Solution: Let $ h(x) = ax^2 + bx + c $. From $ h(1) = a + b + c = 5 $ and $ h(-1) = a - b + c = 3 $, adding gives $ 2a + 2c = 8 $, so $ a + c = 4 $. The sum of roots is $ -rac{b}{a} = 4 $, so $ b = -4a $. Substituting $ b = -4a $ into $ a + b + c = 5 $: $ a - 4a + c = 5 $ â†’ $ -3a + c = 5 $. Since $ a + c = 4 $, subtracting gives $ -4a = 1 $, so $ a = -rac{1}{4} $. Then $ c = 4 - a = 4 + rac{1}{4} = rac{17}{4} $, and $ b = -4a = 1 $. Thus, $ h(x) = -rac{1}{4}x^2 + x + rac{17}{4} $. Multiplying by 4 to eliminate fractions: $ h(x) = -x^2 + 4x + 17 $. Verifying $ h(1) = -1 + 4 + 17 = 20 $? Wait, inconsistency. Rechecking: $ a = -1/4 $, $ c = 17/4 $, $ b = 1 $. Then $ h(1) = -1/4 + 1 + 17/4 = (-1 + 4 + 17)/4 = 20/4 = 5 $, correct. $ h(-1) = -1/4 -1 + 17/4 = ( -1 -4 + 17 )/4 = 12/4 = 3 $, correct. Sum of roots $ -b/a = -1 / (-1/4) = 4 $, correct. Final answer: $ oxed{-x^2 + 4x + \dfrac{17}{4}} $ or $ oxed{-\dfrac{1}{4}x^2 + x + \dfrac{17}{4}} $.

Sun, 03 May 2026 05:20:27 +0000

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