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Let's find the domain for each of the given functions one by one.

1. **First function: \( f(x) = \sqrt{8 - 4x} \)**
   - The domain is determined by the expression inside the square root. For the square root to be defined, the expression inside must be **non-negative**:
     \[
     8 - 4x \geq 0
     \]
     Solving for \( x \):
     \[
     8 \geq 4x \quad \text{or} \quad x \leq 2
     \]
     Therefore, the domain is \( (-\infty, 2] \).

2. **Second function: \( f(x) = \frac{4}{x^2 - 8x - 20} \)**
   - The domain is restricted by the denominator. We must ensure the denominator is **non-zero**:
     \[
     x^2 - 8x - 20 \neq 0
     \]
     Solving this quadratic equation using the quadratic formula:
     \[
     x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-20)}}{2(1)} = \frac{8 \pm \sqrt{64 + 80}}{2} = \frac{8 \pm \sqrt{144}}{2} = \frac{8 \pm 12}{2}
     \]
     So the solutions are \( x = 10 \) and \( x = -2 \). These are the values that make the denominator zero. Hence, the domain is all real numbers **except** \( x = 10 \) and \( x = -2 \):
     \[
     (-\infty, -2) \cup (-2, 10) \cup (10, \infty)
     \]

3. **Third function: \( f(x) = \frac{\sqrt{x - 4}}{2x^2 - 17x + 8} \)**
   - The domain is restricted by both the square root and the denominator:
     - For the square root to be defined, the expression inside must be **non-negative**:
       \[
       x - 4 \geq 0 \quad \Rightarrow \quad x \geq 4
       \]
     - Additionally, the denominator must be **non-zero**:
       \[
       2x^2 - 17x + 8 \neq 0
       \]
       Using the quadratic formula to solve \( 2x^2 - 17x + 8 = 0 \):
       \[
       x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(2)(8)}}{2(2)} = \frac{17 \pm \sqrt{289 - 64}}{4} = \frac{17 \pm \sqrt{225}}{4} = \frac{17 \pm 15}{4}
       \]
       The solutions are \( x = 8 \) and \( x = \frac{1}{2} \). These values make the denominator zero.
     - Since \( x \geq 4 \) from the square root condition, we exclude \( x = 8 \) (since the denominator is zero there). Thus, the domain is:
       \[
       [4, 8) \cup (8, \infty)
       \]

### Summary of Domains:
1. \( f(x) = \sqrt{8 - 4x} \): \( (-\infty, 2] \)
2. \( f(x) = \frac{4}{x^2 - 8x - 20} \): \( (-\infty, -2) \cup (-2, 10) \cup (10, \infty) \)
3. \( f(x) = \frac{\sqrt{x - 4}}{2x^2 - 17x + 8} \): \( [4, 8) \cup (8, \infty) \)

Let me know if you want further details or explanations! 

Here are five related questions to explore:
1. How do we determine the domain of a logarithmic function?
2. What is the difference between finding the domain of rational and radical functions?
3. How do you solve quadratic inequalities?
4. Can a function have an empty domain?
5. How do discontinuities affect the domain of a function?

**Tip**: When working with square roots, always check for values that make the expression inside non-negative before solving further!

Choose the correct domain for the functions: 1) f(x) = √(8 - 4x), 2) f(x) = 4 / (x^2 - 8x - 20), 3) f(x) = √(x - 4) / (2x^2 - 17x + 8).

Learn how to find the domain of various functions, including square roots and rational functions. Explore examples with radical and quadratic expressions to determine valid inputs, excluding values that make the denominator zero or result in negative square roots. Ideal for high school students.

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