Question

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

Answer 1

Vertical: `uarr x=-1 darr`.
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

`( x+1)(y-x+1)=0`

Graph of the hyperbola is inserted.

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