College Algebra, Chapter 4, 4.6, Section 4.6, Problem 52

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

We first factor $r$, so $\displaystyle r(x) = \frac{x - 2}{x (x - 4)}$

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

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

$\displaystyle r(0) = \frac{0 - 2}{0( 0 - 4)} = \frac{-2}{0}$

the $y$-intercept is does not exist.

The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the lines $x = 0$ and $x = 4$ 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 $x = 0$.

$\displaystyle y = \frac{(0 .1) - 2}{(0.1) [(0.1) - 4]}$ whose sign is $\displaystyle \frac{(-)}{(+)(-)}$ (positive)

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

$\displaystyle y = \frac{(-0.1) - 2}{(-0.1)[(-0.1) - 4]}$ whose sign is $\displaystyle \frac{(-)}{(-)(-)}$ (negative)

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 } x \to & 0^+ & 0^- & 4^+ & 4^- \\
\hline\\
\text{Sign of } y = \frac{x - 2}{x(x - 4)} & \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.

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