anonymous
  • anonymous
Give an example of a rational function that has a horizontal asymptote at y = 0 and a vertical asymptote at, x = 2 and x = 1.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
x^2/x-1 i think .
anonymous
  • anonymous
1/[(x-2) (x-1)]
anonymous
  • anonymous

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anonymous
  • anonymous
so i should just write 1/[(x-2) (x-1)]
anonymous
  • anonymous
An example of a rational function that has a horizontal asymptote at y = 0 and a vertical asymptote at, x = 2 and x = 1. horizontal asymptote at y = 0 vertical asymptote at x = 2 and x = 1 meaning that id have to use (x-2) as well as (x-1) and using a one on top of the division would give me the result that i want, which is a rational function that has a horizontal asymptote at y = 0 and a vertical asymptote at, x = 2 and x = 1. 1/[(x-2) (x-1)]
anonymous
  • anonymous
what proof can i give?
anonymous
  • anonymous
Nevermind you are right, the degree of the denoinator has to be greater than the numerator.
anonymous
  • anonymous
you don't need 2 in the numerator. f(x) = 1 / [( x -2)(x-1) ] is fine
anonymous
  • anonymous
yes sorry i misread it . its late :|
anonymous
  • anonymous
what proof do i have, like how can i know that putting a 1 in front will give me my answer
anonymous
  • anonymous
and no worries, thanks for helping me
anonymous
  • anonymous
for large x, the denominator grows to infinity. and 1 / infinity gets close to zero
anonymous
  • anonymous
|dw:1439785307400:dw|
anonymous
  • anonymous
the x^2 term in the denominator goes to infinity much faster than the other terms
anonymous
  • anonymous
An example of a rational function that has a horizontal asymptote at y = 0 and a vertical asymptote at, x = 2 and x = 1. horizontal asymptote at y = 0 vertical asymptote at x = 2 and x = 1 meaning that id have to use (x-2) as well as (x-1) and using a one on top of the division would give me the result that i want, which is a rational function that has a horizontal asymptote at y = 0 and a vertical asymptote at, x = 2 and x = 1. The 1/infinity would be close to zero (horizontal asymptote at y = 0) 1/[(x-2) (x-1)] which equals to 1/(x^2-3x+2)
anonymous
  • anonymous
@jayzdd should i just leave it like that^?
anonymous
  • anonymous
yes, you can leave out lines 3 and 4 since they repeat
anonymous
  • anonymous
thank you

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