Given the formula: f(x)=x^2+ax+b and the vertex point (3,9)
prove that this is the vertex of a minimum parabola using instantaneous rate of change.

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1. y = x^2 + ax + b
dy/dx = 2x + a
= 0 at the vertex
so x = -a/2
but the vertex is (3,9)
so 3 = -a/2
a = -6
also (3,9) satisfies y = x^2 + ax + b, and we know a = -6
9 = 9 - 6(3) + b
b = 18

check:
y = x^2 - 6x + 18
completing the square:
y = x^2 - 6x + 9 - 9 + 18
= (x-3)^2 + 9 , whose vertex is (3,9)

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