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

1 Answer

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.