College Algebra, Chapter 4, 4.6, Section 4.6, Problem 58

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

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

The $x$-intercepts are the zeros of the numerator $x = -2$ and $x = 1$.

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

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

the $y$-intercept does not exist.

The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the lines $x = 0$ 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 $x \to 0^+$, we use a test value close to and to the right of (say $x = 0.1$) to check whether $y$ is positive or negative to the right of $x = 0$.

$\displaystyle y = \frac{2 [(0.1) + 2][(0.1) - 1]}{(0.1)[(0.1) + 1]}$ whose sign is $\displaystyle \frac{(+)(-)}{(+)(+)}$ (negative)

So $y \to - \infty$ as $x \to 0^+$. On the other hand, as $x \to 0^-$, we use a test value close to and to the left of (say $x = -0.1$), to obtain

$\displaystyle y = \frac{2 [(-0.1) + 2][(-0.1) - 1]}{(-0.1)[(-0.1) + 1]}$ whose sign is $\displaystyle \frac{(+)(-)}{(-)(+)}$ (positive)

So $y \to \infty$ as $x \to 0^-$. The other entries in the following table are calculated similarly.

$
\begin{array}{|c|c|c|c|c|}
\hline\\
\text{As } \to & 0^+ & 0^- & -1^+ & -1^- \\
\hline\\
\text{Sign of } \frac{2(x + 2)(x - 1)}{x(x + 1)} & \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 $\displaystyle = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} = \frac{2}{1}$. Thus, the horizontal asymptote is $y = 2$.