To solve this problem, let's analyze the function and the given critical point:

Given: $f(x) = ax^2 + bx + 2$ and that the function $f(x)$ has a critical point at $(1, 4)$.

Steps to Solve

1. Calculate the derivative of $f(x)$: $f'(x) = 2ax + b$

2. Set the derivative to zero at the critical point $x = 1$ since, at a critical point, the derivative is zero: $f'(1) = 2a(1) + b = 0$ 2a + b = 0 \tag{1}

3. Substitute the point $(1, 4)$ into the original function because $f(1) = 4$: $f(1) = a(1)^2 + b(1) + 2 = 4$ $a + b + 2 = 4$ a + b = 2 \tag{2}

4. Solve the system of equations formed by (1) and (2):

   From equation (1): $b = -2a$.

   Substitute $b = -2a$ into equation (2): $a - 2a = 2$ $-a = 2$ $a = -2$

   Now substitute $a = -2$ back into $b = -2a$: $b = -2(-2) = 4$

5. Calculate $a - b$: $a - b = -2 - 4 = -6$

Answer

The correct answer is: (C) -6

Would you like more details on the solution steps or have any questions?

Here are some related questions to deepen your understanding:

1. How would the critical point change if the constant term were different in the quadratic function?
2. Can you determine if this critical point is a maximum or minimum based on the value of $a$?
3. What would be the impact on the solution if the given critical point were at $x = 2$ instead of $x = 1$?
4. How does the derivative of a quadratic function generally indicate the slope at any given point?
5. If $f(x)$ were a cubic function, what additional steps would be necessary to find critical points?

Tip: For quadratic functions, critical points occur at the vertex, which simplifies finding values when given one point of the function and its derivative at that point.