Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 34

Determine the horizontal and vertical asymptotes of the curve $\displaystyle y = \frac{x^2 + 1}{2x^2 - 3x -2}$

Solving for the vertical asymptotes

We set the denominator equal to zero


$
\begin{equation}
\begin{aligned}

2x^2 - 3x - 2 =& 0
\\
\\
(2x + 1)(x - 2) =& 0
\\
\\
2x + 1 =& 0
& x - 2 = 0
\\
\\
2x =& -1
& x = 2
\\
\\
\frac{\cancel{2}x}{\cancel{2}} =& \frac{-1}{2}
\\
\\
x =& \frac{-1}{2}

\end{aligned}
\end{equation}
$


So the horizontal asymptotes are $\displaystyle x = \frac{-1}{2}$ and $x = 2$

Solving for the horizontal asymptotes

In the equation $\displaystyle y = \frac{x^2 + 1}{2x^2 - 3x - 2}$ we remove everything except the biggest exponents of $x$ found in the numerator an denominator.

So we have


$
\begin{equation}
\begin{aligned}

y =& \frac{\cancel{x^2}}{2\cancel{x^2}}
\\
\\
y =& \frac{1}{2}

\end{aligned}
\end{equation}
$



Thus, the horizontal asymptote is $\displaystyle y = \frac{1}{2}$

Therefore,

the vertical asymptotes are $\displaystyle x = \frac{-1}{2}$ and $x = 2$ and the horizontal asymptote is $\displaystyle y = \frac{1}{2}$