College Algebra, Chapter 4, 4.6, Section 4.6, Problem 54

Find the intercepts and asymptotes of the rational function $\displaystyle r(x) = \frac{2x (x + 2)}{(x - 1)(x - 4)}$ and then sketch its graph.

The $x$-intercepts are the zeros of the numerator $x = 0$ and $x = -2$.

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

$\displaystyle r(0) = \frac{2(0) (0 + 2)}{(0 - 1)(0 - 4)} = \frac{0}{4} = 0$

the $y$-intercept is .

The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the lines $x = 1$ 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 $x \to 1^+$, we use a test value close to and to the right of $1$ (say $x = 1.1$) to check whether $y$ is positive or negative to the right of $x = 1$.

$\displaystyle y = \frac{2(1. 1) [(1.1) + 2]}{[(1.1) - 1] [(1.1) - 4]}$ whose sign is $\displaystyle \frac{(+)(+)}{(+)(-)}$ (negative)

So $y \to - \infty$ as $x \to 1^+$. On the other hand, as $x \to 1^-$, we use a test value close to and to the left of $1$ (say $x = 0.9$), to obtain

$\displaystyle y = \frac{2 (0.9) [(0.9) + 2]}{[(0.9) - 1][(0.9) - 4]}$ whose sign is $\displaystyle \frac{(+)(+)}{(-)(-)}$ (positive)

So $y \to \infty$ as $x \to 1^+$. The other entries in the following table are calculated similarly.

$
\begin{array}{|c|c|c|c|c|}
\hline\\
\text{As } x \to & 1^+ & 1^- & 4^+ & 4^- \\
\hline\\
\text{The sign of } \frac{2x (x + 2)}{(x - 1)(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 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 $y = 2$.