Precalculus Examples

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

Step 1

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

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

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 $m$ .

$n = 3$

$m = 2$

Step 5

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

No Horizontal Asymptotes

Step 6

Find the oblique asymptote using polynomial division.

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Step 6.1

Factor $x$ out of $x^{3} - 5 x$ .

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Step 6.1.1

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

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

Step 6.1.2

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

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

Step 6.1.3

Factor $x$ out of $x \cdot x^{2} + x \cdot - 5$ .

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

Step 6.2

Expand $x (x^{2} - 5)$ .

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Step 6.2.1

Apply the distributive property.

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

Step 6.2.2

Reorder $x$ and $- 5$ .

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

Step 6.2.3

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

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

Step 6.2.4

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

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

Step 6.2.5

Add $1$ and $2$ .

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

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 $0$ .

$x^{2}$

$0 x$

$1$

$x^{3}$

$0 x^{2}$

$5 x$

$0$

Step 6.4

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

$x$

$x^{2}$

$0 x$

$1$

$x^{3}$

$0 x^{2}$

$5 x$

$0$

Step 6.5

Multiply the new quotient term by the divisor.

$x$

$x^{2}$

$0 x$

$1$

$x^{3}$

$0 x^{2}$

$5 x$

$0$

$x^{3}$

$0$

$x$

Step 6.6

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} + 0 + x$

$x$

$x^{2}$

$0 x$

$1$

$x^{3}$

$0 x^{2}$

$5 x$

$0$

$x^{3}$

$0$

$x$

Step 6.7

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

$x$

$x^{2}$

$0 x$

$1$

$x^{3}$

$0 x^{2}$

$5 x$

$0$

$x^{3}$

$0$

$x$

$6 x$

Step 6.8

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

$x$

$x^{2}$

$0 x$

$1$

$x^{3}$

$0 x^{2}$

$5 x$

$0$

$x^{3}$

$0$

$x$

$6 x$

$0$

Step 6.9

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

$x + \frac{- 6 x}{x^{2} + 1}$

Step 6.10

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

$y = x$

Step 7

This is the set of all asymptotes.

No Vertical Asymptotes

No Horizontal Asymptotes

Oblique Asymptotes: $y = x$

Step 8