Question

What are the asymptotes of `(x^2 + x + 2)/ (3x +1)`?

Answer 1

`x = -1/3` as vertical asymptote

Explanation:

Generally, to find the vertical asymptotes of any rational function like `f(x)=(P(x))/(Q(x))`, with `P(x), Q(x)` polynomial of any degrees, we have to set the denominator `Q(x)=0`, because, of course, division by zero is undefined.
Once we solve for `x`, we can find the vertical asymptote. So in your case:
`3x+1 = 0`

`3x = -1`

`x = -1/3`.

And here's how your graph looks: graph{(x^2+x+2)/(3x+1) [-10, 10, -5, 5]}
As you can see the asymptote is `x=-1/3`.

The graph suggests there is also a slant asymptote for `x->+-oo`

To find it we need calculus definitions, but we are in Algebra branch, therefore I think it was asked of you to find only vertical asymptotes.