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The function \( h(x) \) is defined as a piecewise function with three different expressions for different intervals of \( x \):

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

We are asked to evaluate \( h(-5) \), \( h(1) \), and \( h(2) \).

### Step 1: Find \( h(-5) \)
Since \( -5 \leq -1 \), we use the first piece of the function:
\[
h(-5) = -\frac{1}{2}(-5) - 1 = \frac{5}{2} - 1 = \frac{3}{2}
\]

### Step 2: Find \( h(1) \)
Since \( -1 < 1 < 2 \), we use the second piece of the function:
\[
h(1) = (1 + 1)^2 - 1 = 2^2 - 1 = 4 - 1 = 3
\]

### Step 3: Find \( h(2) \)
Since \( 2 \geq 2 \), we use the third piece of the function:
\[
h(2) = 2(2) + 1 = 4 + 1 = 5
\]

### Final Results:
\[
h(-5) = \frac{3}{2}, \quad h(1) = 3, \quad h(2) = 5
\]

Would you like further clarification on any of these steps?

Here are 5 additional questions to deepen understanding:
1. What happens to the value of \( h(x) \) as \( x \) approaches 2 from the left?
2. Can you find the value of \( h(0) \)?
3. How does the graph of each piece of the function behave for its respective interval?
4. What is the domain and range of the function \( h(x) \)?
5. Are there any points where the function \( h(x) \) is discontinuous?

**Tip:** Always check the intervals carefully in piecewise functions to ensure you're using the correct piece for a given value of \( x \).

Suppose that the function h is defined, for all real numbers, as follows: h(x) = -1/2x - 1 if x <= -1, (x + 1)^2 - 1 if -1 < x < 2, and 2x + 1 if x >= 2. Find h(-5), h(1), and h(2).

This problem demonstrates how to evaluate a piecewise function with different expressions for various intervals of x. Learn how to compute h(-5), h(1), and h(2) using linear and quadratic functions. Ideal for Grades 9-11 students.

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