# How do you find the vertical, horizontal or slant asymptotes for #f(x)=(2x^3+3x^2+x)/(6x^2+x-1)#?

##### 1 Answer

#### Answer:

The vertical asymptote is

The slant asymptote is

#### Explanation:

Let's start by fatorising the numerator and the denominator

and

So

As we cannot divide by

As the degree of the numerator is greater than the degree of the denominator, we would expect a slant asymptote. So we make a long division.

So

The slant asymptote is

There is no horizontal asymptote as limit of

graph{(x^2+x)/(3x-1) [-7.02, 7.024, -3.51, 3.51]}