College Algebra, Chapter 4, 4.6, Section 4.6, Problem 64

Find the intercepts and asymptotes of the rational function $\displaystyle r(x) = \frac{x^3 - x^2}{x^3 - 3x - 2}$ and then sketch its graph.

We first factor $r$, so $\displaystyle r(x) = \frac{x^2 (x - 1)}{x^3 - 3x - 2}$

Use synthetic Division to factor the denominator, then by trial and error







Thus, $\displaystyle r(x) = \frac{x^2 (x - 1)}{(x - 2)(x^2 + 2x + 1)} = \frac{x^2 (x - 1)}{(x - 2)(x + 1)^2}$

The $x$-intercepts are the zeros of the numerator $x = 0$ and $x = 1$.

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

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

the $y$-intercept is .

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 $x = 2.1$) to check whether $y$ is positive or negative to the right of $x = 2$.

$\displaystyle y = \frac{(2.1)^2 [(2.1) - 1]}{[(2. 1) - 2] [(2.1) + 1]^2}$ 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 $x = 1.9$), to obtain

$\displaystyle y = \frac{(1.9)^2 [(1.9) - 1]}{[(1.9) - 2][(1.9) + 1]^2}$ 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{Sign of } \frac{(x^2) (x - 1)}{(x -2 ) (x + 1)^2} & \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 $\displaystyle = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} = \frac{1}{1}$. Thus, the line $y = 1$ is the horizontal asymptote .