To evaluate the function $f(x) = -x^2 + 5$ when $x = -3$, follow these steps:

1. Substitute $x = -3$ into the function: $f(-3) = -(-3)^2 + 5$

2. First, calculate $(-3)^2$, which is $9$: $f(-3) = -9 + 5$

3. Now, add $-9$ and $5$: $f(-3) = -4$

So, when $x = -3$, the value of the function is $f(-3) = -4$.

Would you like a more detailed explanation or have any questions?

Here are 5 related questions:

1. What is the vertex of the function $f(x) = -x^2 + 5$?
2. How do you determine the domain and range of $f(x) = -x^2 + 5$?
3. What happens to $f(x) = -x^2 + 5$ when $x = 0$?
4. Can you explain how the graph of $f(x) = -x^2 + 5$ looks?
5. How do you solve $f(x) = 0$ for $f(x) = -x^2 + 5$?

Tip: In quadratic functions like $f(x) = -x^2 + 5$, the negative sign in front of $x^2$ flips the parabola downward.