College Algebra, Chapter 4, 4.6, Section 4.6, Problem 44

Find the intercepts and asymptotes of the rational function $\displaystyle r(x) = \frac{1 - 2x}{2x + 3}$ and then sketch its graph.

To determine the $x$ intercept, we set $y = 0$, so


$
\begin{equation}
\begin{aligned}

0 =& \frac{1 - 2x}{2x + 3}
\\
\\
0 =& 1 - 2x
\\
\\
2x =& 1
\\
\\
x =& \frac{1}{2}


\end{aligned}
\end{equation}
$


To determine the $y$ intercept, we set $x = 0$

$\displaystyle y = \frac{1 - 2(0)}{2(0) + 3} = \frac{1}{3}$

Next, if we let $\displaystyle f(x) = \frac{1}{x}$, then we can express $r$ in terms of $f$ as follows:

$\displaystyle r(x) = \frac{1 - 2x}{2x + 3}$

By performing division








$
\begin{equation}
\begin{aligned}
r(x) =& \frac{1 - 2x}{2x + 3}\\
\\
=& -1 + \frac{4}{2x + 3}
&&
\\
\\
=& -1 + \frac{4}{2} \left( \frac{1}{\displaystyle x + \frac{3}{2}} \right)
&& \text{Factor out $4$ and $2$}
\\
\\
=& -1 + 2 f \left(x + \frac{3}{2} \right)
&&

\end{aligned}
\end{equation}
$


It shows that the graph of $r$ is obtained by shifting the graph of $\displaystyle f \frac{3}{2}$ units to the left, and stretching vertically by a factor of $\displaystyle 2$. Then, the result is shifted $1$ unit downward. Thus, $r$ has vertical asymptote at $\displaystyle x = \frac{-3}{2}$ and horizontal asymptote at $\displaystyle y = -1$.