# How do you find vertical, horizontal and oblique asymptotes for #(4x^2-x+2)/(x+1)#?

##### 1 Answer

Aug 10, 2017

**Vertical asymptote is at**

**Slant asymptote is**

#### Explanation:

Vertical asymptote is fomed when deniminator is zero.

Here degree in numerator is

denominator . So we have a slant asymptote which is found by

doing long division . By long division we find Quotient as

graph{(4x^2-x+2)/(x+1) [-101.25, 101.2, -50.6, 50.7]} [Ans]