Find the Asymptotes (x^3-9x)/(3x^2-6x-9)

$\frac{x^{3} - 9 x}{3 x^{2} - 6 x - 9}$

Step 1

Find where the expression $\frac{x^{3} - 9 x}{3 x^{2} - 6 x - 9}$ is undefined.

$x = - 1, x = 3$

Step 2

Since $\frac{x^{3} - 9 x}{3 x^{2} - 6 x - 9}$ $\to$ $\infty$ as $x$ $\to$ $- 1$ from the left and $\frac{x^{3} - 9 x}{3 x^{2} - 6 x - 9}$ $\to$ $- \infty$ as $x$ $\to$ $- 1$ from the right, then $x = - 1$ is a vertical asymptote.

$x = - 1$

Step 3

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 4

Find $n$ and $m$ .

$n = 3$

$m = 2$

Step 5

Since $n > m$ , there is no horizontal asymptote.

No Horizontal Asymptotes

Step 6

Find the oblique asymptote using polynomial division.

Simplify the expression.

Simplify the numerator.

Factor $x$ out of $x^{3} - 9 x$ .

Factor $x$ out of $x^{3}$ .

$\frac{x \cdot x^{2} - 9 x}{3 x^{2} - 6 x - 9}$

Factor $x$ out of $- 9 x$ .

$\frac{x \cdot x^{2} + x \cdot - 9}{3 x^{2} - 6 x - 9}$

Factor $x$ out of $x \cdot x^{2} + x \cdot - 9$ .

$\frac{x (x^{2} - 9)}{3 x^{2} - 6 x - 9}$

Rewrite $9$ as $3^{2}$ .

$\frac{x (x^{2} - 3^{2})}{3 x^{2} - 6 x - 9}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$\frac{x (x + 3) (x - 3)}{3 x^{2} - 6 x - 9}$

Simplify the denominator.

Factor $3$ out of $3 x^{2} - 6 x - 9$ .

Factor $3$ out of $3 x^{2}$ .

$\frac{x (x + 3) (x - 3)}{3 (x^{2}) - 6 x - 9}$

Factor $3$ out of $- 6 x$ .

$\frac{x (x + 3) (x - 3)}{3 (x^{2}) + 3 (- 2 x) - 9}$

Factor $3$ out of $- 9$ .

$\frac{x (x + 3) (x - 3)}{3 x^{2} + 3 (- 2 x) + 3 \cdot - 3}$

Factor $3$ out of $3 x^{2} + 3 (- 2 x)$ .

$\frac{x (x + 3) (x - 3)}{3 (x^{2} - 2 x) + 3 \cdot - 3}$

Factor $3$ out of $3 (x^{2} - 2 x) + 3 \cdot - 3$ .

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

Factor $x^{2} - 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$\frac{x (x + 3) (x - 3)}{3 ((x - 3) (x + 1))}$

$\frac{x (x + 3) (x - 3)}{3 (x - 3) (x + 1)}$

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$\frac{x (x + 3) (x - 3)}{3 (x - 3) (x + 1)}$

Rewrite the expression.

$\frac{x (x + 3)}{3 (x + 1)}$

Expand $x (x + 3)$ .

Apply the distributive property.

$\frac{x \cdot x + x \cdot 3}{3 (x + 1)}$

Reorder $x$ and $3$ .

$\frac{x \cdot x + 3 x}{3 (x + 1)}$

Raise $x$ to the power of $1$ .

$\frac{x \cdot x + 3 x}{3 (x + 1)}$

Raise $x$ to the power of $1$ .

$\frac{x \cdot x + 3 x}{3 (x + 1)}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{x^{1 + 1} + 3 x}{3 (x + 1)}$

Add $1$ and $1$ .

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

Expand $3 (x + 1)$ .

Apply the distributive property.

$\frac{x^{2} + 3 x}{3 x + 3 \cdot 1}$

Multiply $3$ by $1$ .

$\frac{x^{2} + 3 x}{3 x + 3}$

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$3 x$

$3$

$x^{2}$

$3 x$

$0$

Divide the highest order term in the dividend $x^{2}$ by the highest order term in divisor $3 x$ .

					$\frac{x}{3}$
	$3 x$	+	$3$		$x^{2}$	+	$3 x$	+	$0$

Multiply the new quotient term by the divisor.

				$\frac{x}{3}$
$3 x$	+	$3$		$x^{2}$	+	$3 x$	+	$0$
			+	$x^{2}$	+	$x$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{2} + x$

				$\frac{x}{3}$
$3 x$	+	$3$		$x^{2}$	+	$3 x$	+	$0$
			-	$x^{2}$	-	$x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$\frac{x}{3}$
$3 x$	+	$3$		$x^{2}$	+	$3 x$	+	$0$
			-	$x^{2}$	-	$x$
					+	$2 x$

Pull the next terms from the original dividend down into the current dividend.

				$\frac{x}{3}$
$3 x$	+	$3$		$x^{2}$	+	$3 x$	+	$0$
			-	$x^{2}$	-	$x$
					+	$2 x$	+	$0$

Divide the highest order term in the dividend $2 x$ by the highest order term in divisor $3 x$ .

				$\frac{x}{3}$	+	$\frac{2}{3}$
$3 x$	+	$3$		$x^{2}$	+	$3 x$	+	$0$
			-	$x^{2}$	-	$x$
					+	$2 x$	+	$0$

Multiply the new quotient term by the divisor.

				$\frac{x}{3}$	+	$\frac{2}{3}$
$3 x$	+	$3$		$x^{2}$	+	$3 x$	+	$0$
			-	$x^{2}$	-	$x$
					+	$2 x$	+	$0$
					+	$2 x$	+	$2$

The expression needs to be subtracted from the dividend, so change all the signs in $2 x + 2$

				$\frac{x}{3}$	+	$\frac{2}{3}$
$3 x$	+	$3$		$x^{2}$	+	$3 x$	+	$0$
			-	$x^{2}$	-	$x$
					+	$2 x$	+	$0$
					-	$2 x$	-	$2$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$\frac{x}{3}$	+	$\frac{2}{3}$
$3 x$	+	$3$		$x^{2}$	+	$3 x$	+	$0$
			-	$x^{2}$	-	$x$
					+	$2 x$	+	$0$
					-	$2 x$	-	$2$
							-	$2$

The final answer is the quotient plus the remainder over the divisor.

$\frac{x}{3} + \frac{2}{3} - \frac{2}{3 x + 3}$

Split the solution into the polynomial portion and the remainder.

$\frac{x}{3} + \frac{2}{3} + \frac{- 2 x + 6}{3 x^{2} - 6 x - 9}$

The oblique asymptote is the polynomial portion of the long division result.

$y = \frac{x}{3} + \frac{2}{3}$

Step 7

This is the set of all asymptotes.

Vertical Asymptotes: $x = - 1$

No Horizontal Asymptotes

Oblique Asymptotes: $y = \frac{x}{3} + \frac{2}{3}$

Step 8

Find the Asymptotes (x^3-9x)/(3x^2-6x-9)

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