Find the Domain and Range y^2(x^2-4)=x+2

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

Step 1

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

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

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

Simplify the left side.

Cancel the common factor of $x^{2} - 4$ .

Cancel the common factor.

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

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{x}{x^{2} - 4} + \frac{2}{x^{2} - 4}$

Simplify the right side.

Combine the numerators over the common denominator.

$y^{2} = \frac{x + 2}{x^{2} - 4}$

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$y^{2} = \frac{x + 2}{x^{2} - 2^{2}}$

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 $b = 2$ .

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

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 2

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\frac{1}{x - 2}}$

Step 3

Simplify $\pm \sqrt{\frac{1}{x - 2}}$ .

Rewrite $\sqrt{\frac{1}{x - 2}}$ as $\frac{\sqrt{1}}{\sqrt{x - 2}}$ .

$y = \pm \frac{\sqrt{1}}{\sqrt{x - 2}}$

Any root of $1$ is $1$ .

$y = \pm \frac{1}{\sqrt{x - 2}}$

Multiply $\frac{1}{\sqrt{x - 2}}$ by $\frac{\sqrt{x - 2}}{\sqrt{x - 2}}$ .

$y = \pm \frac{1}{\sqrt{x - 2}} \cdot \frac{\sqrt{x - 2}}{\sqrt{x - 2}}$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{x - 2}}$ by $\frac{\sqrt{x - 2}}{\sqrt{x - 2}}$ .

$y = \pm \frac{\sqrt{x - 2}}{\sqrt{x - 2} \sqrt{x - 2}}$

Raise $\sqrt{x - 2}$ to the power of $1$ .

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

Raise $\sqrt{x - 2}$ to the power of $1$ .

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

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

$y = \pm \frac{\sqrt{x - 2}}{{\sqrt{x - 2}}^{1 + 1}}$

Add $1$ and $1$ .

$y = \pm \frac{\sqrt{x - 2}}{{\sqrt{x - 2}}^{2}}$

Rewrite ${\sqrt{x - 2}}^{2}$ as $x - 2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x - 2}$ as ${(x - 2)}^{\frac{1}{2}}$ .

$y = \pm \frac{\sqrt{x - 2}}{{({(x - 2)}^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = \pm \frac{\sqrt{x - 2}}{{(x - 2)}^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$y = \pm \frac{\sqrt{x - 2}}{{(x - 2)}^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y = \pm \frac{\sqrt{x - 2}}{{(x - 2)}^{\frac{2}{2}}}$

Rewrite the expression.

$y = \pm \frac{\sqrt{x - 2}}{{(x - 2)}^{1}}$

Simplify.

$y = \pm \frac{\sqrt{x - 2}}{x - 2}$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = \frac{\sqrt{x - 2}}{x - 2}$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \frac{\sqrt{x - 2}}{x - 2}$

The complete solution is the result of both the positive and negative portions of the solution.

$y = \frac{\sqrt{x - 2}}{x - 2}$

$y = - \frac{\sqrt{x - 2}}{x - 2}$

$y = \frac{\sqrt{x - 2}}{x - 2}$

$y = - \frac{\sqrt{x - 2}}{x - 2}$

Step 5

Set the radicand in $\sqrt{x - 2}$ greater than or equal to $0$ to find where the expression is defined.

$x - 2 \geq 0$

Step 6

Add $2$ to both sides of the inequality.

$x \geq 2$

Step 7

Set the denominator in $\frac{\sqrt{x - 2}}{x - 2}$ equal to $0$ to find where the expression is undefined.

$x - 2 = 0$

Step 8

Add $2$ to both sides of the equation.

$x = 2$

Step 9

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

Interval Notation:

$(2, \infty)$

Set-Builder Notation:

${x | x > 2}$

Step 10

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

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${y | y \in ℝ}$

Step 11

Determine the domain and range.

Domain: $(2, \infty), {x | x > 2}$

Range: $(- \infty, \infty), {y | y \in ℝ}$

Step 12

Find the Domain and Range y^2(x^2-4)=x+2

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