Answer
horizontal asymptote: $y=5$
Work Step by Step
We have the function:
$r(x)=\dfrac{5x^3}{x^3+2x^2+5x}=\dfrac{5x^2}{x^2+2x+5}$
Vertical asymptotes occur where the denominator is zero.
We cannot factor the denominator further because the polynomial with degree $2$ has a negative discriminant ($2^2-4(1)(5)=-16)$. This means that the denominator is never zero (for real numbers). Hence, this function has no vertical asymptotes.
Here, the degrees of the numerator and the denominator are the same. Thus, the horizontal asymptote is the quotient of the leading coefficients: $y=\frac{5}{1}=5$
