# How do you find vertical, horizontal and oblique asymptotes for #(x^4 -81) /(x^3+3x^2 - x - 3)#?

##### 2 Answers

#### Answer:

Invisible hole at (-3, -27/2). Slant asymptote :

#### Explanation:

Sans the hole at

graph{(x-3)(1+10/(x^2-1)) [-164, 164.1, -82.1, 82]}

graph{((x-3)(1+10/(x^2-1))-y)(y-x-3)(x-1+.01y)(x+1-.01y)=0 [-164, 164.1, -82.1, 82]}

graph{(x-3)(1+10/(x^2-1)) [-16, 16, -160, 160]}

#### Answer:

We have vertical asymptote at

#### Explanation:

First factorize numerator and denominators

and

Hence

=

Observe that although we have cancelled out

Further at

we have vertical asymptotes at

Here as degree of polynomial in numerator is just one more than that of denominator, we do not have a horizontal asymptote, but we do have a oblique or slant asymptote.

As

=

and as

we have an oblique or slant asymptote

graph{(x^4-81)/(x^3+3x^2-x-3) [-20, 20, -40, 40]}