptotes or holes and intercepts for `f(x)=1/(x-2)^2`?

Answer 1

The `y`-intercept is `(0,1/4)`; `y=0` is the horizontal asymptote; `x=2` is the vertical asymptote.

Explanation:

When `x=0` we have `f(0)=1/(0-2)^2=1/4`, so the `y`-intercept is `(0,1/4)`.

The equation `f(x)=0` has no solutions so there are no `x`-intercepts.

Since the degree of the numerator is less than the degree of the denominator we know that `y=0` is the horizontal asymptote. (Or you could justify it by saying `lim_(x\to oo)f(x)=0` if you've gotten to that concept yet.)

There are no holes because there are no common factors in the numerator and denominator.

`x=2` is the vertical asymptote because it is the zero of the denominator. (Or you could use limits to justify that as well.)