College Algebra, Chapter 4, 4.6, Section 4.6, Problem 72

Find the slant asymptote, the vertical asymptotes, and sketch a graph of the function $\displaystyle r(x) = \frac{2x^3 + 2x}{x^2 - 1}$.

By applying Long Division,







By factoring,

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

Thus,

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

Therefore, the line $y = 2x$ is the slant asymptote.

The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the lines $\displaystyle x = 1$ and $x = -1$ are the vertical asymptotes.

To sketch the graph of the function, we must first determine the intercepts.

$x$-intercepts: The $x$-intercepts are the zeros of the numerator, in our case, $x = 0$ is the only real $x$-intercept.

$y$-intercept: To find $y$-intercept, we set $x = 0$ into the original form of the function

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

The $y$-intercept is .

Next, we must determine the end behavior of the function near the vertical asymptote. By using test values, we found out that $y \to \infty$ as $\displaystyle x \to 1^+$ and $x \to - 1^+$. On the other hand as $y \to -\infty$ as $x \to -1^-$ and $\displaystyle x \to 1^-$. So the graph is