The function provided in the image is $f(x) = -12 + \sqrt{x + 33}$. We need to calculate the following values: $f(-4)$, $f(-1)$, $f(0)$, and $f(3)$.

Let's calculate each one step by step.

1. $f(-4)$

Substitute $x = -4$ into the function: $f(-4) = -12 + \sqrt{-4 + 33} = -12 + \sqrt{29}$ $f(-4) = -12 + 5.385 \approx -6.615$

2. $f(-1)$

Substitute $x = -1$ into the function: $f(-1) = -12 + \sqrt{-1 + 33} = -12 + \sqrt{32}$ $f(-1) = -12 + 5.657 \approx -6.343$

3. $f(0)$

Substitute $x = 0$ into the function: $f(0) = -12 + \sqrt{0 + 33} = -12 + \sqrt{33}$ $f(0) = -12 + 5.745 \approx -6.255$

4. $f(3)$

Substitute $x = 3$ into the function: $f(3) = -12 + \sqrt{3 + 33} = -12 + \sqrt{36}$ $f(3) = -12 + 6 = -6$

Summary of results:

• $f(-4) \approx -6.615$
• $f(-1) \approx -6.343$
• $f(0) \approx -6.255$
• $f(3) = -6$

Let me know if you need further details or explanations.

Here are 5 questions to expand this:

1. How do you handle negative inputs inside a square root?
2. How does changing the constant inside the square root affect the function?
3. What is the domain of the function $f(x) = -12 + \sqrt{x + 33}$?
4. How would you find the range of this function?
5. What transformations are applied to the parent function $f(x) = \sqrt{x}$?

Tip: Always ensure the value under the square root is non-negative to maintain real-valued results.