College Algebra, Chapter 4, 4.6, Section 4.6, Problem 68

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

By applying Long Division,







By factoring,

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

Thus,

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

Therefore, $\displaystyle y = \frac{-1}{2}x + 1$ 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 $x = 3$

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

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

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