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