# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= (x^2+1)/(x+1)?

Dec 18, 2016

Vertical: $\uparrow x = - 1 \downarrow$.
Slant : $y = x - 1$

#### Explanation:

Let y = f(x) = x-1+2/(x+1). Then, reorganizing

(x+1)(y-x+1)=2 and this represents a hyperbola having the pair of

asymptotes given by

$\left(x + 1\right) \left(y - x + 1\right) = 0$

Graph of the hyperbola is inserted.

graph{x^2+1-(x+2)(y-x+1)=0 [-80, 80, -80, 80]}