Question: 9. For the functionf(x) = x2 - 4x + 1, evaluatef(x + h). (A) x2 - 4x + 1 + h (C) x2 +

9. For the functionf(x) = x2 - 4x + 1, evaluatef(x +

9. For the functionf(x) = x2 - 4x + 1, evaluatef(x + h). (A) x2 - 4x + 1 + h (C) x2 + h2 + 2xh - 4x - 4h + 1 (D) x2 + h2 - 4x - 4h + 1

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