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We are given the function \( h(x) = \frac{-2x + 4}{x - 5} \), and we need to evaluate it for three values:

1. \( h(2) \)
2. \( h(-10) \)
3. \( h\left( \frac{1}{4} \right) \)

### Let's evaluate each one:

1. **For \( h(2) \):**
   \[
   h(2) = \frac{-2(2) + 4}{2 - 5} = \frac{-4 + 4}{-3} = \frac{0}{-3} = 0
   \]
   So, \( h(2) = 0 \).

2. **For \( h(-10) \):**
   \[
   h(-10) = \frac{-2(-10) + 4}{-10 - 5} = \frac{20 + 4}{-15} = \frac{24}{-15} = -\frac{8}{5}
   \]
   So, \( h(-10) = -\frac{8}{5} \).

3. **For \( h\left( \frac{1}{4} \right) \):**
   \[
   h\left( \frac{1}{4} \right) = \frac{-2\left( \frac{1}{4} \right) + 4}{\frac{1}{4} - 5} = \frac{-\frac{2}{4} + 4}{\frac{1}{4} - \frac{20}{4}} = \frac{-\frac{1}{2} + 4}{-\frac{19}{4}} = \frac{\frac{7}{2}}{-\frac{19}{4}} = \frac{7}{2} \times \frac{4}{-19} = -\frac{28}{19}
   \]
   So, \( h\left( \frac{1}{4} \right) = -\frac{28}{19} \).

### Final Answers:
- \( h(2) = 0 \)
- \( h(-10) = -\frac{8}{5} \)
- \( h\left( \frac{1}{4} \right) = -\frac{28}{19} \)

Would you like more details on any step or have any questions?

Here are five related questions:
1. What is the domain of the function \( h(x) \)?
2. How would the function behave as \( x \to 5 \)?
3. What happens to \( h(x) \) as \( x \to \infty \)?
4. Can we find the x-intercept of the function \( h(x) \)?
5. How do we determine if the function has any vertical or horizontal asymptotes?

**Tip:** When working with rational functions, always check if the denominator could ever be zero to avoid undefined points.

Given that h(x) = (-2x + 4) / (x - 5), find h(2), h(-10), and h(1/4).

Learn how to evaluate the rational function h(x) = (-2x + 4) / (x - 5) for specific values: h(2), h(-10), and h(1/4). This solution includes detailed steps, covering key concepts in algebra and rational functions. Suitable for students in Grades 9-12.

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