College Algebra, Chapter 4, 4.6, Section 4.6, Problem 30

Find all horizontal and vertical asymptotes of the rational function $\displaystyle r(x) = \frac{5x^3}{x^3 + 2x^2 + 5x}$.

Since the degree of the numerator is equal to the degree of the denominator, then the horizontal asymptote = $\displaystyle \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} = \frac{5}{1}$. Thus, the horizontal asymptote is $\displaystyle y = 5$.

To determine the vertical asymptotes, we set the denominator equal to .


$
\begin{equation}
\begin{aligned}

& x^3 + 2x^2 + 5x = 0
&& \text{Given}
\\
\\
& x(x^2 + 2x + 5) = 0
&& \text{Factor out } x
\\
\\
& x = 0 \text{ and } x^2 + 2x + 5 = 0
&& \text{Zero Product Property}

\end{aligned}
\end{equation}
$


Since the discriminant of $x^2 + 2x + 5 < 0$, then the function has complex roots. Therefore, the vertical asymptote is only at $x = 0$.