Find the Domain and Range 9x^2+16y^2=144

$9 x^{2} + 16 y^{2} = 144$

Step 1

Subtract $9 x^{2}$ from both sides of the equation.

$16 y^{2} = 144 - 9 x^{2}$

Step 2

Divide each term in $16 y^{2} = 144 - 9 x^{2}$ by $16$ and simplify.

Divide each term in $16 y^{2} = 144 - 9 x^{2}$ by $16$ .

$\frac{16 y^{2}}{16} = \frac{144}{16} + \frac{- 9 x^{2}}{16}$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{16 y^{2}}{16} = \frac{144}{16} + \frac{- 9 x^{2}}{16}$

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

$y^{2} = \frac{144}{16} + \frac{- 9 x^{2}}{16}$

Simplify the right side.

Simplify each term.

Divide $144$ by $16$ .

$y^{2} = 9 + \frac{- 9 x^{2}}{16}$

Move the negative in front of the fraction.

$y^{2} = 9 - \frac{9 x^{2}}{16}$

Step 3

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

$y = \pm \sqrt{9 - \frac{9 x^{2}}{16}}$

Step 4

Simplify $\pm \sqrt{9 - \frac{9 x^{2}}{16}}$ .

Write the expression using exponents.

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

$y = \pm \sqrt{3^{2} - \frac{9 x^{2}}{16}}$

Rewrite $\frac{9 x^{2}}{16}$ as ${(\frac{3 x}{4})}^{2}$ .

$y = \pm \sqrt{3^{2} - {(\frac{3 x}{4})}^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = \frac{3 x}{4}$ .

$y = \pm \sqrt{(3 + \frac{3 x}{4}) (3 - \frac{3 x}{4})}$

To write $3$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$y = \pm \sqrt{(3 \cdot \frac{4}{4} + \frac{3 x}{4}) (3 - \frac{3 x}{4})}$

Combine $3$ and $\frac{4}{4}$ .

$y = \pm \sqrt{(\frac{3 \cdot 4}{4} + \frac{3 x}{4}) (3 - \frac{3 x}{4})}$

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{3 \cdot 4 + 3 x}{4} (3 - \frac{3 x}{4})}$

Factor $3$ out of $3 \cdot 4 + 3 x$ .

Factor $3$ out of $3 \cdot 4$ .

$y = \pm \sqrt{\frac{3 (4) + 3 x}{4} (3 - \frac{3 x}{4})}$

Factor $3$ out of $3 x$ .

$y = \pm \sqrt{\frac{3 (4) + 3 (x)}{4} (3 - \frac{3 x}{4})}$

Factor $3$ out of $3 (4) + 3 (x)$ .

$y = \pm \sqrt{\frac{3 (4 + x)}{4} (3 - \frac{3 x}{4})}$

To write $3$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$y = \pm \sqrt{\frac{3 (4 + x)}{4} (3 \cdot \frac{4}{4} - \frac{3 x}{4})}$

Combine $3$ and $\frac{4}{4}$ .

$y = \pm \sqrt{\frac{3 (4 + x)}{4} (\frac{3 \cdot 4}{4} - \frac{3 x}{4})}$

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{3 (4 + x)}{4} \cdot \frac{3 \cdot 4 - 3 x}{4}}$

Factor $3$ out of $3 \cdot 4 - 3 x$ .

Factor $3$ out of $3 \cdot 4$ .

$y = \pm \sqrt{\frac{3 (4 + x)}{4} \cdot \frac{3 (4) - 3 x}{4}}$

Factor $3$ out of $- 3 x$ .

$y = \pm \sqrt{\frac{3 (4 + x)}{4} \cdot \frac{3 (4) + 3 (- x)}{4}}$

Factor $3$ out of $3 (4) + 3 (- x)$ .

$y = \pm \sqrt{\frac{3 (4 + x)}{4} \cdot \frac{3 (4 - x)}{4}}$

Multiply $\frac{3 (4 + x)}{4}$ by $\frac{3 (4 - x)}{4}$ .

$y = \pm \sqrt{\frac{3 (4 + x) (3 (4 - x))}{4 \cdot 4}}$

Multiply.

Multiply $3$ by $3$ .

$y = \pm \sqrt{\frac{9 (4 + x) (4 - x)}{4 \cdot 4}}$

Multiply $4$ by $4$ .

$y = \pm \sqrt{\frac{9 (4 + x) (4 - x)}{16}}$

Rewrite $\frac{9 (4 + x) (4 - x)}{16}$ as ${(\frac{3}{4})}^{2} ((2^{2} + x) (4 - x))$ .

Factor the perfect power $3^{2}$ out of $9 (4 + x) (4 - x)$ .

$y = \pm \sqrt{\frac{3^{2} ((2^{2} + x) (4 - x))}{16}}$

Factor the perfect power $4^{2}$ out of $16$ .

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

Rearrange the fraction $\frac{3^{2} ((2^{2} + x) (4 - x))}{4^{2} \cdot 1}$ .

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

Pull terms out from under the radical.

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

Raise $2$ to the power of $2$ .

$y = \pm \frac{3}{4} \sqrt{(4 + x) (4 - x)}$

Combine $\frac{3}{4}$ and $\sqrt{(4 + x) (4 - x)}$ .

$y = \pm \frac{3 \sqrt{(4 + x) (4 - x)}}{4}$

Step 5

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.

$y = \frac{3 \sqrt{(4 + x) (4 - x)}}{4}$

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

$y = - \frac{3 \sqrt{(4 + x) (4 - x)}}{4}$

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

$y = \frac{3 \sqrt{(4 + x) (4 - x)}}{4}$

$y = - \frac{3 \sqrt{(4 + x) (4 - x)}}{4}$

$y = \frac{3 \sqrt{(4 + x) (4 - x)}}{4}$

$y = - \frac{3 \sqrt{(4 + x) (4 - x)}}{4}$

Step 6

Set the radicand in $\sqrt{(4 + x) (4 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(4 + x) (4 - x) \geq 0$

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 $0$ .

$4 + x = 0$

$4 - x = 0$

Set $4 + x$ equal to $0$ and solve for $x$ .

Set $4 + x$ equal to $0$ .

$4 + x = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Set $4 - x$ equal to $0$ and solve for $x$ .

Set $4 - x$ equal to $0$ .

$4 - x = 0$

Solve $4 - x = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$- x = - 4$

Divide each term in $- x = - 4$ by $- 1$ and simplify.

Divide each term in $- x = - 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 4}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x = 4$

The final solution is all the values that make $(4 + x) (4 - x) \geq 0$ true.

$x = - 4, 4$

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 4$

$x > 4$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 4$ to see if it makes the inequality true.

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$(4 - 6) (4 - (- 6)) \geq 0$

The left side $- 20$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 4 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$(4 + 0) (4 - (0)) \geq 0$

The left side $16$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$(4 + 6) (4 - (6)) \geq 0$

The left side $- 20$ is less than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 4$ False

$- 4 < x < 4$ True

$x > 4$ False

$x < - 4$ False

$- 4 < x < 4$ True

$x > 4$ False

The solution consists of all of the true intervals.

$- 4 \leq x \leq 4$

Step 8

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

Interval Notation:

$[- 4, 4]$

Set-Builder Notation:

${x | - 4 \leq x \leq 4}$

Step 9

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

Interval Notation:

$[- 3, 3]$

Set-Builder Notation:

${y | - 3 \leq y \leq 3}$

Step 10

Determine the domain and range.

Domain: $[- 4, 4], {x | - 4 \leq x \leq 4}$

Range: $[- 3, 3], {y | - 3 \leq y \leq 3}$

Step 11

Find the Domain and Range 9x^2+16y^2=144

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