0 0 10 we see rx 0 on 0 10 and since rx

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Unformatted text preview: on, we have g (x) = 2 − x2x−7 6 . For g (x) = 2, we would need x2x−7 6 = 0. −x− −x− This gives x − 7 = 0, or x = 7. Note that x − 7 is the remainder when 2x2 − 3x − 5 is divided by x2 − x − 6, and so it makes sense that for g (x) to equal the quotient 2, the remainder from the division must be 0. Sure enough, we find g (7) = 2. Moreover, it stands to reason that g must attain a relative minimum at some point past x = 7. Calculus verifies that at x = 13, we have such a minimum at exactly (13, 1.96). The reader is challenged to find calculator windows which show the graph crossing its horizontal asymptote on one window, and the relative minimum in the other. 11 In the denominator, we would have (1billion)2 − 1billion − 6. It’s easy to see why the 6 is insignificant, but to ignore the 1 billion seems criminal. However, compared to (1 billion)2 , it’s on the insignificant side; it’s 1018 versus 109 . We are once again using the fact that for polynomial...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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