Discussion
Let \(f(x) = x^{2} + \alpha x + \beta\) . If \(f\) has a local minimum at (2, 6), then \(f\)(0) is equal to
Answer: A
\(f(0) = \) 10.
A local minimum at \(x = 2\) needs \(f'(2) = 0\), and the point (2, 6) lies on the curve:
\[ \begin{aligned} f'(2) &= 4 + \alpha = 0 \Rightarrow \alpha = -4 \\ f(2) &= 4 - 8 + \beta = 6 \Rightarrow \beta = 10 \end{aligned} \]
So \(f(x) = x^{2} - 4x + 10 = (x - 2)^{2} + 6\) and \(f(0) = \beta = 10\). 6 is the minimum value itself, not \(f(0)\); -6 and the other signs come from slips in \(\alpha\).
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