Unformatted text preview: f (1) = a + b + c = 2 f (2) = a + 2 b + 4 c =1 f (3) = a + 3 b + 9 c =8 . Solving the system, we get a + b + c = 2 a + 2 b + 4 c =1 a + 3 b + 9 c =8 = ⇒ a + b + c = 2 b + 3 c =3 2 b + 8 c =10 = ⇒ a + b + c = 2 b + 3 c =3 2 c =4 a + b = 4 b = 3 c =2 = ⇒ a = 1 b = 3 c =2 . Answer: f ( t ) = 1 + 3 t2 t 2 . 1...
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This note was uploaded on 12/26/2011 for the course MATH 214 taught by Professor Conger during the Fall '08 term at University of Michigan.
 Fall '08
 Conger
 Linear Algebra, Algebra

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