(A1)(ft) OR = 2x 3 f′(x) – (M1) Note: (M1) for correct differentiation = 1.5 x (A1)(ft) = y –2.25 (A1)(ft) OR for finding the midpoint of 0 and 3 (M1) = 1.5 x (A1)(ft) = y –2.25 (A1)(ft) (C3) Note: If final answer is given as (1.5, –2.25) award a maximum of (M1)(A1)(A0) [6] 2. (a) (0) = 25 (A1) Q (C1) (b) = 0.003(20)2 0.625(20) + 25 = 13.7 (A1) Q(20) – (C1)
(A1)(ft) OR f′(x) = 2x – 3 (M1) Note: (M1) for correct differentiation x = 1.5 (A1)(ft) y = –2.25 (A1)(ft) OR for finding the midpoint of 0 and 3 (M1) x = 1.5 (A1)(ft) y = –2.25 (A1)(ft) (C3) Note: If final answer is given as (1.5, –2.25) award a maximum of (M1)(A1)(A0) [6] 2. (a) Q (0) = 25 (A1) (C1) (b) Q(20) = 0.003(20)2 – 0.625(20) + 25 = 13.7 (A1) (C1)
ms_1_review_package_for_quadratic_functions2.docx