2.3 - Math 2940 Solutions, Fall 2011 Section 2.3 12) 987...

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Math 2940 Solutions, Fall 2011 Section 2 . 3 12 ) 0 0 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 0 0 1 0 1 0 1 0 0 = 7 8 9 4 5 6 1 2 3 0 0 1 0 1 0 1 0 0 = 9 8 7 6 5 4 3 2 1 and 1 0 0 - 1 1 0 - 1 0 1 1 2 3 1 3 1 1 4 0 = 1 2 3 0 1 - 2 0 2 - 3 17 ) We have a + b + c = 4, a + 2 b + 4 c = 8 and a + 3 b + 9 c = 14, that is 1 1 1 1 2 4 1 3 9 a b c = 4 8 14 . Adding - 1 times the first row to the second and third rows gives 1 1 1 0 1 3 0 2 8 a b c = 4 4 10 . Adding - 2 times the new second row to the third rows gives 1 1 1 0 1 3 0 0 2 a b c = 4 4 2 . So c = 1, and back substituting gives b = 1 and a = 2. Our quadratic is y = 2 + x + x 2 . 19
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This note was uploaded on 01/02/2012 for the course MATH 2940 at Cornell University (Engineering School).

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2.3 - Math 2940 Solutions, Fall 2011 Section 2.3 12) 987...

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