Precalculus Examples

$f (x) = \frac{4 x^{2} - 16}{- 5 x + 10}$

Step 1

Find where the expression $\frac{4 x^{2} - 16}{- 5 x + 10}$ is undefined.

$x = 2$

Step 2

The vertical asymptotes occur at areas of infinite discontinuity.

No Vertical Asymptotes

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 $m$ .

$n = 2$

$m = 1$

Step 5

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

No Horizontal Asymptotes

Step 6

Find the oblique asymptote using polynomial division.

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Step 6.1

Simplify the expression.

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Step 6.1.1

Simplify the numerator.

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Step 6.1.1.1

Factor $4$ out of $4 x^{2} - 16$ .

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Step 6.1.1.1.1

Factor $4$ out of $4 x^{2}$ .

$\frac{4 (x^{2}) - 16}{- 5 x + 10}$

Step 6.1.1.1.2

Factor $4$ out of $- 16$ .

$\frac{4 x^{2} + 4 \cdot - 4}{- 5 x + 10}$

Step 6.1.1.1.3

Factor $4$ out of $4 x^{2} + 4 \cdot - 4$ .

$\frac{4 (x^{2} - 4)}{- 5 x + 10}$

Step 6.1.1.2

Rewrite $4$ as $2^{2}$ .

$\frac{4 (x^{2} - 2^{2})}{- 5 x + 10}$

Step 6.1.1.3

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 $b = 2$ .

$\frac{4 (x + 2) (x - 2)}{- 5 x + 10}$

Step 6.1.2

Simplify terms.

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Step 6.1.2.1

Factor $5$ out of $- 5 x + 10$ .

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Step 6.1.2.1.1

Factor $5$ out of $- 5 x$ .

$\frac{4 (x + 2) (x - 2)}{5 (- x) + 10}$

Step 6.1.2.1.2

Factor $5$ out of $10$ .

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

Step 6.1.2.1.3

Factor $5$ out of $5 (- x) + 5 (2)$ .

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

Step 6.1.2.2

Cancel the common factor of $x - 2$ and $- x + 2$ .

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Step 6.1.2.2.1

Factor $- 1$ out of $x$ .

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

Step 6.1.2.2.2

Rewrite $- 2$ as $- 1 (2)$ .

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

Step 6.1.2.2.3

Factor $- 1$ out of $- 1 (- x) - 1 (2)$ .

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

Step 6.1.2.2.4

Cancel the common factor.

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

Step 6.1.2.2.5

Rewrite the expression.

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

Step 6.1.2.3

Simplify the expression.

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Step 6.1.2.3.1

Multiply $- 1$ by $4$ .

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

Step 6.1.2.3.2

Move the negative in front of the fraction.

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

Step 6.2

Expand $- 4 (x + 2)$ .

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Step 6.2.1

Start expanding.

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

Step 6.2.2

Apply the distributive property.

$\frac{- 4 x - 4 \cdot 2}{5}$

Step 6.2.3

Multiply $- 4$ by $2$ .

$\frac{- 4 x - 8}{5}$

Step 6.3

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

$5$

$4 x$

$8$

Step 6.4

Divide the highest order term in the dividend $- 4 x$ by the highest order term in divisor $5$ .

		-	$\frac{4 x}{5}$
	$5$	-	$4 x$	-	$8$

Step 6.5

Multiply the new quotient term by the divisor.

	-	$\frac{4 x}{5}$
$5$	-	$4 x$	-	$8$
	-	$4 x$

Step 6.6

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x$

	-	$\frac{4 x}{5}$
$5$	-	$4 x$	-	$8$
	+	$4 x$

Step 6.7

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

	-	$\frac{4 x}{5}$
$5$	-	$4 x$	-	$8$
	+	$4 x$
		$0$

Step 6.8

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

	-	$\frac{4 x}{5}$
$5$	-	$4 x$	-	$8$
	+	$4 x$
		$0$	-	$8$

Step 6.9

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

$- \frac{4 x}{5} - \frac{8}{5}$

Step 6.10

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

$y = - \frac{4 x}{5}$

Step 7

This is the set of all asymptotes.

No Vertical Asymptotes

No Horizontal Asymptotes

Oblique Asymptotes: $y = - \frac{4 x}{5}$

Step 8