College Algebra, Chapter 4, Chapter Review, Section Review, Problem 68

Graph the rational function $\displaystyle r(x) = \frac{1}{(x + 2)^2}$. Show clearly all $x$ and $y$ intercepts and asymptotes.

To graph the rational function, we first determine the $x$ and $y$ intercepts. So the $x$ intercept is undefined and does not exist based from the equation. To find the $y$ intercept we set $x = 0$. Then $\displaystyle r(0) = \frac{1}{( 0 + 2)^2} = \frac{1}{2^2} = \frac{1}{4}$

Thus, the $y$ intercept is $\displaystyle \frac{1}{4}$. Next, the vertical asymptotes are the zeros of the denominator so the line $x = -2$ is the vertical asymptote and since the degree of the numerator is less than the degree of the denominator, then the function has horizontal asymptote at $y =0$. After obtaining all the necessary informations, we want to know the behavior of the graph near the vertical asymptotes whether $y \to \infty$ or $y \to \infty$ on each side of the vertical asymptote. For instance, as $x \to -2^+$, we use a test value to chose to and to the right of $-2$ (say $x = -1.9$) to check whether $y$ is positive or negative to the right of $x = -2$.

$\displaystyle y = \frac{1}{(-1.9 + 2)^2}$ whose sign is $\displaystyle \frac{1}{(+)}$ (positive)

So $y \to \infty$ as $x \to -2^+$. Similarly, by using test values to the left of $-2$ (say $x= -2.1$), we found out that $y \to \infty$ as $x \to -2^-$. Thus, the graph of $r(x)$ is