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We are tasked with evaluating the piecewise-defined function \( h(x) \) at three points: \( h(-2) \), \( h(0) \), and \( h(3) \). The function is defined as follows:

\[
h(x) =
\begin{cases} 
\frac{1}{4}x - 1 & \text{if } x < -2, \\
(x+1)^2 - 1 & \text{if } -2 \leq x < 2, \\
-\frac{1}{2}x + 2 & \text{if } x \geq 2.
\end{cases}
\]

### Step-by-step evaluation:
1. **Evaluate \( h(-2) \):**
   - Since \( -2 \) falls into the interval \( -2 \leq x < 2 \), we use the second case of the piecewise function: \( h(x) = (x+1)^2 - 1 \).
   - Substituting \( x = -2 \) into this expression:
     \[
     h(-2) = (-2+1)^2 - 1 = (-1)^2 - 1 = 1 - 1 = 0.
     \]
   Therefore, \( h(-2) = 0 \).

2. **Evaluate \( h(0) \):**
   - Since \( 0 \) also falls into the interval \( -2 \leq x < 2 \), we use the same expression as before: \( h(x) = (x+1)^2 - 1 \).
   - Substituting \( x = 0 \):
     \[
     h(0) = (0+1)^2 - 1 = 1^2 - 1 = 1 - 1 = 0.
     \]
   Therefore, \( h(0) = 0 \).

3. **Evaluate \( h(3) \):**
   - Since \( 3 \geq 2 \), we use the third case of the function: \( h(x) = -\frac{1}{2}x + 2 \).
   - Substituting \( x = 3 \):
     \[
     h(3) = -\frac{1}{2}(3) + 2 = -\frac{3}{2} + 2 = -\frac{3}{2} + \frac{4}{2} = \frac{1}{2}.
     \]
   Therefore, \( h(3) = \frac{1}{2} \).

### Final answers:
- \( h(-2) = 0 \),
- \( h(0) = 0 \),
- \( h(3) = \frac{1}{2} \).

Would you like any further explanation or additional details on this?

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Here are 5 related questions to explore:
1. How do you determine which piece of a piecewise function to use for a given \( x \)?
2. What happens if a piecewise function has more than three cases?
3. How do you graph a piecewise-defined function?
4. Can a piecewise function be discontinuous? If so, how can you tell?
5. What are the applications of piecewise functions in real-life problems?

**Tip:** Always check the intervals of the piecewise function carefully to ensure you are using the correct expression for the given input.

Suppose the function h(x) is defined as a piecewise function with three cases. Find h(-2), h(0), and h(3).

Learn how to evaluate a piecewise-defined function at specific points such as h(-2), h(0), and h(3). This step-by-step solution explains how to choose the correct function case based on the input value. Ideal for Grades 9-11 students.

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