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

1 Answer
Dec 18, 2016

Answer:

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]}