College Algebra, Chapter 4, 4.6, Section 4.6, Problem 26

Find all horizontal and vertical asymptotes of the rational function $\displaystyle r(x) = \frac{8x^2+ 1}{4x^2 + 2x - 6}$.

Since the degree of the numerator is equal to the degree of the denominator, then the horizontal asymptote = $\displaystyle \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} = \frac{8}{4} = 2$. Thus, the horizontal asymptote is $y = 2$.

To determine the vertical asymptotes, we set $4x^2 + 2x - 6 = 0$.


$\begin{equation} \begin{aligned} 4x^2 + 2x - 6 =& 0 && \\ \\ x^2 + \frac{1}{2} x - \frac{3}{2} =& 0 && \text{Divide by } 4 \\ \\ (x - 1)\left( x + \frac{3}{2} \right) =& 0 && \text{Factor} \\ \\ x - 1 =& 0 \text{ and } x + \frac{3}{2} = 0 && \text{Zero Product Property} \end{aligned} \end{equation}$


Thus, the vertical asymptotes are $x = 1$ and $\displaystyle x = - \frac{3}{2}$.