College Algebra, Chapter 4, 4.6, Section 4.6, Problem 62

Find the intercepts and asymptotes of the rational function $\displaystyle r(x) = \frac{5x^2 + 5}{x^2 + 4x + 4}$ and then sketch its graph.

We first factor $r$, so $\displaystyle r(x) = \frac{5 (x^2 + 1)}{(x + 2)^2}$

The $x$-intercepts are the zeros of the numerator. In our case, the $x$-intercept does not exist.

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

$\displaystyle r(0) = \frac{5 [(0^2) + 1]}{(0 + 2)^2} = \frac{5(1)}{2^2} = \frac{5}{4}$

the $y$-intercept is $\displaystyle \frac{5}{4}$.

The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the lines $x = -2$ is 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 $x \to -2^+$, we use a test value close to and to the right of $-2$ (say $x = -1.9$) to check whether $y$ is positive or negative to the right of $x = -2$.

$\displaystyle y = \frac{5 [(-1.9)^2 + 1]}{[(-1.9) + 2]^2}$ whose sign is $\displaystyle \frac{(+)}{(+)}$ (positive)

So $y \to \infty$ as $x \to -2^+$. On the other hand, as $x \to -2^-$, we use a test value close to and to the left of $-2$ (say $x = -2.1$), to obtain

$\displaystyle y = \frac{5 [(-2.1)^2 + 1]}{[(-2.1) + 2]^2}$ whose sign is $\displaystyle \frac{(+)}{(+)}$ (positive)

So $y \to \infty$ as $x \to -2^-$.

Horizontal Asymptote. Since the degree of the numerator is equal to the degree of the denominator, then the horizontal asymptote $\displaystyle = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} = \frac{5}{1}$. Thus, the line $y = 5$ is the horizontal asymptote .