Question

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

Answer 1

The vertical asymptote is `x=1/3`
The slant asymptote is `y=x/3+4/9`

Explanation:

Let's start by fatorising the numerator and the denominator
`2x^3+3x^2+x=x(2x^2+3x+1)=x(2x+1)(x+1)`
and `6x^2+x-1=(3x-1)(2x+1)`

So `f(x)=(2x^3+3x^2+x)/(6x^2+x-1)=(xcancel(2x+1)(x+1))/((3x-1)cancel(2x+1))=(x(x+1))/((3x-1))`

As we cannot divide by `0`, so the vertical asymptote is `x=1/3`

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.
`x^2+x` `color(white)(aaaa)` `∣` `3x-1`
`x^2-x/3` `color(white)(aaa)` `∣` `x/3+4/9`
`0+(4x)/3`
`color(white)(aaaa)` `(4x)/3-4/9`
`color(white)(aaaaaa)` `0+4/9`

So `f(x)=x/3+4/9+(4/9)/(3x-1)`
The slant asymptote is `y=x/3+4/9`
There is no horizontal asymptote as limit of `f(x)` as `x->+-oo` is `+-oo`
graph{(x^2+x)/(3x-1) [-7.02, 7.024, -3.51, 3.51]}