How does one find the oblique asymptotes of a rational function? For example in
f(x) = (2x^2-3x-1)/(x-2)

Question

How does one find the oblique asymptotes of a rational function? For example in 
f(x) = (2x^2-3x-1)/(x-2)

anonymous · Answer

divide it and find the remainder

anonymous · Answer

i'm looking at here but i can check ur answer < http://answers.yahoo.com/question/index?qid=20090522120106AA53hNZ >

campbell_st · Answer

use polynomial division and don't worry about the remainder... the quotient will be the asymptote

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so y = 2x + 1 is the oblique asymptote.

anonymous · Answer

I was looking more for why you do that division

shubhamsrg · Answer

asymptotes are the lines which are tangent to the curve at infinity(+ or -) 

f(x) =  (2x^2-3x-1)/(x-2)
let px + q be the asymptote,,

so we can see
x->infinity(+,-) (f(x) - (px + q)) = 0

=> (2x^2-3x-1)/(x-2) - px-q =0
=>(2x+1) + 1/(x-2) -px -q =0

=>1/(x-2) + x(2-p) + 1-q =0

=>now on putting x= infintiy, you can easily see, why p=2 and q=1
or asymp. is 2x + 1..

hope you understand why we did long division..