College Algebra, Chapter 4, 4.6, Section 4.6, Problem 60

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

We first factor $r$, so $\displaystyle r(x) = \frac{x (x + 3)}{(x - 3)(x + 2)}$

The $x$-intercepts are the zeros of the numerator $x = 0$ and $x = -3$.

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

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

the $y$-intercept is .

The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the lines $x = 3$ and $x = -2$ 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 $x = 3.1$) to check whether $y$ is positive or negative to the right of $x = 3$.

$\displaystyle y = \frac{(3.1) [(3.1) + 3]}{[(3.1) - 3][(3.1) + 2]}$ 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 the left of $3$ (say $x = 2.9$), to obtain

$\displaystyle y = \frac{(2.9) [(2.9) + 3]}{[(2.9) - 3][(2.9) + 2]}$ 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 } & 3^+ & 3^- & -2^+ & -2^- \\
\hline\\
\text{Sign of } \frac{x (x + 3)}{(x - 3)(x + 2)}& \frac{(+)(+)}{(+)(+)} & \frac{(+)(+)}{(-)(+)} & \frac{(-)(+)}{(-)(+)} & \frac{(-)(+)}{(-)(-)} \\
\hline\\
y \to & \infty & - \infty & \infty & - \infty\\
\hline
\end{array} $

Horizontal Asymptote. Since the degree of the numerator is equal to the degree of the denominator, then the horizontal asymptote $\displaystyle = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} = \frac{1}{1}$. Thus, the horizontal asymptote is $y = 1$.