Question

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

Answer 1

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(x+1)`

`-2x^2-6x-4=-2(x^2+3x+2)=-2(x+1)(x+2)`

Therefore,

`f(x)=(-2x^2-6x-4)/(x^2+x)=(-2cancel(x+1)(x+2))/(xcancel(x+1))`

There is a hole at `x=-1`

As we cannot divide by `0`, `x!=0`

The vertical asymptote is `x=0`

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

`lim_(x->+-oo)f(x)=lim_(x->+-oo)(-2x)/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]}