College Algebra, Chapter 4, 4.6, Section 4.6, Problem 70

Find the slant asymptote, the vertical asymptotes, and sketch a graph of the function $\displaystyle r(x) = \frac{x^3 + 4}{2x^2 + x - 1}$.

By applying Long Division,







By factoring,

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

Thus,

$\displaystyle r(x) = \frac{x^3 + 4}{2x^2 + x - 1} = \frac{1}{2} x + \frac{1}{4} + \frac{\displaystyle \frac{-3}{4} x + \frac{9}{4}}{2x^2 + x - 1}$

Therefore, $\displaystyle y = \frac{1}{2}x + \frac{1}{4}$ is the slant asymptote.

The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the lines $\displaystyle x = \frac{1}{2}$ 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, $x = \sqrt[3]{-4}$

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

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

The $y$-intercept is $-4$.

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 $\displaystyle x \to \frac{1}{2}^+$ and $x \to - 1^-$. On the other hand as $y \to -\infty$ as $x \to -1^+$ and $\displaystyle x \to \frac{1}{2}^-$. So the graph is