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March 31, 2015

March 31, 2015

Posted by **Anonymous** on Monday, February 16, 2009 at 6:11am.

Thanks!

- Math: Integration -
**drwls**, Monday, February 16, 2009 at 6:44amThese conditions must be satisfied:

6 = a*2^2 + b*2 + c = 4a + 2b + c

16 = 9a + 3b + c

dy/dx @ P = 2a*2 + b = 4a + b = 7

Solve those three equations in 3 unknowns. It does not require integral calculus.

4a + 2b + c = 6

4a + b = 7

Combine these two to get b + c = -1

36a + 12 b + 4c = 64

36a + 18 b + 9c = 54

Combine the last 2 to get

6b + 5c = -10

Now you have two equations in two unknowns. Use substitution to eliminate one of the variables.

6b + 5(-1 -b) = -10

b = -5

4a = 7-b = 12 ; a = 3

5c = -10 -6b = -10 + 30 = 20

c = 4

y = 3x^2 -5x + 4

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