Graph y=x+1/x

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

Find where the expression $x + \frac{1}{x}$ is undefined.

$x = 0$

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).

Find $n$ and $m$ .

$n = 2$

$m = 1$

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

No Horizontal Asymptotes

Find the oblique asymptote using polynomial division.

Combine.

To write $x$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

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

Combine the numerators over the common denominator.

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

Multiply $x$ by $x$ .

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

Simplify.

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

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

$x$

$0$

$x^{2}$

$0 x$

$1$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

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

$y = x$

This is the set of all asymptotes.

Vertical Asymptotes: $x = 0$

No Horizontal Asymptotes

Oblique Asymptotes: $y = x$

Graph y=x+1/x

Related Questions

Support LCMGCF.com by disabling your adblocker.