# How do you identify all asymptotes or holes for f(x)=(2x-2)/(x^2-2x-3)?

Nov 24, 2016

The vertical asymptotes are $x = - 1$ and $x = 3$
No slant asymptote
The horizontal asymptote is $y = 0$
No holes

#### Explanation:

Let's factorise the denominator

${x}^{2} - 2 x - 3 = \left(x + 1\right) \left(x - 3\right)$

The domain of $f \left(x\right)$ is ${D}_{f \left(x\right)} = \mathbb{R} - \left\{- 1 , 3\right\}$

As you cannot divide by $0$, $x \ne - 1$ and $x \ne 3$

The vertical asymptotes are $x = - 1$ and $x = 3$

The degree of the numerator $<$ than the degree of the denominator, thereis no slant asymptote

We calculate the limits of $f \left(x\right)$, we take only the terms of highest degree.

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{2 x}{x} ^ 2 = {\lim}_{x \to - \infty} \frac{2}{x} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{2 x}{x} ^ 2 = {\lim}_{x \to + \infty} \frac{2}{x} = {0}^{+}$

The horizontal asymptote is $y = 0$

graph{(2x-2)/(x^2-2x-3) [-8.89, 8.89, -4.444, 4.445]}