College Algebra, Chapter 4, 4.6, Section 4.6, Problem 56

Find the intercepts and asymptotes of the rational function $\displaystyle r(x) = \frac{4x^2}{x^2 - 2x - 3}$ and then sketch its graph.

We first factor $r$, so $\displaystyle r(x) = \frac{4x^2}{(x - 3)(x + 1)}$

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

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

$\displaystyle r(0) = \frac{4 (0)^2}{(0 - 3)(0 + 1)} = \frac{0}{-3} = 0$

the $y$-intercept is .

The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the lines $x = 3$ 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 $x \to 3^+$, we use a test value close to and to the right of $3$ (say $x = 3.1$) to check whether $y$ is positive or negative to the right of $x = 3$.

$\displaystyle y = \frac{4(3.1)^2}{[(3.1) - 3][(3.1) + 1]}$ whose sign is $\displaystyle \frac{(+)}{(+)(+)}$ (positive)

So $y \to \infty$ as $x \to 3^+$. On the other hand, as $x \to 3^-$, we use a test value close to and to the left of $3$ (say $x = 2.9$), to obtain

$\displaystyle y = \frac{4 (2.9)^2}{[(2.9) - 3][(2.9) + 1]}$ whose sign is $\displaystyle \frac{(+)}{(-)(+)}$ (negative)

So $y \to - \infty$ as $x \to 3^-$. The other entries in the following table are calculated similarly.

$
\begin{array}{|c|c|c|c|c|}
\hline\\
\text{As } x \to & 3^+ & 3^- & -1^+ & -1^- \\
\hline\\
\text{Sign of } \frac{4x^2}{(x - 3)(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 $\displaystyle = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} = \frac{4}{1}$. Thus, the horizontal asymptote is $y = 4$.