2.6.8-eg-1 - information about the way (shown above) that...

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Example Find all vertical, horizontal, and oblique asymptotes of the curve y = f ( x ), where f ( x ) = 4 x 2 - 3 2 x + 1 . Solution: Vertical asymptotes: The function f is undefined at x = - 1 / 2. Using the fact that lim x ( - 1 / 2) - (2 x + 1) = 0 - , and lim x ( - 1 / 2) + (2 x + 1) = 0 + , and that the numerator of f ( x ) is negative at x = - 1 / 2, it follows that lim x ( - 1 / 2) - 4 x 2 - 3 2 x + 1 = + , and lim x ( - 1 / 2) + 4 x 2 - 3 2 x + 1 = -∞ , or that the curve y = f ( x ) has a single vertical asymptote at the line x = - 1 / 2 . Horizontal asymptotes: Looking at the behavior of f ( x ) as x → ±∞ , lim x →∞ 4 x 2 - 3 2 x + 1 = lim x →∞ 4 x - (3 /x ) 2 + (1 /x ) = + , and lim x →-∞ 4 x 2 - 3 2 x + 1 = lim x →∞ 4 x - (3 /x ) 2 + (1 /x ) = -∞ . so the curve does not approach a constant (i.e., a horizontal line) as x → ±∞ . Thus there are no horizontal asymptotes for this curve. Oblique asymptotes: An oblique asymptote for the curve y = f ( x ) gives even more
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Unformatted text preview: information about the way (shown above) that the curve y = f ( x ) becomes unbounded as x . To determine such an asymptote, we rewrite the function f ( x ) = 4 x 2-3 2 x + 1 using long division: 2 x-1 2 x + 1 ) 4 x 2 + 0 x-3 4 x 2 + 2 x-2 x-3-2 x-1-2 Thus, f ( x ) = 4 x 2-3 2 x + 1 = 2 x-1-2 2 x + 1 , where lim x 2 2 x + 1 = 0 , and lim x - 2 2 x + 1 = 0 . So as x , the part of y = f ( x ) that does not go to zero is the line y = 2 x-1 . It follows that the curve y = f ( x ) approaches this (non-horizontal) line as x , or that the line y = 2 x-1 is an oblique asymptote for the curve. 2...
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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2.6.8-eg-1 - information about the way (shown above) that...

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