Find the Asymptotes f(x)=(x^2-9)/x

$f (x) = \frac{x^{2} - 9}{x}$

Step 1

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

$x = 0$

Step 2

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 3

Find $n$ and $m$ .

$n = 2$

$m = 1$

Step 4

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

No Horizontal Asymptotes

Step 5

Find the oblique asymptote using polynomial division.

Simplify the numerator.

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

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

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 $b = 3$ .

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

Expand $(x + 3) (x - 3)$ .

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Reorder $x$ and $- 3$ .

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Multiply $3$ by $- 3$ .

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

Add $- 3 x$ and $3 x$ .

$\frac{x^{2} + 0 - 9}{x}$

Subtract $9$ from $0$ .

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

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

$x$

$0$

$x^{2}$

$0 x$

$9$

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

					$x$
	$x$	+	$0$		$x^{2}$	+	$0 x$	-	$9$

Multiply the new quotient term by the divisor.

				$x$
$x$	+	$0$		$x^{2}$	+	$0 x$	-	$9$
			+	$x^{2}$	+	$0$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{2} + 0$

				$x$
$x$	+	$0$		$x^{2}$	+	$0 x$	-	$9$
			-	$x^{2}$	-	$0$

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

				$x$
$x$	+	$0$		$x^{2}$	+	$0 x$	-	$9$
			-	$x^{2}$	-	$0$
						$0$

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

				$x$
$x$	+	$0$		$x^{2}$	+	$0 x$	-	$9$
			-	$x^{2}$	-	$0$
						$0$	-	$9$

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

$x - \frac{9}{x}$

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

$y = x$

Step 6

This is the set of all asymptotes.

Vertical Asymptotes: $x = 0$

No Horizontal Asymptotes

Oblique Asymptotes: $y = x$

Step 7

Find the Asymptotes f(x)=(x^2-9)/x

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