College Algebra, Chapter 4, 4.6, Section 4.6, Problem 48

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

The $x$-intercepts are the zeros of the numerator $x = -2$.

To find the $y$-intercept, we set $x = 0$ then

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

the $y$-intercept is $\displaystyle \frac{-2}{3}$.

The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the lines $x = -3$ and $x = 1$ are the vertical asymptotes.

We need to know whether $y \to \infty$ or $y \to - \infty$ on each side of each vertical asymptote. We use test values to determine the sign of $y$ for $x$- values near the vertical asymptotes. For instance, as $x \to -3^+$, we use a test value close to and to the right of $-3$ (say $x = -2.9$) to check whether $y$ is positive or negative to the right of $x = -3$.

$\displaystyle y = \frac{(-2.9) + 2}{[(-2.9) + 3][(-2.9)] - 1}$ whose sign is $\displaystyle \frac{(-)}{(+)(-)}$ (positive)

So $y \to \infty$ as $x \to -3^+$. On the other hand, as $x \to -3^-$, we use a test value close to and to left of $-3$ (say $x = -3.1$), to obtain

$\displaystyle y = \frac{(-3.1) + 2}{[(-3.1) + 3][(-3.1) - 1]}$ whose sign is $\displaystyle \frac{(-)}{(-)(-)}$ (negative)

So $y \to - \infty$ as $x \to -3^-$. The other entries in the following table are calculated similarly.

$\begin{array}{|c|c|c|c|c|}
\hline\\
\text{As } x \to & -3^+ & -3^- & 1^+ & 1^- \\
\hline\\
\displaystyle \text{the sign of } y = \frac{x + 2}{(x + 3)(x - 1)} \text{ is} & \frac{(-)}{(+)(-)} & \frac{(-)}{(-)(-)} & \frac{(+)}{(+)(+)} & \frac{(+)}{(+)(-)} \\
\hline\\
\text{So } y \to & \infty & - \infty & \infty & - \infty\\
\hline
\end{array} $

Horizontal Asymptote. Since the degree of the numerator is less than the degree of the denominator, then $y = 0$ is the horizontal asymptote.