Question

How do you identify all asymptotes or holes for `g(x)=(2x^2+3x-1)/(x+1)`?

Answer 1

There is no hole. We have vertical asymptote at `x=-1` and oblique asymptote at `y=2x+1`

Explanation:

Let us examine `g(x)=(2x^2+3x-1)/(x+1)`.

As for the `x=-1`, although denominator `x+1=0` but `2x^2+3x-1!=0`, there is no common factor between numerator and denominator and hence we do not have any hole.

It is also apparent that as `x->-1`, `(2x^2+3x-1)/(x+1)->oo`

and as we have `f(x)=(x^2+11x+18)/(2x+1)`, `f(x)->oo` or `f(x)->-oo`, as we approach `x=-1` from negative or positive side.

Hence, we have a vertical asymptote at `x=-1/2`.

Further as `2x^2+3x-1=2x(x+1)+1(x+1)-2`

As such, `f(x)=(2x^2+3x-1)/(x+1)=2x+1-2/(x+1)`

Hence, as `x->oo`. `f(x)->2x+1`

and we have an oblique asymptote at `y=2x+1`
graph{(2x^2+3x-1)/(x+1) [-10, 10, -5, 5]}