College Algebra, Chapter 4, 4.6, Section 4.6, Problem 66

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

By applying Long Division,







By factoring,

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

Thus,

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

Therefore, $y = x + 3$ is the slant asymptote.


The vertical asymptotes occur where the denominator is , that is, where the function is undefined. Hence the lines $x = 1$ is 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, $x = 0$ and $x = -2$

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

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

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 $x \to 1^+$ and $y \to - \infty$ as $x \to 1^-$. So the graph is