Question

Find the inverse of the translated quadratic function y=x^2+11
with domain restricted to x≥0
.(1 point)

Answer 1

To find the inverse of the function \( y = x^2 + 11 \) with the domain restricted to \( x \geq 0 \), we can follow these steps:

1. **Express \( x \) in terms of \( y \)**:
Start with the original equation:
\[
y = x^2 + 11.
\]

2. **Subtract 11 from both sides**:
\[
y - 11 = x^2.
\]

3. **Take the square root of both sides**:
Since we are only considering the domain where \( x \geq 0 \), we take the positive square root:
\[
x = \sqrt{y - 11}.
\]

4. **Solve for \( y \)**:
Now we express \( y \) in terms of \( x \):
\[
y = x^2 + 11.
\]

To find the inverse, we swap \( x \) and \( y \):
\[
x = \sqrt{y - 11}.
\]

5. **Solve for \( y \)** in terms of \( x \):
Rearranging the expression gives:
\[
y - 11 = x^2.
\]
\[
y = x^2 + 11.
\]

Since we need the inverse function, we write it as:
\[
y = x^2 - 11.
\]

Thus, the inverse function is:
\[
f^{-1}(x) = \sqrt{x - 11}.
\]

6. **Specifying the domain**:
Since we originally had \( x^2 + 11 \) and restricted \( x \) to \( x \geq 0 \), the range of the original function is \( [11, \infty) \). Therefore, the domain of the inverse function will also be \( [11, \infty) \).

In conclusion, the inverse of the translated quadratic function \( y = x^2 + 11 \) for \( x \geq 0 \) is:
\[
f^{-1}(x) = \sqrt{x - 11}, \quad \text{for } x \geq 11.
\]