# How do you identify all asymptotes or holes for f(x)=(-2x^3-12x^2-16x)/(x^3+x^2-2x)?

Dec 13, 2016

The vertical asymptote is $x = 1$
The holes are when $x = 0$ and $x = - 2$

The horizontal asymptote is $y = - 2$

#### Explanation:

Let's factorise the numerator

$- 2 {x}^{3} - 12 {x}^{2} - 16 x = - 2 x \left({x}^{2} + 6 x + 8\right)$
$= - 2 x \left(x + 2\right) \left(x + 4\right)$

Lets factorise the denominator

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

So,

$f \left(x\right) = \frac{- 2 {x}^{3} - 12 {x}^{2} - 16 x}{{x}^{3} + {x}^{2} - 2 x}$

$= \frac{- 2 \cancel{x} \cancel{x + 2} \left(x + 4\right)}{\cancel{x} \cancel{x + 2} \left(x - 1\right)}$

$= \frac{- 2 \left(x + 4\right)}{x - 1}$

There are holes when $x = 0$ and $x = - 2$

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

So, the vertical asymptote is $x = 1$

Now, we calculate the limits as $x \to \infty$

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

The horizontal asymptote is $y = - 2$

graph{(y-(-2(x+4))/(x-1))=0 [-41.1, 41.1, -20.56, 20.56]}