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To solve problem 28, we need to determine the quadratic function \( f : \mathbb{R} \to \mathbb{R} \), given by \( f(x) = ax^2 + bx + c \), with the following conditions:

1. The point \( A(1, -6) \) is on the graph of \( f(x) \), so \( f(1) = -6 \).
2. The point \( B(4, -6) \) is on the graph, so \( f(4) = -6 \).
3. The function has a minimum value of 2 when \( x = 1 \), so \( f(1) = 2 \) and \( f'(1) = 0 \) (because the derivative is zero at a minimum point).

Let's break down the solution:

### Step 1: Apply the conditions for the function at specific points

1. **Condition \( f(1) = -6 \):**
   \[
   a(1)^2 + b(1) + c = -6
   \]
   This simplifies to:
   \[
   a + b + c = -6
   \]

2. **Condition \( f(4) = -6 \):**
   \[
   a(4)^2 + b(4) + c = -6
   \]
   Which simplifies to:
   \[
   16a + 4b + c = -6
   \]

3. **Minimum condition \( f(1) = 2 \):**
   Since \( f(1) = 2 \), we actually have:
   \[
   a + b + c = 2
   \]

We have the following equations from the conditions:

1. \( a + b + c = -6 \)
2. \( 16a + 4b + c = -6 \)
3. \( f(1) = 2 \) 

Now

Learn how to determine the quadratic function f(x) = ax^2 + bx + c given points A(1, -6) and B(4, -6) and knowing the minimum value at x = 1. This solution explores key concepts in algebra and graph analysis, ideal for high school math students.

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