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

1 Answer

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.