Pre-Algebra Examples

$\frac{\frac{x^{2} + x - 6}{x^{2} - 4 x}}{\frac{4 x - 8}{x - 3}}$

Step 1

Find where the expression $\frac{(x + 3) (x - 3)}{4 x (x - 4)}$ is undefined.

$x = 0, x = 4$

Step 2

Since $\frac{(x + 3) (x - 3)}{4 x (x - 4)}$ $\to$ $- \infty$ as $x$ $\to$ $0$ from the left and $\frac{(x + 3) (x - 3)}{4 x (x - 4)}$ $\to$ $\infty$ as $x$ $\to$ $0$ from the right, then $x = 0$ is a vertical asymptote.

$x = 0$

Step 3

Since $\frac{(x + 3) (x - 3)}{4 x (x - 4)}$ $\to$ $- \infty$ as $x$ $\to$ $4$ from the left and $\frac{(x + 3) (x - 3)}{4 x (x - 4)}$ $\to$ $\infty$ as $x$ $\to$ $4$ from the right, then $x = 4$ is a vertical asymptote.

$x = 4$

Step 4

List all of the vertical asymptotes:

$x = 0, 4$

Step 5

Consider the rational function $R (x) = \frac{a x^{n}}{b x^{m}}$ where $n$ is the degree of the numerator and $m$ is the degree of the denominator.

1. If $n < m$ , then the x-axis, $y = 0$ , is the horizontal asymptote.

2. If $n = m$ , then the horizontal asymptote is the line $y = \frac{a}{b}$ .

3. If $n > m$ , then there is no horizontal asymptote (there is an oblique asymptote).

Step 6

Find $n$ and $m$ .

$n = 2$

$m = 2$

Step 7

Since $n = m$ , the horizontal asymptote is the line $y = \frac{a}{b}$ where $a = 1$ and $b = 4$ .

$y = \frac{1}{4}$

Step 8

There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.

No Oblique Asymptotes

Step 9

This is the set of all asymptotes.

Vertical Asymptotes: $x = 0, 4$

Horizontal Asymptotes: $y = \frac{1}{4}$

No Oblique Asymptotes

Step 10