Graph 6x^2+3xy-8x+y-12=0

$6 x^{2} + 3 x y - 8 x + y - 12 = 0$

Step 1

Simplify.

Move all terms not containing $y$ to the right side of the equation.

Subtract $6 x^{2}$ from both sides of the equation.

$3 x y - 8 x + y - 12 = - 6 x^{2}$

Add $8 x$ to both sides of the equation.

$3 x y + y - 12 = - 6 x^{2} + 8 x$

Add $12$ to both sides of the equation.

$3 x y + y = - 6 x^{2} + 8 x + 12$

Factor $y$ out of $3 x y + y$ .

Factor $y$ out of $3 x y$ .

$y (3 x) + y = - 6 x^{2} + 8 x + 12$

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

$y (3 x) + y = - 6 x^{2} + 8 x + 12$

Factor $y$ out of $y^{1}$ .

$y (3 x) + y \cdot 1 = - 6 x^{2} + 8 x + 12$

Factor $y$ out of $y (3 x) + y \cdot 1$ .

$y (3 x + 1) = - 6 x^{2} + 8 x + 12$

Divide each term in $y (3 x + 1) = - 6 x^{2} + 8 x + 12$ by $3 x + 1$ and simplify.

Divide each term in $y (3 x + 1) = - 6 x^{2} + 8 x + 12$ by $3 x + 1$ .

$\frac{y (3 x + 1)}{3 x + 1} = \frac{- 6 x^{2}}{3 x + 1} + \frac{8 x}{3 x + 1} + \frac{12}{3 x + 1}$

Simplify the left side.

Cancel the common factor of $3 x + 1$ .

Cancel the common factor.

$\frac{y (3 x + 1)}{3 x + 1} = \frac{- 6 x^{2}}{3 x + 1} + \frac{8 x}{3 x + 1} + \frac{12}{3 x + 1}$

Divide $y$ by $1$ .

$y = \frac{- 6 x^{2}}{3 x + 1} + \frac{8 x}{3 x + 1} + \frac{12}{3 x + 1}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{- 6 x^{2} + 8 x + 12}{3 x + 1}$

Factor $2$ out of $- 6 x^{2} + 8 x + 12$ .

Factor $2$ out of $- 6 x^{2}$ .

$y = \frac{2 (- 3 x^{2}) + 8 x + 12}{3 x + 1}$

Factor $2$ out of $8 x$ .

$y = \frac{2 (- 3 x^{2}) + 2 (4 x) + 12}{3 x + 1}$

Factor $2$ out of $12$ .

$y = \frac{2 (- 3 x^{2}) + 2 (4 x) + 2 (6)}{3 x + 1}$

Factor $2$ out of $2 (- 3 x^{2}) + 2 (4 x)$ .

$y = \frac{2 (- 3 x^{2} + 4 x) + 2 (6)}{3 x + 1}$

Factor $2$ out of $2 (- 3 x^{2} + 4 x) + 2 (6)$ .

$y = \frac{2 (- 3 x^{2} + 4 x + 6)}{3 x + 1}$

Factor $- 1$ out of $- 3 x^{2}$ .

$y = \frac{2 (- (3 x^{2}) + 4 x + 6)}{3 x + 1}$

Factor $- 1$ out of $4 x$ .

$y = \frac{2 (- (3 x^{2}) - (- 4 x) + 6)}{3 x + 1}$

Factor $- 1$ out of $- (3 x^{2}) - (- 4 x)$ .

$y = \frac{2 (- (3 x^{2} - 4 x) + 6)}{3 x + 1}$

Rewrite $6$ as $- 1 (- 6)$ .

$y = \frac{2 (- (3 x^{2} - 4 x) - 1 (- 6))}{3 x + 1}$

Factor $- 1$ out of $- (3 x^{2} - 4 x) - 1 (- 6)$ .

$y = \frac{2 (- (3 x^{2} - 4 x - 6))}{3 x + 1}$

Simplify the expression.

Rewrite $- (3 x^{2} - 4 x - 6)$ as $- 1 (3 x^{2} - 4 x - 6)$ .

$y = \frac{2 (- 1 (3 x^{2} - 4 x - 6))}{3 x + 1}$

Move the negative in front of the fraction.

$y = - \frac{2 (3 x^{2} - 4 x - 6)}{3 x + 1}$

Step 2

Find where the expression $- \frac{2 (3 x^{2} - 4 x - 6)}{3 x + 1}$ is undefined.

$x = - \frac{1}{3}$

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.

Simplify the expression.

Factor $2$ out of $- 6 x^{2} + 8 x + 12$ .

Factor $2$ out of $- 6 x^{2}$ .

$\frac{2 (- 3 x^{2}) + 8 x + 12}{3 x + 1}$

Factor $2$ out of $8 x$ .

$\frac{2 (- 3 x^{2}) + 2 (4 x) + 12}{3 x + 1}$

Factor $2$ out of $12$ .

$\frac{2 (- 3 x^{2}) + 2 (4 x) + 2 (6)}{3 x + 1}$

Factor $2$ out of $2 (- 3 x^{2}) + 2 (4 x)$ .

$\frac{2 (- 3 x^{2} + 4 x) + 2 (6)}{3 x + 1}$

Factor $2$ out of $2 (- 3 x^{2} + 4 x) + 2 (6)$ .

$\frac{2 (- 3 x^{2} + 4 x + 6)}{3 x + 1}$

Factor $- 1$ out of $- 3 x^{2}$ .

$\frac{2 (- (3 x^{2}) + 4 x + 6)}{3 x + 1}$

Factor $- 1$ out of $4 x$ .

$\frac{2 (- (3 x^{2}) - (- 4 x) + 6)}{3 x + 1}$

Factor $- 1$ out of $- (3 x^{2}) - (- 4 x)$ .

$\frac{2 (- (3 x^{2} - 4 x) + 6)}{3 x + 1}$

Rewrite $6$ as $- 1 (- 6)$ .

$\frac{2 (- (3 x^{2} - 4 x) - 1 (- 6))}{3 x + 1}$

Factor $- 1$ out of $- (3 x^{2} - 4 x) - 1 (- 6)$ .

$\frac{2 (- (3 x^{2} - 4 x - 6))}{3 x + 1}$

Simplify the expression.

Rewrite $- (3 x^{2} - 4 x - 6)$ as $- 1 (3 x^{2} - 4 x - 6)$ .

$\frac{2 (- 1 (3 x^{2} - 4 x - 6))}{3 x + 1}$

Move the negative in front of the fraction.

$- \frac{2 (3 x^{2} - 4 x - 6)}{3 x + 1}$

Expand $- 2 (3 x^{2} - 4 x - 6)$ .

Start expanding.

$\frac{- 2 (3 x^{2} - 4 x - 6)}{3 x + 1}$

Apply the distributive property.

$\frac{- 2 (3 x^{2} - 4 x) - 2 \cdot - 6}{3 x + 1}$

Apply the distributive property.

$\frac{- 2 (3 x^{2}) - 2 (- 4 x) - 2 \cdot - 6}{3 x + 1}$

Remove parentheses.

$\frac{- 2 \cdot (3 x^{2}) - 2 (- 4 x) - 2 \cdot - 6}{3 x + 1}$

Remove parentheses.

$\frac{- 2 \cdot (3 x^{2}) - 2 \cdot (- 4 x) - 2 \cdot - 6}{3 x + 1}$

Multiply $- 2$ by $3$ .

$\frac{- 6 x^{2} - 2 \cdot (- 4 x) - 2 \cdot - 6}{3 x + 1}$

Multiply $- 2$ by $- 4$ .

$\frac{- 6 x^{2} + 8 x - 2 \cdot - 6}{3 x + 1}$

Multiply $- 2$ by $- 6$ .

$\frac{- 6 x^{2} + 8 x + 12}{3 x + 1}$

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

$3 x$

$1$

$6 x^{2}$

$8 x$

$12$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $3 x$ .

				-	$2 x$
	$3 x$	+	$1$	-	$6 x^{2}$	+	$8 x$	+	$12$

Multiply the new quotient term by the divisor.

			-	$2 x$
$3 x$	+	$1$	-	$6 x^{2}$	+	$8 x$	+	$12$
			-	$6 x^{2}$	-	$2 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x^{2} - 2 x$

			-	$2 x$
$3 x$	+	$1$	-	$6 x^{2}$	+	$8 x$	+	$12$
			+	$6 x^{2}$	+	$2 x$

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

			-	$2 x$
$3 x$	+	$1$	-	$6 x^{2}$	+	$8 x$	+	$12$
			+	$6 x^{2}$	+	$2 x$
					+	$10 x$

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

			-	$2 x$
$3 x$	+	$1$	-	$6 x^{2}$	+	$8 x$	+	$12$
			+	$6 x^{2}$	+	$2 x$
					+	$10 x$	+	$12$

Divide the highest order term in the dividend $10 x$ by the highest order term in divisor $3 x$ .

			-	$2 x$	+	$\frac{10}{3}$
$3 x$	+	$1$	-	$6 x^{2}$	+	$8 x$	+	$12$
			+	$6 x^{2}$	+	$2 x$
					+	$10 x$	+	$12$

Multiply the new quotient term by the divisor.

			-	$2 x$	+	$\frac{10}{3}$
$3 x$	+	$1$	-	$6 x^{2}$	+	$8 x$	+	$12$
			+	$6 x^{2}$	+	$2 x$
					+	$10 x$	+	$12$
					+	$10 x$	+	$\frac{10}{3}$

The expression needs to be subtracted from the dividend, so change all the signs in $10 x + \frac{10}{3}$

			-	$2 x$	+	$\frac{10}{3}$
$3 x$	+	$1$	-	$6 x^{2}$	+	$8 x$	+	$12$
			+	$6 x^{2}$	+	$2 x$
					+	$10 x$	+	$12$
					-	$10 x$	-	$\frac{10}{3}$

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

			-	$2 x$	+	$\frac{10}{3}$
$3 x$	+	$1$	-	$6 x^{2}$	+	$8 x$	+	$12$
			+	$6 x^{2}$	+	$2 x$
					+	$10 x$	+	$12$
					-	$10 x$	-	$\frac{10}{3}$
							+	$\frac{26}{3}$

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

$- 2 x + \frac{10}{3} + \frac{26}{3 (3 x + 1)}$

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

$y = - 2 x + \frac{10}{3}$

Step 7

This is the set of all asymptotes.

Vertical Asymptotes: $x = - \frac{1}{3}$

No Horizontal Asymptotes

Oblique Asymptotes: $y = - 2 x + \frac{10}{3}$

Step 8

Graph 6x^2+3xy-8x+y-12=0

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