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

##### 1 Answer

Dec 17, 2016

There is no hole. We have vertical asymptote at

#### Explanation:

Let us examine

As for the

It is also apparent that as

and as we have

Hence, we have a vertical asymptote at

Further as

As such,

Hence, as

and we have an oblique asymptote at

graph{(2x^2+3x-1)/(x+1) [-10, 10, -5, 5]}