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### Problem 12: Evaluate \( f(x) = -2 |x - 5| \) at each value and simplify.

#### a. \( f(4) \)
You already started solving this:
- \( f(4) = -2 |4 - 5| \)
- \( = -2 |-1| \)
- The absolute value of \(-1\) is \(1\), so \( f(4) = -2 \times 1 = -2 \).

Thus, the value of \( f(4) \) is **-2**.

#### b. \( f(x^2 + 5) \)
Here, we evaluate the same function with \( x^2 + 5 \) inside the absolute value:
- \( f(x^2 + 5) = -2 |x^2 + 5 - 5| \)
- Simplifying inside the absolute value: \( f(x^2 + 5) = -2 |x^2| \)
- Since \( |x^2| = x^2 \) (because squaring a number always gives a non-negative result), we have \( f(x^2 + 5) = -2 x^2 \).

Thus, the simplified expression for \( f(x^2 + 5) \) is **-2x²**.

Would you like further clarification on any part of this?

Here are 5 questions related to this topic:
1. What happens if we evaluate \( f(x) \) for \( x > 5 \)?
2. How would the graph of \( f(x) = -2 |x - 5| \) look like?
3. How does the value of \( x \) affect the absolute value expression in \( f(x) = -2 |x - 5| \)?
4. What would \( f(10) \) be, using the same function \( f(x) = -2 |x - 5| \)?
5. How would the result differ if the coefficient outside the absolute value were positive instead of negative?

**Tip:** Always simplify the expression inside absolute values before applying any coefficients outside.

Evaluate f(x) = -2 |x - 5| at each value and simplify. f(4) and f(x^2 + 5)

Learn how to evaluate and simplify the function f(x) = -2 |x - 5| at different values like f(4) and f(x^2 + 5). This problem covers key concepts of absolute values, algebraic functions, and simplification. Suitable for students in Grades 9-11.

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