Solution: Assume $ h(x) = ax^2 + bx + c $. Using $ h(1) = a + b + c = 4 $, $ h(2) = 4a + 2b + c = 11 $, $ h(3) = 9a + 3b + c = 22 $. Subtract first equation from second: $ 3a + b = 7 $. Subtract second from third: $ 5a + b = 11 $. Subtract these: $ 2a = 4 \Rightarrow a = 2 $. Then $ 3(2) + b = 7 \Rightarrow b = 1 $. From $ 2 + 1 + c = 4 \Rightarrow c = 1 $. Thus, $ h(x) = 2x^2 + x + 1 $, and $ h(4) = 32 + 4 + 1 = 37 $. \boxed{37}

Thu, 07 May 2026 13:03:20 +0000

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