{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 2.3 - Math 2940 Solutions Fall 2011 Section 2.3 12 987 001...

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

2.3 - Math 2940 Solutions Fall 2011 Section 2.3 12 987 001...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online