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

Oct 2, 2016

$x = - \frac{1}{3}$ as vertical asymptote

#### Explanation:

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

$3 x = - 1$

$x = - \frac{1}{3}$.

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 $x = - \frac{1}{3}$.

The graph suggests there is also a slant asymptote for $x \to \pm \infty$

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.