Precalculus Examples

$\frac{- 4 x^{2} - 8 x + 5}{2 x + 1}$

Step 1

Find where the expression $\frac{- 4 x^{2} - 8 x + 5}{2 x + 1}$ is undefined.

$x = - \frac{1}{2}$

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.

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Step 5.1

Simplify the expression.

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Step 5.1.1

Factor by grouping.

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Step 5.1.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 4 \cdot 5 = - 20$ and whose sum is $b = - 8$ .

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Step 5.1.1.1.1

Factor $- 8$ out of $- 8 x$ .

$\frac{- 4 x^{2} - 8 (x) + 5}{2 x + 1}$

Step 5.1.1.1.2

Rewrite $- 8$ as $2$ plus $- 10$

$\frac{- 4 x^{2} + (2 - 10) x + 5}{2 x + 1}$

Step 5.1.1.1.3

Apply the distributive property.

$\frac{- 4 x^{2} + 2 x - 10 x + 5}{2 x + 1}$

Step 5.1.1.2

Factor out the greatest common factor from each group.

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Step 5.1.1.2.1

Group the first two terms and the last two terms.

$\frac{(- 4 x^{2} + 2 x) - 10 x + 5}{2 x + 1}$

Step 5.1.1.2.2

Factor out the greatest common factor (GCF) from each group.

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

Step 5.1.1.3

Factor the polynomial by factoring out the greatest common factor, $- 2 x + 1$ .

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

Step 5.1.2

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

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

Step 5.1.3

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

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

Step 5.1.4

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

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

Step 5.1.5

Simplify the expression.

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Step 5.1.5.1

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

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

Step 5.1.5.2

Move the negative in front of the fraction.

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

Step 5.2

Expand $- (2 x - 1) (2 x + 5)$ .

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Step 5.2.1

Negate $2 x - 1$ .

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

Step 5.2.2

Apply the distributive property.

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

Step 5.2.3

Apply the distributive property.

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

Step 5.2.4

Apply the distributive property.

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

Step 5.2.5

Apply the distributive property.

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

Step 5.2.6

Remove parentheses.

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

Step 5.2.7

Remove parentheses.

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

Step 5.2.8

Move $x$ .

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

Step 5.2.9

Remove parentheses.

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

Step 5.2.10

Move $x$ .

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

Step 5.2.11

Remove parentheses.

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

Step 5.2.12

Multiply $- 1$ by $2$ .

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

Step 5.2.13

Multiply $- 2$ by $2$ .

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

Step 5.2.14

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

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

Step 5.2.15

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

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

Step 5.2.16

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

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

Step 5.2.17

Add $1$ and $1$ .

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

Step 5.2.18

Multiply $- 1$ by $2$ .

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

Step 5.2.19

Multiply $- 2$ by $5$ .

$\frac{- 4 x^{2} - 10 x + 1 \cdot (2 x) + 1 \cdot 5}{2 x + 1}$

Step 5.2.20

Multiply $- 1$ by $- 1$ .

$\frac{- 4 x^{2} - 10 x + 1 \cdot (2 x) + 1 \cdot 5}{2 x + 1}$

Step 5.2.21

Multiply $2$ by $1$ .

$\frac{- 4 x^{2} - 10 x + 2 x + 1 \cdot 5}{2 x + 1}$

Step 5.2.22

Multiply $- 1$ by $- 1$ .

$\frac{- 4 x^{2} - 10 x + 2 x + 1 \cdot 5}{2 x + 1}$

Step 5.2.23

Multiply $5$ by $1$ .

$\frac{- 4 x^{2} - 10 x + 2 x + 5}{2 x + 1}$

Step 5.2.24

Add $- 10 x$ and $2 x$ .

$\frac{- 4 x^{2} - 8 x + 5}{2 x + 1}$

Step 5.3

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

$2 x$

$1$

$4 x^{2}$

$8 x$

$5$

Step 5.4

Divide the highest order term in the dividend $- 4 x^{2}$ by the highest order term in divisor $2 x$ .

				-	$2 x$
	$2 x$	+	$1$	-	$4 x^{2}$	-	$8 x$	+	$5$

Step 5.5

Multiply the new quotient term by the divisor.

			-	$2 x$
$2 x$	+	$1$	-	$4 x^{2}$	-	$8 x$	+	$5$
			-	$4 x^{2}$	-	$2 x$

Step 5.6

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x^{2} - 2 x$

			-	$2 x$
$2 x$	+	$1$	-	$4 x^{2}$	-	$8 x$	+	$5$
			+	$4 x^{2}$	+	$2 x$

Step 5.7

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

			-	$2 x$
$2 x$	+	$1$	-	$4 x^{2}$	-	$8 x$	+	$5$
			+	$4 x^{2}$	+	$2 x$
					-	$6 x$

Step 5.8

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

			-	$2 x$
$2 x$	+	$1$	-	$4 x^{2}$	-	$8 x$	+	$5$
			+	$4 x^{2}$	+	$2 x$
					-	$6 x$	+	$5$

Step 5.9

Divide the highest order term in the dividend $- 6 x$ by the highest order term in divisor $2 x$ .

			-	$2 x$	-	$3$
$2 x$	+	$1$	-	$4 x^{2}$	-	$8 x$	+	$5$
			+	$4 x^{2}$	+	$2 x$
					-	$6 x$	+	$5$

Step 5.10

Multiply the new quotient term by the divisor.

			-	$2 x$	-	$3$
$2 x$	+	$1$	-	$4 x^{2}$	-	$8 x$	+	$5$
			+	$4 x^{2}$	+	$2 x$
					-	$6 x$	+	$5$
					-	$6 x$	-	$3$

Step 5.11

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x - 3$

			-	$2 x$	-	$3$
$2 x$	+	$1$	-	$4 x^{2}$	-	$8 x$	+	$5$
			+	$4 x^{2}$	+	$2 x$
					-	$6 x$	+	$5$
					+	$6 x$	+	$3$

Step 5.12

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

			-	$2 x$	-	$3$
$2 x$	+	$1$	-	$4 x^{2}$	-	$8 x$	+	$5$
			+	$4 x^{2}$	+	$2 x$
					-	$6 x$	+	$5$
					+	$6 x$	+	$3$
							+	$8$

Step 5.13

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

$- 2 x - 3 + \frac{8}{2 x + 1}$

Step 5.14

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

$y = - 2 x - 3$

Step 6

This is the set of all asymptotes.

Vertical Asymptotes: $x = - \frac{1}{2}$

No Horizontal Asymptotes

Oblique Asymptotes: $y = - 2 x - 3$

Step 7