College Algebra, Chapter 4, 4.6, Section 4.6, Problem 50

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

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

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

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

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

the $y$-intercept is $2$.

The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the lines $x = -2$ 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 $x \to -2^+$, 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 $x = -2$.

$\displaystyle y = \frac{2 (-1.9 - 2)}{(-1.9 + 2)(-1.9 - 1)}$ 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

$\displaystyle y = \frac{2 (-2.1 - 2)}{(-2.1 + 2)(-2.1 - 1)}$ whose sign is $\displaystyle \frac{(-)}{(-)(-)}$ (negative)

So $y \to - \infty$ as $x \to -2^-$. The other entries in the following table are calculated similarly.

$\begin{array}{|c|c|c|c|c|}
\hline\\
\text{As } x \to & -2^+ & -2^- & 1^+ & 1^- \\
\hline\\
\text{The sign of } y = \frac{2(x - 2)}{(x + 2)(x - 1)} & \frac{(-)}{(+)(-)} & \frac{(-)}{(-)(-)} & \frac{(-)}{(+)(+)} & \frac{(-)}{(+)(-)} \\
\hline\\
\text{So } y \to & \infty & - \infty & - \infty & \infty \\
\hline
\end{array} $

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