Question

How do you identify all the asymptote of `f(x)= (x^2+1)/(x+1)`?

Answer 1

Vertical: `x=-1`
Horizontal: none
Oblique (skew, slant) `y = x-1`

Explanation:

Vertical asymptotes occur where we have the reduced form denominator `=0`.

`f(x) = (x^2+1)/(x+1)` cannot be reduced so there is a vertical asymptote where the denominator is `0`. The denominator is `0` at `-1`, so the equation of the vertical asymptote is `x=-1`

Because the degree on the numerator is greater than that of the denominator, the values of `f(x)` increase without bound as `x` increases without bound. (`f(x) rarroo` as `xrarroo`).

So, there is no horizontal asymptote.

The degree of the numerator is `1` greater than that of the denominator, so there is an oblique asymptote. (also called a slant or skew asymptote).

Divide to rewrite the function:

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

The line `y = x-1` is an asymptote. (On both sides.)