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next part of this question asks compute f^-1 solve for that

To compute the inverse of the given quadratic function \( f(x) = -5x^2 + 30x - 7 \), we need to attempt to solve for \( f^{-1}(x) \).

However, note that a quadratic function typically does not have an inverse over its entire domain since it is not one-to-one (it fails the horizontal line test). We can, however, find the inverse on a restricted domain where the function is one-to-one, typically by considering only one branch of the parabola (either the increasing or decreasing part).

### Steps to compute the inverse \( f^{-1}(x) \):

1. **Start with the original function:**
   \[
   y = -5x^2 + 30x - 7
   \]
   To solve for \(x\), first rewrite the equation in terms of \(x\).

2. **Complete the square** (this is similar to finding the vertex form):
   \[
   y = -5(x^2 - 6x) - 7
   \]
   Completing the square inside the parentheses:
   \[
   x^2 - 6x = (x - 3)^2 - 9
   \]
   Substituting this back into the equation:
   \[
   y = -5((x - 3)^2 - 9) - 7
   \]
   Simplify:
   \[
   y = -5(x - 3)^2 + 45 - 7
   \]
   \[
   y = -5(x - 3)^2 + 38
   \]
   Now, we have the vertex form:
   \[
   f(x) = -5(x - 3)^2 + 38
   \]
   The vertex is at \( (3, 38) \), and this shows that the function is symmetric about \( x = 3 \).

3. **Solve for \(x\) in terms of \(y\):**
   To find the inverse, we now solve for \(x\) in terms of \(y\).
   \[
   y = -5(x - 3)^2 + 38
   \]
   Subtract 38 from both sides:
   \[
   y - 38 = -5(x - 3)^2
   \]
   Divide by \(-5\):
   \[
   \frac{y - 38}{-5} = (x - 3)^2
   \]
   Take the square root of both sides (considering both the positive and negative roots):
   \[
   x - 3 = \pm \sqrt{\frac{38 - y}{5}}
   \]
   Finally, solve for \(x\):
   \[
   x = 3 \pm \sqrt{\frac{38 - y}{5}}
   \]

4. **Interpretation**:
   Since the original quadratic is downward-opening, we typically take the negative square root for the inverse over the range where the function is decreasing (for \(x \leq 3\)). Hence, the inverse function is:
   \[
   f^{-1}(y) = 3 - \sqrt{\frac{38 - y}{5}}
   \]
   This inverse is valid for \(y \leq 38\), corresponding to the range of the original function.

---

### Summary:
The inverse of the quadratic function \( f(x) = -5x^2 + 30x - 7 \) (on the appropriate domain) is:

\[
f^{-1}(y) = 3 - \sqrt{\frac{38 - y}{5}}
\]

### Additional Questions:
1. Why do we restrict the domain when finding the inverse of a quadratic function?
2. What is the significance of the vertex in determining the inverse of a quadratic?
3. How does completing the square help in finding the inverse of a quadratic function?
4. Can you always find an inverse for quadratic functions? Why or why not?
5. How do you verify that a function and its inverse are correct?

**Tip:** When finding the inverse of a function, always ensure the original function is one-to-one on the domain you're considering; otherwise, no inverse exists on that domain.

Compute the vertex form of the quadratic function f(x) = -5x^2 + 30x - 7 and solve for the inverse of f.

Learn how to convert the quadratic function f(x) = -5x^2 + 30x - 7 to vertex form and solve for its inverse. This problem demonstrates completing the square and determining the inverse function for a restricted domain. Suitable for Grades 10-12.

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