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

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Answer 1 · 2016-12-18T08:59:18

Vertical: $uarr x=-1 darr$ .
Slant : $y=x-1$

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

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

Graph of the hyperbola is inserted.

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

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