Trigonometry Examples

$y = x^{2} - 4 x - 15$

Step 1

Interchange the variables.

$x = y^{2} - 4 y - 15$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $y^{2} - 4 y - 15 = x$ .

$y^{2} - 4 y - 15 = x$

Step 2.2

Subtract $x$ from both sides of the equation.

$y^{2} - 4 y - 15 - x = 0$

Step 2.3

Use the quadratic formula to find the solutions.

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

Step 2.4

Substitute the values $a = 1$ , $b = - 4$ , and $c = - 15 - x$ into the quadratic formula and solve for $y$ .

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

Step 2.5

Simplify.

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Step 2.5.1

Simplify the numerator.

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Step 2.5.1.1

Factor $- 4$ out of ${(- 4)}^{2} - 4 \cdot 1 \cdot (- 15 - x)$ .

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Step 2.5.1.1.1

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

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

Step 2.5.1.1.2

Factor $- 4$ out of $- 4 \cdot 1 \cdot (- 15 - x)$ .

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

Step 2.5.1.1.3

Factor $- 4$ out of $- 4 \cdot - 4 - 4 (1 \cdot (- 15 - x))$ .

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

Step 2.5.1.2

Multiply $- 15 - x$ by $1$ .

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

Step 2.5.1.3

Subtract $15$ from $- 4$ .

$y = \frac{4 \pm \sqrt{- 4 (- 19 - x)}}{2 \cdot 1}$

Step 2.5.1.4

Rewrite $- 4 (- 19 - x)$ as $2^{2} \cdot (- 1 (- 19 - x))$ .

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Step 2.5.1.4.1

Factor $4$ out of $- 4$ .

$y = \frac{4 \pm \sqrt{4 (- 1) (- 19 - x)}}{2 \cdot 1}$

Step 2.5.1.4.2

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

$y = \frac{4 \pm \sqrt{2^{2} \cdot (- 1 (- 19 - x))}}{2 \cdot 1}$

Step 2.5.1.4.3

Add parentheses.

$y = \frac{4 \pm \sqrt{2^{2} \cdot (- 1 (- 19 - x))}}{2 \cdot 1}$

Step 2.5.1.5

Pull terms out from under the radical.

$y = \frac{4 \pm 2 \sqrt{- 1 (- 19 - x)}}{2 \cdot 1}$

Step 2.5.2

Multiply $2$ by $1$ .

$y = \frac{4 \pm 2 \sqrt{- 1 (- 19 - x)}}{2}$

Step 2.5.3

Simplify $\frac{4 \pm 2 \sqrt{- 1 (- 19 - x)}}{2}$ .

$y = 2 \pm \sqrt{- 1 (- 19 - x)}$

Step 2.6

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

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Step 2.6.1

Simplify the numerator.

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Step 2.6.1.1

Factor $- 4$ out of ${(- 4)}^{2} - 4 \cdot 1 \cdot (- 15 - x)$ .

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Step 2.6.1.1.1

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

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

Step 2.6.1.1.2

Factor $- 4$ out of $- 4 \cdot 1 \cdot (- 15 - x)$ .

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

Step 2.6.1.1.3

Factor $- 4$ out of $- 4 \cdot - 4 - 4 (1 \cdot (- 15 - x))$ .

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

Step 2.6.1.2

Multiply $- 15 - x$ by $1$ .

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

Step 2.6.1.3

Subtract $15$ from $- 4$ .

$y = \frac{4 \pm \sqrt{- 4 (- 19 - x)}}{2 \cdot 1}$

Step 2.6.1.4

Rewrite $- 4 (- 19 - x)$ as $2^{2} \cdot (- 1 (- 19 - x))$ .

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Step 2.6.1.4.1

Factor $4$ out of $- 4$ .

$y = \frac{4 \pm \sqrt{4 (- 1) (- 19 - x)}}{2 \cdot 1}$

Step 2.6.1.4.2

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

$y = \frac{4 \pm \sqrt{2^{2} \cdot (- 1 (- 19 - x))}}{2 \cdot 1}$

Step 2.6.1.4.3

Add parentheses.

$y = \frac{4 \pm \sqrt{2^{2} \cdot (- 1 (- 19 - x))}}{2 \cdot 1}$

Step 2.6.1.5

Pull terms out from under the radical.

$y = \frac{4 \pm 2 \sqrt{- 1 (- 19 - x)}}{2 \cdot 1}$

Step 2.6.2

Multiply $2$ by $1$ .

$y = \frac{4 \pm 2 \sqrt{- 1 (- 19 - x)}}{2}$

Step 2.6.3

Simplify $\frac{4 \pm 2 \sqrt{- 1 (- 19 - x)}}{2}$ .

$y = 2 \pm \sqrt{- 1 (- 19 - x)}$

Step 2.6.4

Change the $\pm$ to $+$ .

$y = 2 + \sqrt{- 1 (- 19 - x)}$

Step 2.7

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

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Step 2.7.1

Simplify the numerator.

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Step 2.7.1.1

Factor $- 4$ out of ${(- 4)}^{2} - 4 \cdot 1 \cdot (- 15 - x)$ .

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Step 2.7.1.1.1

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

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

Step 2.7.1.1.2

Factor $- 4$ out of $- 4 \cdot 1 \cdot (- 15 - x)$ .

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

Step 2.7.1.1.3

Factor $- 4$ out of $- 4 \cdot - 4 - 4 (1 \cdot (- 15 - x))$ .

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

Step 2.7.1.2

Multiply $- 15 - x$ by $1$ .

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

Step 2.7.1.3

Subtract $15$ from $- 4$ .

$y = \frac{4 \pm \sqrt{- 4 (- 19 - x)}}{2 \cdot 1}$

Step 2.7.1.4

Rewrite $- 4 (- 19 - x)$ as $2^{2} \cdot (- 1 (- 19 - x))$ .

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Step 2.7.1.4.1

Factor $4$ out of $- 4$ .

$y = \frac{4 \pm \sqrt{4 (- 1) (- 19 - x)}}{2 \cdot 1}$

Step 2.7.1.4.2

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

$y = \frac{4 \pm \sqrt{2^{2} \cdot (- 1 (- 19 - x))}}{2 \cdot 1}$

Step 2.7.1.4.3

Add parentheses.

$y = \frac{4 \pm \sqrt{2^{2} \cdot (- 1 (- 19 - x))}}{2 \cdot 1}$

Step 2.7.1.5

Pull terms out from under the radical.

$y = \frac{4 \pm 2 \sqrt{- 1 (- 19 - x)}}{2 \cdot 1}$

Step 2.7.2

Multiply $2$ by $1$ .

$y = \frac{4 \pm 2 \sqrt{- 1 (- 19 - x)}}{2}$

Step 2.7.3

Simplify $\frac{4 \pm 2 \sqrt{- 1 (- 19 - x)}}{2}$ .

$y = 2 \pm \sqrt{- 1 (- 19 - x)}$

Step 2.7.4

Change the $\pm$ to $-$ .

$y = 2 - \sqrt{- 1 (- 19 - x)}$

Step 2.8

The final answer is the combination of both solutions.

$y = 2 + \sqrt{- 1 (- 19 - x)}$

$y = 2 - \sqrt{- 1 (- 19 - x)}$

$y = 2 + \sqrt{- 1 (- 19 - x)}$

$y = 2 - \sqrt{- 1 (- 19 - x)}$

Step 3

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = 2 + \sqrt{- 1 (- 19 - x)}, 2 - \sqrt{- 1 (- 19 - x)}$

Step 4

Verify if $f^{- 1} (x) = 2 + \sqrt{- 1 (- 19 - x)}, 2 - \sqrt{- 1 (- 19 - x)}$ is the inverse of $f (x) = x^{2} - 4 x - 15$ .

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Step 4.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = x^{2} - 4 x - 15$ and $f^{- 1} (x) = 2 + \sqrt{- 1 (- 19 - x)}, 2 - \sqrt{- 1 (- 19 - x)}$ and compare them.

Step 4.2

Find the range of $f (x) = x^{2} - 4 x - 15$ .

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Step 4.2.1

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

Interval Notation:

$[- 19, \infty)$

Step 4.3

Find the domain of $2 + \sqrt{- 1 (- 19 - x)}$ .

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Step 4.3.1

Set the radicand in $\sqrt{- 1 (- 19 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$- 1 (- 19 - x) \geq 0$

Step 4.3.2

Solve for $x$ .

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Step 4.3.2.1

Divide each term in $- 1 (- 19 - x) \geq 0$ by $- 1$ and simplify.

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Step 4.3.2.1.1

Divide each term in $- 1 (- 19 - x) \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 1 (- 19 - x)}{- 1} \leq \frac{0}{- 1}$

Step 4.3.2.1.2

Simplify the left side.

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Step 4.3.2.1.2.1

Dividing two negative values results in a positive value.

$\frac{- 19 - x}{1} \leq \frac{0}{- 1}$

Step 4.3.2.1.2.2

Divide $- 19 - x$ by $1$ .

$- 19 - x \leq \frac{0}{- 1}$

Step 4.3.2.1.3

Simplify the right side.

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Step 4.3.2.1.3.1

Divide $0$ by $- 1$ .

$- 19 - x \leq 0$

Step 4.3.2.2

Add $19$ to both sides of the inequality.

$- x \leq 19$

Step 4.3.2.3

Divide each term in $- x \leq 19$ by $- 1$ and simplify.

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Step 4.3.2.3.1

Divide each term in $- x \leq 19$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{19}{- 1}$

Step 4.3.2.3.2

Simplify the left side.

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Step 4.3.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{19}{- 1}$

Step 4.3.2.3.2.2

Divide $x$ by $1$ .

$x \geq \frac{19}{- 1}$

Step 4.3.2.3.3

Simplify the right side.

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Step 4.3.2.3.3.1

Divide $19$ by $- 1$ .

$x \geq - 19$

Step 4.3.3

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

$[- 19, \infty)$

Step 4.4

Find the domain of $f (x) = x^{2} - 4 x - 15$ .

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Step 4.4.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

Step 4.5

Since the domain of $f^{- 1} (x) = 2 + \sqrt{- 1 (- 19 - x)}, 2 - \sqrt{- 1 (- 19 - x)}$ is the range of $f (x) = x^{2} - 4 x - 15$ and the range of $f^{- 1} (x) = 2 + \sqrt{- 1 (- 19 - x)}, 2 - \sqrt{- 1 (- 19 - x)}$ is the domain of $f (x) = x^{2} - 4 x - 15$ , then $f^{- 1} (x) = 2 + \sqrt{- 1 (- 19 - x)}, 2 - \sqrt{- 1 (- 19 - x)}$ is the inverse of $f (x) = x^{2} - 4 x - 15$ .

$f^{- 1} (x) = 2 + \sqrt{- 1 (- 19 - x)}, 2 - \sqrt{- 1 (- 19 - x)}$

Step 5