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To complete the function table, we need to substitute the given values of \( x \) into the function \( h(x) = 2x - 5 \).

### Let's calculate \( h(x) \) for each value of \( x \):

1. For \( x = -4 \):
   \[
   h(-4) = 2(-4) - 5 = -8 - 5 = -13
   \]

2. For \( x = 0 \):
   \[
   h(0) = 2(0) - 5 = 0 - 5 = -5
   \]

3. For \( x = 1 \):
   \[
   h(1) = 2(1) - 5 = 2 - 5 = -3
   \]

4. For \( x = 3 \):
   \[
   h(3) = 2(3) - 5 = 6 - 5 = 1
   \]

5. For \( x = 4 \):
   \[
   h(4) = 2(4) - 5 = 8 - 5 = 3
   \]

### Complete function table:

| \( x \)  | \( h(x) \) |
|:-------:|:----------:|
|   -4    |    -13     |
|    0    |    -5      |
|    1    |    -3      |
|    3    |     1      |
|    4    |     3      |

Let me know if you'd like further explanations!

---

### 5 Related Questions:
1. How do you interpret the function \( h(x) = 2x - 5 \) graphically?
2. What is the slope and y-intercept of the linear function \( h(x) \)?
3. How would the table change if the function was \( h(x) = 3x - 5 \)?
4. What happens to the values of \( h(x) \) as \( x \) increases by 1 each time?
5. Can you determine when \( h(x) = 0 \)?

#### Tip:
Always substitute values carefully into functions and follow the order of operations to ensure accurate results!

Complete the function table for the rule h(x) = 2x - 5.

Learn how to complete a function table for the linear function h(x) = 2x - 5 by substituting different values of x and solving. This problem demonstrates key algebra concepts and is suitable for students in Grades 6-8.

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