# What are the values and types of the critical points, if any, of f(x) = (x-5)/(x-3)^2?

Nov 11, 2016

The vertical asymptote is $x = 3$
The horizontal asymptote is $y = 0$
The intercepts are $\left(0 , - \frac{5}{9}\right)$ and $\left(5 , 0\right)$

#### Explanation:

As we cannot divide by $0$, the vertical asymptote is $x = 3$

The degree of the numerator is $<$ the dergee of the denominator, there is no oblique asymptote.
${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{x}{x} ^ 2 = {\lim}_{x \to - \infty} \frac{1}{x} = {0}^{-}$
${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{x}{x} ^ 2 = {\lim}_{x \to + \infty} \frac{1}{x} = {0}^{+}$
The horizontal asymptote is $y = 0$

The intercepts are
on the y-axis, $x = 0$$\implies$$f \left(0\right) = - \frac{5}{9}$
on the x-axis, $y = 0$$\implies$$0 = x - 5$$\implies$$x = 5$
graph{(x-5)/(x-3)^2 [-10, 10, -5.005, 4.995]}