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December 20, 2014

December 20, 2014

Posted by **Lost One** on Monday, December 31, 2012 at 2:38pm.

- Calculus -
**Steve**, Tuesday, January 1, 2013 at 4:13amassuming a polynomial, we see that y' is a quadratic, so

y = ax^3 + bx^2 + cx + d

(0,0) => d=0

(3,-5) => 27a+9b+3c=-5

y' = 3ax^2 + 2bx+c

y'(0) = 0 => c=0

y'(3)=0 => 27a+6b=0

solving for a and b, we get

y = 10/27 x^3 - 5/3 x^2

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