# How do you find the asymptotes for f(x) =(x^3-1)/(x^2-6)?

Jan 10, 2017

The vertical asymptotes are $x = - \sqrt{6}$ and $x = \sqrt{6}$
The slant asymptote is $y = x$
No horizontal asymptote

#### Explanation:

Let's factorise the denominator

${x}^{2} - 6 = \left(x - \sqrt{6}\right) \left(x + \sqrt{6}\right)$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- \sqrt{6} , \sqrt{6}\right\}$

As we cannot divide by $0$, $x \ne - \sqrt{6}$ and $x \ne \sqrt{6}$

The vertical asymptotes are $x = - \sqrt{6}$ and $x = \sqrt{6}$

As the degree of the numerator is $>$ than the degree of the denominator, there is a slant asymptote.

We perform a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3}$$\textcolor{w h i t e}{a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a}$∣${x}^{2} - 6$

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 6 x$$\textcolor{w h i t e}{a a a a}$color(white)(aaaa)∣$x$

$\textcolor{w h i t e}{a a a a a}$$0 + 6 x - 1$$\textcolor{w h i t e}{a a a a}$

Therefore,

$f \left(x\right) = \left(x\right) + \frac{6 x - 1}{{x}^{2} - 6}$

${\lim}_{x \to - \infty} \left(f \left(x\right) - x\right) = {\lim}_{x \to - \infty} \frac{6 x - 1}{{x}^{2} - 6} = {\lim}_{x \to - \infty} \frac{6 x}{x} ^ 2 = {\lim}_{x \to - \infty} \frac{6}{x} = {0}^{-}$

${\lim}_{x \to + \infty} \left(f \left(x\right) - x\right) = {\lim}_{x \to + \infty} \frac{6 x - 1}{{x}^{2} - 6} = {\lim}_{x \to + \infty} \frac{6 x}{x} ^ 2 = {\lim}_{x \to + \infty} \frac{6}{x} = {0}^{+}$

So, the slant asymptote is $y = x$

graph{(y-(x^3-1)/(x^2-6))(y-x)=0 [-28.86, 28.86, -14.43, 14.45]}