Given the function \( f(x) = x^2 - 6x + 8 \), we are interested in analyzing the behavior of the function over the interval from \( a = 1 \) to \( b = 3 \).

### Let's go step by step:

1. **Find the value of \( f(x) \) at the endpoints:**
   - \( f(1) = (1)^2 - 6(1) + 8 = 1 - 6 + 8 = 3 \)
   - \( f(3) = (3)^2 - 6(3) + 8 = 9 - 18 + 8 = -1 \)

So, \( f(1) = 3 \) and \( f(3) = -1 \).

2. **Find the derivative of \( f(x) \) to check for critical points:**
   The derivative of \( f(x) \) is:
   \[
   f'(x) = 2x - 6
   \]
   Setting the derivative equal to zero to find critical points:
   \[
   2x - 6 = 0 \quad \Rightarrow \quad x = 3
   \]
   
   The critical point occurs at \( x = 3 \), but since this is an endpoint of the interval, we don't have any additional critical points inside the interval \( [1, 3] \).

3. **Analyze the behavior of \( f(x) \) over the interval:**
   - At \( x = 1 \), the function value is 3.
   - At \( x = 3 \), the function value is -1.
   
   The function decreases from \( f(1) = 3 \) to \( f(3) = -1 \).

Would you like to explore more details on this or perform further calculations?

Here are some follow-up questions:
1. What is the vertex of the parabola represented by \( f(x) \)?
2. How does the function behave beyond the interval \( [1, 3] \)?
3. What are the roots of the function \( f(x) \)?
4. Can you rewrite \( f(x) \) in vertex form?
5. How can we determine if the function has a maximum or minimum within the interval?

**Tip:** Finding the vertex of a quadratic function helps you understand where the function reaches its maximum or minimum value!

Explore the behavior of the quadratic function f(x) = x^2 - 6x + 8 on the interval [1, 3]. Learn how to calculate function values at endpoints, find the derivative, and analyze critical points. This problem is ideal for students in Grades 9-11, covering key concepts like derivatives and critical points.

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