Find the Asymptotes y=(x^3-1)/(x^2-1)

$y = \frac{x^{3} - 1}{x^{2} - 1}$

Step 1

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

$x = - 1, x = 1$

Step 2

Since $\frac{x^{3} - 1}{x^{2} - 1}$ $\to$ $- \infty$ as $x$ $\to$ $- 1$ from the left and $\frac{x^{3} - 1}{x^{2} - 1}$ $\to$ $\infty$ 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.

Rewrite $1$ as $1^{3}$ .

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

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 1$ .

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

Simplify.

Multiply $x$ by $1$ .

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

One to any power is one.

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

Simplify the denominator.

Rewrite $1$ as $1^{2}$ .

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

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 $b = 1$ .

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

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x$

$1$

$x^{2}$

$x$

$1$

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

					$x$
	$x$	+	$1$		$x^{2}$	+	$x$	+	$1$

Multiply the new quotient term by the divisor.

				$x$
$x$	+	$1$		$x^{2}$	+	$x$	+	$1$
			+	$x^{2}$	+	$x$

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

				$x$
$x$	+	$1$		$x^{2}$	+	$x$	+	$1$
			-	$x^{2}$	-	$x$

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

				$x$
$x$	+	$1$		$x^{2}$	+	$x$	+	$1$
			-	$x^{2}$	-	$x$
						$0$

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

				$x$
$x$	+	$1$		$x^{2}$	+	$x$	+	$1$
			-	$x^{2}$	-	$x$
						$0$	+	$1$

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

$x + \frac{1}{x + 1}$

Split the solution into the polynomial portion and the remainder.

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

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

$y = x$

Step 7

This is the set of all asymptotes.

Vertical Asymptotes: $x = - 1$

No Horizontal Asymptotes

Oblique Asymptotes: $y = x$

Step 8

Find the Asymptotes y=(x^3-1)/(x^2-1)

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