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We are given a piecewise function \( f(x) \) and are asked to find \( f(-6) \).

The piecewise function is defined as:

\[
f(x) =
\begin{cases}
x + 6 & \text{for } x < -5 \\
-5 & \text{for } x = -5 \\
-(x + 5)^2 + 1 & \text{for } x > -5
\end{cases}
\]

Since \( -6 \) is less than \( -5 \), we use the first piece of the function, \( f(x) = x + 6 \).

Now substitute \( x = -6 \) into this expression:

\[
f(-6) = -6 + 6 = 0
\]

Thus, \( f(-6) = 0 \).

Would you like more details or explanations on this? Here are some related questions you might find helpful:

1. What is the value of \( f(-5) \) using the same function?
2. How would you find \( f(0) \) for this function?
3. What is the behavior of the function at \( x = -5 \)?
4. How can you determine whether the function is continuous at \( x = -5 \)?
5. What would the graph of this piecewise function look like?

**Tip:** When evaluating a piecewise function, always check which condition the given value satisfies before applying the correct expression.

Find f(-6) given the piecewise function: f(x) = { x + 6 for x < -5, -5 for x = -5, -(x + 5)^2 + 1 for x > -5 }.

Learn how to evaluate the piecewise function f(x) at x = -6. This algebra problem includes three cases, covering linear and quadratic expressions. Suitable for Grades 9-11 students studying piecewise functions.

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