To find the inverse function of $f(x) = (x-1)^2 - 6$, follow these steps:

1. Rewrite the function: $y = (x-1)^2 - 6$.

2. Solve for $x$ in terms of $y$:

   • Start by isolating the squared term: $y + 6 = (x-1)^2$.
   • Take the square root of both sides: $\sqrt{y + 6} = x - 1$.
   • Note that there are two possible solutions for $x$: $x - 1 = \sqrt{y + 6}$ and $x - 1 = -\sqrt{y + 6}$.

3. Express $x$ explicitly:

   • $x = 1 + \sqrt{y + 6}$ and $x = 1 - \sqrt{y + 6}$.

Since a function must have only one output for each input, the inverse must be restricted to a domain where it is one-to-one. Here, we'll assume that $f(x)$ is restricted to $x \geq 1$, ensuring $f(x)$ is increasing and one-to-one. Thus, the inverse function is:

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

So, the inverse function of $f(x) = (x-1)^2 - 6$, assuming $x \geq 1$, is:

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

Would you like more details or have any questions about this solution?

Here are 5 questions you might want to ask:

1. How do you determine the domain and range of the inverse function?
2. What is the process for verifying that two functions are inverses?
3. Can we graph the original function and its inverse?
4. How do we handle inverses for functions that are not one-to-one on their entire domain?
5. Can you explain how to find inverses for trigonometric functions?

Tip: When finding the inverse of a function, make sure the original function is one-to-one over its domain. If not, restrict the domain appropriately.