College Algebra, Chapter 4, 4.6, Section 4.6, Problem 42

Find the intercepts and asymptotes of the rational function $\displaystyle r(x) = \frac{2x + 6}{-6x + 3}$ and then sketch its graph.

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


$
\begin{equation}
\begin{aligned}

0 =& \frac{2x + 6}{-6x + 3}
\\
\\
0 =& 2x + 6
\\
\\
2x =& -6
\\
\\
x =& -3

\end{aligned}
\end{equation}
$


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

$\displaystyle y = \frac{2(0) + 6}{-6 (0) + 3} = \frac{6}{3} = 2$

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

$\displaystyle r(x) = \frac{2x + 6}{-6x + 3}$

By performing division








$
\begin{equation}
\begin{aligned}
r(x) =& \frac{2x + 6}{-6x + 3}\\
\\
=& \frac{-1}{3} + \frac{7}{-6x + 3}
&&
\\
\\
=& \frac{-1}{3} + \frac{7}{-6} \left( \frac{1}{\displaystyle x - \frac{1}{2}} \right)
&& \text{Factor out $7$ and $-6$}
\\
\\
=& \frac{-1}{3} - \frac{7}{6} f \left( x - \frac{1}{2} \right)
&&

\end{aligned}
\end{equation}
$


It shows that the graph of $r$ is obtained by shifting the graph of $\displaystyle f \frac{1}{2}$ units to the right, reflecting the graph of $f$ about $x$-axis and stretching vertically by a factor of $\displaystyle \frac{7}{6}$. Then, the result is shifted $\displaystyle \frac{1}{3}$ units downward. Thus, $r$ has vertical asymptote at $\displaystyle x = \frac{1}{2}$ and horizontal asymptote at $\displaystyle y = \frac{-1}{3}$.