CalcNotes0203-page21

# CalcNotes0203-page21 - (Section 2.3: Limits and Infinity I)...

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(Section 2.3: Limits and Infinity I) 2.3.21 How can we find an equation for the slant asymptote (SA)? We can use the Long Division technique from Precalculus to reexpress fx () = ± 5 + 3 x 2 + 6 x 3 1 + 3 x 2 as: = 2 x + 1 polynomial part, px ±² ³´ ³ + ± 2 x ± 6 3 x 2 + 1 proper rational part, rx ³ = 2 x + 1 ± 2 x + 6 3 x 2 + 1 Here’s the work: We can now stop the division process here, because the degree of the new dividend is less than the degree of the divisor. The degree of ± 2 x ± 6 is 1, which is less than the degree of 3 x 2 + 0 x + 1 , which is 2; this means that ± 2 x ± 6 is our remainder, and = ± 2 x ± 6 3 x 2 + 1 corresponds to a
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## This note was uploaded on 12/29/2011 for the course MATH 150 taught by Professor Sturst during the Fall '10 term at SUNY Stony Brook.

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