Find the Domain and Range x^2+6x+y^2-2y-15=0

$x^{2} + 6 x + y^{2} - 2 y - 15 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 2$ , and $c = x^{2} + 6 x - 15$ into the quadratic formula and solve for $y$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot (x^{2} + 6 x - 15))}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (x^{2} + 6 x - 15)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot (x^{2} + 6 x - 15)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 4 (6 x) - 4 \cdot - 15}}{2 \cdot 1}$

Simplify.

Multiply $6$ by $- 4$ .

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 24 x - 4 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4$ by $- 15$ .

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 24 x + 60}}{2 \cdot 1}$

Add $4$ and $60$ .

$y = \frac{2 \pm \sqrt{- 4 x^{2} - 24 x + 64}}{2 \cdot 1}$

Rewrite $- 4 x^{2} - 24 x + 64$ in a factored form.

Factor $4$ out of $- 4 x^{2} - 24 x + 64$ .

Factor $4$ out of $- 4 x^{2}$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2}) - 24 x + 64}}{2 \cdot 1}$

Factor $4$ out of $- 24 x$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2}) + 4 (- 6 x) + 64}}{2 \cdot 1}$

Factor $4$ out of $64$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2}) + 4 (- 6 x) + 4 (16)}}{2 \cdot 1}$

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

$y = \frac{2 \pm \sqrt{4 (- x^{2} - 6 x) + 4 (16)}}{2 \cdot 1}$

Factor $4$ out of $4 (- x^{2} - 6 x) + 4 (16)$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2} - 6 x + 16)}}{2 \cdot 1}$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 16 = - 16$ and whose sum is $b = - 6$ .

Factor $- 6$ out of $- 6 x$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2} - 6 x + 16)}}{2 \cdot 1}$

Rewrite $- 6$ as $2$ plus $- 8$

$y = \frac{2 \pm \sqrt{4 (- x^{2} + (2 - 8) x + 16)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 (- x^{2} + 2 x - 8 x + 16)}}{2 \cdot 1}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$y = \frac{2 \pm \sqrt{4 ((- x^{2} + 2 x) - 8 x + 16)}}{2 \cdot 1}$

Factor out the greatest common factor (GCF) from each group.

$y = \frac{2 \pm \sqrt{4 (x (- x + 2) + 8 (- x + 2))}}{2 \cdot 1}$

Factor the polynomial by factoring out the greatest common factor, $- x + 2$ .

$y = \frac{2 \pm \sqrt{4 ((- x + 2) (x + 8))}}{2 \cdot 1}$

$y = \frac{2 \pm \sqrt{4 (- x + 2) (x + 8)}}{2 \cdot 1}$

Rewrite $4 (- x + 2) (x + 8)$ as $2^{2} ((- x + 2) (x + 8))$ .

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

$y = \frac{2 \pm \sqrt{2^{2} (- x + 2) (x + 8)}}{2 \cdot 1}$

Add parentheses.

$y = \frac{2 \pm \sqrt{2^{2} ((- x + 2) (x + 8))}}{2 \cdot 1}$

Pull terms out from under the radical.

$y = \frac{2 \pm 2 \sqrt{(- x + 2) (x + 8)}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{2 \pm 2 \sqrt{(- x + 2) (x + 8)}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{(- x + 2) (x + 8)}}{2}$ .

$y = 1 \pm \sqrt{(- x + 2) (x + 8)}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

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

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (x^{2} + 6 x - 15)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot (x^{2} + 6 x - 15)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 4 (6 x) - 4 \cdot - 15}}{2 \cdot 1}$

Simplify.

Multiply $6$ by $- 4$ .

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 24 x - 4 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4$ by $- 15$ .

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 24 x + 60}}{2 \cdot 1}$

Add $4$ and $60$ .

$y = \frac{2 \pm \sqrt{- 4 x^{2} - 24 x + 64}}{2 \cdot 1}$

Rewrite $- 4 x^{2} - 24 x + 64$ in a factored form.

Factor $4$ out of $- 4 x^{2} - 24 x + 64$ .

Factor $4$ out of $- 4 x^{2}$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2}) - 24 x + 64}}{2 \cdot 1}$

Factor $4$ out of $- 24 x$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2}) + 4 (- 6 x) + 64}}{2 \cdot 1}$

Factor $4$ out of $64$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2}) + 4 (- 6 x) + 4 (16)}}{2 \cdot 1}$

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

$y = \frac{2 \pm \sqrt{4 (- x^{2} - 6 x) + 4 (16)}}{2 \cdot 1}$

Factor $4$ out of $4 (- x^{2} - 6 x) + 4 (16)$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2} - 6 x + 16)}}{2 \cdot 1}$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 16 = - 16$ and whose sum is $b = - 6$ .

Factor $- 6$ out of $- 6 x$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2} - 6 x + 16)}}{2 \cdot 1}$

Rewrite $- 6$ as $2$ plus $- 8$

$y = \frac{2 \pm \sqrt{4 (- x^{2} + (2 - 8) x + 16)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 (- x^{2} + 2 x - 8 x + 16)}}{2 \cdot 1}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$y = \frac{2 \pm \sqrt{4 ((- x^{2} + 2 x) - 8 x + 16)}}{2 \cdot 1}$

Factor out the greatest common factor (GCF) from each group.

$y = \frac{2 \pm \sqrt{4 (x (- x + 2) + 8 (- x + 2))}}{2 \cdot 1}$

Factor the polynomial by factoring out the greatest common factor, $- x + 2$ .

$y = \frac{2 \pm \sqrt{4 ((- x + 2) (x + 8))}}{2 \cdot 1}$

$y = \frac{2 \pm \sqrt{4 (- x + 2) (x + 8)}}{2 \cdot 1}$

Rewrite $4 (- x + 2) (x + 8)$ as $2^{2} ((- x + 2) (x + 8))$ .

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

$y = \frac{2 \pm \sqrt{2^{2} (- x + 2) (x + 8)}}{2 \cdot 1}$

Add parentheses.

$y = \frac{2 \pm \sqrt{2^{2} ((- x + 2) (x + 8))}}{2 \cdot 1}$

Pull terms out from under the radical.

$y = \frac{2 \pm 2 \sqrt{(- x + 2) (x + 8)}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{2 \pm 2 \sqrt{(- x + 2) (x + 8)}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{(- x + 2) (x + 8)}}{2}$ .

$y = 1 \pm \sqrt{(- x + 2) (x + 8)}$

Change the $\pm$ to $+$ .

$y = 1 + \sqrt{(- x + 2) (x + 8)}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

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

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (x^{2} + 6 x - 15)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot (x^{2} + 6 x - 15)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 4 (6 x) - 4 \cdot - 15}}{2 \cdot 1}$

Simplify.

Multiply $6$ by $- 4$ .

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 24 x - 4 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4$ by $- 15$ .

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 24 x + 60}}{2 \cdot 1}$

Add $4$ and $60$ .

$y = \frac{2 \pm \sqrt{- 4 x^{2} - 24 x + 64}}{2 \cdot 1}$

Rewrite $- 4 x^{2} - 24 x + 64$ in a factored form.

Factor $4$ out of $- 4 x^{2} - 24 x + 64$ .

Factor $4$ out of $- 4 x^{2}$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2}) - 24 x + 64}}{2 \cdot 1}$

Factor $4$ out of $- 24 x$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2}) + 4 (- 6 x) + 64}}{2 \cdot 1}$

Factor $4$ out of $64$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2}) + 4 (- 6 x) + 4 (16)}}{2 \cdot 1}$

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

$y = \frac{2 \pm \sqrt{4 (- x^{2} - 6 x) + 4 (16)}}{2 \cdot 1}$

Factor $4$ out of $4 (- x^{2} - 6 x) + 4 (16)$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2} - 6 x + 16)}}{2 \cdot 1}$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 16 = - 16$ and whose sum is $b = - 6$ .

Factor $- 6$ out of $- 6 x$ .

$y = \frac{2 \pm \sqrt{4 (- x^{2} - 6 x + 16)}}{2 \cdot 1}$

Rewrite $- 6$ as $2$ plus $- 8$

$y = \frac{2 \pm \sqrt{4 (- x^{2} + (2 - 8) x + 16)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 (- x^{2} + 2 x - 8 x + 16)}}{2 \cdot 1}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$y = \frac{2 \pm \sqrt{4 ((- x^{2} + 2 x) - 8 x + 16)}}{2 \cdot 1}$

Factor out the greatest common factor (GCF) from each group.

$y = \frac{2 \pm \sqrt{4 (x (- x + 2) + 8 (- x + 2))}}{2 \cdot 1}$

Factor the polynomial by factoring out the greatest common factor, $- x + 2$ .

$y = \frac{2 \pm \sqrt{4 ((- x + 2) (x + 8))}}{2 \cdot 1}$

$y = \frac{2 \pm \sqrt{4 (- x + 2) (x + 8)}}{2 \cdot 1}$

Rewrite $4 (- x + 2) (x + 8)$ as $2^{2} ((- x + 2) (x + 8))$ .

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

$y = \frac{2 \pm \sqrt{2^{2} (- x + 2) (x + 8)}}{2 \cdot 1}$

Add parentheses.

$y = \frac{2 \pm \sqrt{2^{2} ((- x + 2) (x + 8))}}{2 \cdot 1}$

Pull terms out from under the radical.

$y = \frac{2 \pm 2 \sqrt{(- x + 2) (x + 8)}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{2 \pm 2 \sqrt{(- x + 2) (x + 8)}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{(- x + 2) (x + 8)}}{2}$ .

$y = 1 \pm \sqrt{(- x + 2) (x + 8)}$

Change the $\pm$ to $-$ .

$y = 1 - \sqrt{(- x + 2) (x + 8)}$

The final answer is the combination of both solutions.

$y = 1 + \sqrt{(- x + 2) (x + 8)}$

$y = 1 - \sqrt{(- x + 2) (x + 8)}$

Set the radicand in $\sqrt{(- x + 2) (x + 8)}$ greater than or equal to $0$ to find where the expression is defined.

$(- x + 2) (x + 8) \geq 0$

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 $0$ .

$- x + 2 = 0$

$x + 8 = 0$

Set $- x + 2$ equal to $0$ and solve for $x$ .

Set $- x + 2$ equal to $0$ .

$- x + 2 = 0$

Solve $- x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$- x = - 2$

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Divide each term in $- x = - 2$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x = 2$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $(- x + 2) (x + 8) \geq 0$ true.

$x = 2, - 8$

Use each root to create test intervals.

$x < - 8$

$- 8 < x < 2$

$x > 2$

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 $x < - 8$ to see if it makes the inequality true.

Choose a value on the interval $x < - 8$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

$(- (- 10) + 2) ((- 10) + 8) \geq 0$

The left side $- 24$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 8 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 8 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$(- (0) + 2) ((0) + 8) \geq 0$

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 $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$(- (4) + 2) ((4) + 8) \geq 0$

The left side $- 24$ 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.

$x < - 8$ False

$- 8 < x < 2$ True

$x > 2$ False

$x < - 8$ False

$- 8 < x < 2$ True

$x > 2$ False

The solution consists of all of the true intervals.

$- 8 \leq x \leq 2$

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

Interval Notation:

$[- 8, 2]$

Set-Builder Notation:

${x | - 8 \leq x \leq 2}$

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

Interval Notation:

$[- 4, 6]$

Set-Builder Notation:

${y | - 4 \leq y \leq 6}$

Determine the domain and range.

Domain: $[- 8, 2], {x | - 8 \leq x \leq 2}$

Range: $[- 4, 6], {y | - 4 \leq y \leq 6}$

Find the Domain and Range x^2+6x+y^2-2y-15=0

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