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

Dec 16, 2017

The $y$-intercept is $\left(0 , \frac{1}{4}\right)$; $y = 0$ is the horizontal asymptote; $x = 2$ is the vertical asymptote.

#### Explanation:

When $x = 0$ we have $f \left(0\right) = \frac{1}{0 - 2} ^ 2 = \frac{1}{4}$, so the $y$-intercept is $\left(0 , \frac{1}{4}\right)$.

The equation $f \left(x\right) = 0$ 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 ${\lim}_{x \setminus \to \infty} f \left(x\right) = 0$ 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.)