# How do you identify all asymptotes or holes for f(x)=(-2x^2-6x-4)/(x^2+x)?

Feb 25, 2017

The vertical asymptote is $x = 0$
The horizontal asymptote is $y = - 2$
There is a hole at $x = - 1$
No slant asymptote

#### Explanation:

Let's factorise the denominator and the numerator

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

$- 2 {x}^{2} - 6 x - 4 = - 2 \left({x}^{2} + 3 x + 2\right) = - 2 \left(x + 1\right) \left(x + 2\right)$

Therefore,

$f \left(x\right) = \frac{- 2 {x}^{2} - 6 x - 4}{{x}^{2} + x} = \frac{- 2 \cancel{x + 1} \left(x + 2\right)}{x \cancel{x + 1}}$

There is a hole at $x = - 1$

As we cannot divide by $0$, $x \ne 0$

The vertical asymptote is $x = 0$

The degree of the numerator $=$ the degree of the denominator, there is no slant asymptote.

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

The horizontal asymptote is $y = - 2$

graph{(y+2(x+2)/(x))(y+2)=0 [-20.28, 20.27, -10.14, 10.14]}