Precalculus Examples

$\frac{(x^{3} + 5) (2 x - 1)}{(x^{2} - 4) (x^{2} + 4)}$

Step 1

Find where the expression $\frac{(x^{3} + 5) (2 x - 1)}{(x^{2} - 4) (x^{2} + 4)}$ is undefined.

$x = - 2, x = 2$

Step 2

Since $\frac{(x^{3} + 5) (2 x - 1)}{(x^{2} - 4) (x^{2} + 4)}$ $\to$ $\infty$ as $x$ $\to$ $- 2$ from the left and $\frac{(x^{3} + 5) (2 x - 1)}{(x^{2} - 4) (x^{2} + 4)}$ $\to$ $- \infty$ as $x$ $\to$ $- 2$ from the right, then $x = - 2$ is a vertical asymptote.

$x = - 2$

Step 3

Since $\frac{(x^{3} + 5) (2 x - 1)}{(x^{2} - 4) (x^{2} + 4)}$ $\to$ $- \infty$ as $x$ $\to$ $2$ from the left and $\frac{(x^{3} + 5) (2 x - 1)}{(x^{2} - 4) (x^{2} + 4)}$ $\to$ $\infty$ as $x$ $\to$ $2$ from the right, then $x = 2$ is a vertical asymptote.

$x = 2$

Step 4

List all of the vertical asymptotes:

$x = - 2, 2$

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 $m$ .

$n = 4$

$m = 4$

Step 7

Since $n = m$ , the horizontal asymptote is the line $y = \frac{a}{b}$ where $a = 2$ and $b = 1$ .

$y = 2$

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: $x = - 2, 2$

Horizontal Asymptotes: $y = 2$

No Oblique Asymptotes

Step 10