Find the Domain and Range x^2y-4x^2-5x-9y-6=0

$x^{2} y - 4 x^{2} - 5 x - 9 y - 6 = 0$

Step 1

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

Add $4 x^{2}$ to both sides of the equation.

$x^{2} y - 5 x - 9 y - 6 = 4 x^{2}$

Add $5 x$ to both sides of the equation.

$x^{2} y - 9 y - 6 = 4 x^{2} + 5 x$

Add $6$ to both sides of the equation.

$x^{2} y - 9 y = 4 x^{2} + 5 x + 6$

Step 2

Factor $y$ out of $x^{2} y - 9 y$ .

Factor $y$ out of $x^{2} y$ .

$y x^{2} - 9 y = 4 x^{2} + 5 x + 6$

Factor $y$ out of $- 9 y$ .

$y x^{2} + y \cdot - 9 = 4 x^{2} + 5 x + 6$

Factor $y$ out of $y x^{2} + y \cdot - 9$ .

$y (x^{2} - 9) = 4 x^{2} + 5 x + 6$

Step 3

Rewrite $9$ as $3^{2}$ .

$y (x^{2} - 3^{2}) = 4 x^{2} + 5 x + 6$

Step 4

Factor.

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 $b = 3$ .

$y ((x + 3) (x - 3)) = 4 x^{2} + 5 x + 6$

Remove unnecessary parentheses.

$y (x + 3) (x - 3) = 4 x^{2} + 5 x + 6$

Step 5

Divide each term in $y (x + 3) (x - 3) = 4 x^{2} + 5 x + 6$ by $(x + 3) (x - 3)$ and simplify.

Divide each term in $y (x + 3) (x - 3) = 4 x^{2} + 5 x + 6$ by $(x + 3) (x - 3)$ .

$\frac{y (x + 3) (x - 3)}{(x + 3) (x - 3)} = \frac{4 x^{2}}{(x + 3) (x - 3)} + \frac{5 x}{(x + 3) (x - 3)} + \frac{6}{(x + 3) (x - 3)}$

Simplify the left side.

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$\frac{y (x + 3) (x - 3)}{(x + 3) (x - 3)} = \frac{4 x^{2}}{(x + 3) (x - 3)} + \frac{5 x}{(x + 3) (x - 3)} + \frac{6}{(x + 3) (x - 3)}$

Rewrite the expression.

$\frac{y (x - 3)}{x - 3} = \frac{4 x^{2}}{(x + 3) (x - 3)} + \frac{5 x}{(x + 3) (x - 3)} + \frac{6}{(x + 3) (x - 3)}$

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$\frac{y (x - 3)}{x - 3} = \frac{4 x^{2}}{(x + 3) (x - 3)} + \frac{5 x}{(x + 3) (x - 3)} + \frac{6}{(x + 3) (x - 3)}$

Divide $y$ by $1$ .

$y = \frac{4 x^{2}}{(x + 3) (x - 3)} + \frac{5 x}{(x + 3) (x - 3)} + \frac{6}{(x + 3) (x - 3)}$

Step 6

Set the denominator in $\frac{4 x^{2}}{(x + 3) (x - 3)}$ equal to $0$ to find where the expression is undefined.

$(x + 3) (x - 3) = 0$

Step 7

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 3 = 0$

$x - 3 = 0$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(x + 3) (x - 3) = 0$ true.

$x = - 3, 3$

Step 8

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq 3, - 3}$

Step 9

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, \frac{10 - 3 \sqrt{19}}{6}] \cup [\frac{10 + 3 \sqrt{19}}{6}, \infty)$

Set-Builder Notation:

${y | y \leq \frac{10 - 3 \sqrt{19}}{6}, y \geq \frac{10 + 3 \sqrt{19}}{6}}$

Step 10

Determine the domain and range.

Domain: $(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty), {x | x \neq 3, - 3}$

Range: $(- \infty, \frac{10 - 3 \sqrt{19}}{6}] \cup [\frac{10 + 3 \sqrt{19}}{6}, \infty), {y | y \leq \frac{10 - 3 \sqrt{19}}{6}, y \geq \frac{10 + 3 \sqrt{19}}{6}}$

Step 11

Find the Domain and Range x^2y-4x^2-5x-9y-6=0

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