To solve for the inverse function $f^{-1}(x)$ for the given function $f(x) = x^2 + 8x - 2$, we follow these steps:

Step 1: Replace $f(x)$ with $y$

Let $y = x^2 + 8x - 2$.

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

Rewrite the equation: $y = x^2 + 8x - 2$

Rearranging: $x^2 + 8x - (y + 2) = 0$

This is a quadratic equation in $x$. We solve it using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 8$, and $c = -(y + 2)$. Substituting: $x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-(y + 2))}}{2(1)}$

Simplify: $x = \frac{-8 \pm \sqrt{64 + 4(y + 2)}}{2}$ $x = \frac{-8 \pm \sqrt{64 + 4y + 8}}{2}$ $x = \frac{-8 \pm \sqrt{72 + 4y}}{2}$ $x = \frac{-8 \pm \sqrt{4(18 + y)}}{2}$ $x = \frac{-8 \pm 2\sqrt{18 + y}}{2}$ $x = -4 \pm \sqrt{18 + y}$

Step 3: Swap $x$ and $y$

The inverse function is: $f^{-1}(x) = -4 \pm \sqrt{x + 18}$

Final Answer:

The correct option is: A. $-4 \pm \sqrt{x + 18}$

Would you like further details or clarifications?

Here are 5 follow-up questions you might find interesting:

1. How do we restrict the domain to ensure the function is invertible?
2. What are the conditions for a function to have an inverse?
3. Can the process of finding an inverse be generalized for any quadratic function?
4. How do the coefficients of the quadratic equation affect the inverse function?
5. What is the graphical interpretation of the inverse function?

Tip: Always check the domain and range of a function and its inverse to ensure they correspond appropriately.