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August 2, 2014

August 2, 2014

Posted by **Anonymous** on Monday, October 17, 2011 at 11:32pm.

- calculus -
**Reiny**, Monday, October 17, 2011 at 11:52pmFirst of all, both given points are on the curve, so

for

(0,2) ---> 2 = 0+0+0+d

so we know d=2

(-1,-2) ---> -2 = a(-1) + b(1) + c(1) + d

-a + b + c + 2 = -2

a - b - c = 4

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

since (0,2) is a max, we know y' = 0 when x = 0

0 = 3a(0) + 2b(0) + c

c = 0

y'' = 6ax + 2b

we know y'' = 0 when x = -1

0 = -6a + 2b

2b = 6a

b = 3a

so in a-b-c=4

a - 3a - 0 = 4

-2a = 4

a = -2 and b=-6 , c=0 , d=2

**then the function is**

y = -2x^3 - 6x^2 + 2

check my arithmetic.

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