Question

How do you find vertical, horizontal and oblique asymptotes for `(x^2+1)/(x+1)`?

Answer 1

`(x^2+1)/(x+1)` has a vertical asymptote `x=-1` and an oblique asymptote `y = x-1`

Explanation:

`f(x) = (x^2+1)/(x+1)`

`=(x^2-1+2)/(x+1)`

`=((x-1)(x+1)+2)/(x+1)`

`=x-1+2/(x+1)`

As `x->+-oo`, `2/(x+1) -> 0`

So `f(x)` has an oblique asymptote `y=(x-1)`

When `x=-1` the denominator `(x+1)` is zero, but the numerator `(x^2+1)` is non-zero.

Hence `f(x)` has a vertical asymptote at `x=-1`

graph{(y-(x^2+1)/(x+1))(y-x+1) = 0 [-20, 20, -10, 10]}