1. The problem states a quadratic function $f(x) = ax^2 + bx + 2$ has a critical point at $(1,4)$. 2. A critical point occurs where the derivative of $f(x)$ equals zero. 3. Find the derivative of $f(x)$: $$f'(x) = 2ax + b$$ 4. Substitute the critical point $x = 1$ into $f'(x)$ and set it equal to zero: $$f'(1) = 2a(1) + b = 0 \implies 2a + b = 0$$ 5. Also, since $(1,4)$ lies on the graph, substitute $x=1$ into $f(x)$ and set equal to 4: $$f(1) = a(1)^2 + b(1) + 2 = a + b + 2 = 4$$ 6. Simplify this equality: $$a + b = 2$$ 7. From step 4, express $b$ in terms of $a$: $$b = -2a$$ 8. Substitute $b = -2a$ into the equation from step 6: $$a + (-2a) = 2 \implies -a = 2 \implies a = -2$$ 9. Now find $b$ using $b = -2a$: $$b = -2(-2) = 4$$ 10. Finally, calculate $a - b$: $$a - b = -2 - 4 = -6$$ Answer: $-6$, which corresponds to option (c).