College Algebra, Chapter 4, 4.6, Section 4.6, Problem 46

Find the intercepts and asymptotes of the rational function $\displaystyle r(x) = \frac{x - 2}{(x + 1)^2}$ 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

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

the $y$-intercept is $2$.

The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the line $x = -1$ is the vertical asymptote.

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 $x \to -1^+$, we use a test value close to and to the right of $-1$ (say $x = -0.9$) to check whether $y$ is positive or negative to the right of $x = -1$.

$\displaystyle y = \frac{(-0.9) - 2}{(-0.9 + 1)^2}$ whose sign is $\displaystyle \frac{(-)}{(+)}$ (negative)

So $y \to - \infty$ as $x \to -1^+$. On the other hand, as $x \to -1^-$, we use a test value close to and to left of $-1$ (say $x = -1.1$), to obtain

$\displaystyle y = \frac{(-1.1) -2}{(-1.1 + 1)^2}$ whose sign is $\displaystyle \frac{(-)}{(+)}$ (negative)

So $y \to - \infty$ as $x \to -1^-$.

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

Graph. Use the information we have found together with some additional values in the table to sketch the graph of the function.

$\begin{array}{|c|c|}
\hline\\
x & r(x) \\
\hline\\
-6 & -0.32 \\
-4 & -0.666 \\
-2 & -4 \\
2 & 0 \\
4 & 0.08 \\
6 & 0.0816\\
\hline
\end{array} $