Answer
vertical asymptotes: $x=0$
horizontal asymptote: $y=3$
Work Step by Step
Vertical asymptotes occur where the denominator is zero. Thus, we set the denominator to zero and solve for the $x$ values to find the vertical asymptotes:
$2x^3+5x^2+6x=0\\x(2x^2+5x+6)=0$
We cannot factor further because this polynomial with degree $2$ has a negative discriminant ($5^2-4(2)(6)=-23)$. Hence, the vertical asymptote is $x=0$
Here, the degrees of the numerator and the denominator are the same. Thus, the horizontal asymptote is the quotient of the leading coefficients:
$y=\frac{6}{2}=3$
