SOLUTION: what is the equation of the quadratic function described of the minimum value of f is f(3)= -5 and f(1) = 2

Question 210791: what is the equation of the quadratic function described of the minimum value of f is f(3)= -5 and f(1) = 2
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
the minimum lies on the axis of symmetry, so f(5) = f(1) = 2

f(x) = ax^2 + bx + c

f(1) = a(1^2) + b(1) + c ___ 2 = a + b + c ___ 2 - a - b = c

f(3) = a(3^2) + b(3) + c ___ -5 = 9a + 3b + c
substituting ___ -5 = 9a + 3b + 2 - a - b ___ -7 = 8a + 2b ___ -14 = 16a + 4b

f(5) = a(5^2) + b(5) + c ___ 2 = 25a + 5b + c
substituting ___ 2 = 25a + 5b + 2 - a - b ___ 0 = 24a + 4b

subtracting equations ___ 14 = 8a ___ 7/4 = a

substituting ___ 0 = 24(7/4) + 4b ___ -42 = 4b ___ -21/2 = b

substituting ___ 2 - 7/4 - -21/2 = c ___ 43/4 = c

f(x) = (7/4)x^2 - (21/2)x + 43/4
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