Find the Inverse y=-6+(x+10)^2

$y = - 6 + {(x + 10)}^{2}$

Step 1

Interchange the variables.

$x = - 6 + {(y + 10)}^{2}$

Step 2

Solve for $y$ .

Rewrite the equation as $- 6 + {(y + 10)}^{2} = x$ .

$- 6 + {(y + 10)}^{2} = x$

Add $6$ to both sides of the equation.

${(y + 10)}^{2} = x + 6$

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

$y + 10 = \pm \sqrt{x + 6}$

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.

$y + 10 = \sqrt{x + 6}$

Subtract $10$ from both sides of the equation.

$y = \sqrt{x + 6} - 10$

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

$y + 10 = - \sqrt{x + 6}$

Subtract $10$ from both sides of the equation.

$y = - \sqrt{x + 6} - 10$

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

$y = \sqrt{x + 6} - 10$

$y = - \sqrt{x + 6} - 10$

$y = \sqrt{x + 6} - 10$

$y = - \sqrt{x + 6} - 10$

$y = \sqrt{x + 6} - 10$

$y = - \sqrt{x + 6} - 10$

Step 3

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

$f^{- 1} (x) = \sqrt{x + 6} - 10, - \sqrt{x + 6} - 10$

Step 4

Verify if $f^{- 1} (x) = \sqrt{x + 6} - 10, - \sqrt{x + 6} - 10$ is the inverse of $f (x) = - 6 + {(x + 10)}^{2}$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = - 6 + {(x + 10)}^{2}$ and $f^{- 1} (x) = \sqrt{x + 6} - 10, - \sqrt{x + 6} - 10$ and compare them.

Find the range of $f (x) = - 6 + {(x + 10)}^{2}$ .

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

Interval Notation:

$[- 6, \infty)$

Find the domain of $\sqrt{x + 6} - 10$ .

Set the radicand in $\sqrt{x + 6}$ greater than or equal to $0$ to find where the expression is defined.

$x + 6 \geq 0$

Subtract $6$ from both sides of the inequality.

$x \geq - 6$

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

$[- 6, \infty)$

Find the domain of $f (x) = - 6 + {(x + 10)}^{2}$ .

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.

$(- \infty, \infty)$

Since the domain of $f^{- 1} (x) = \sqrt{x + 6} - 10, - \sqrt{x + 6} - 10$ is the range of $f (x) = - 6 + {(x + 10)}^{2}$ and the range of $f^{- 1} (x) = \sqrt{x + 6} - 10, - \sqrt{x + 6} - 10$ is the domain of $f (x) = - 6 + {(x + 10)}^{2}$ , then $f^{- 1} (x) = \sqrt{x + 6} - 10, - \sqrt{x + 6} - 10$ is the inverse of $f (x) = - 6 + {(x + 10)}^{2}$ .

$f^{- 1} (x) = \sqrt{x + 6} - 10, - \sqrt{x + 6} - 10$

Step 5

Find the Inverse y=-6+(x+10)^2

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