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anonymous

  • one year ago

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.

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  1. anonymous
    • one year ago
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    x^2/x-1 i think .

  2. anonymous
    • one year ago
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    1/[(x-2) (x-1)]

  3. anonymous
    • one year ago
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  4. anonymous
    • one year ago
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    so i should just write 1/[(x-2) (x-1)]

  5. anonymous
    • one year ago
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    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)]

  6. anonymous
    • one year ago
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    what proof can i give?

  7. anonymous
    • one year ago
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    Nevermind you are right, the degree of the denoinator has to be greater than the numerator.

  8. anonymous
    • one year ago
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    you don't need 2 in the numerator. f(x) = 1 / [( x -2)(x-1) ] is fine

  9. anonymous
    • one year ago
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    yes sorry i misread it . its late :|

  10. anonymous
    • one year ago
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    what proof do i have, like how can i know that putting a 1 in front will give me my answer

  11. anonymous
    • one year ago
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    and no worries, thanks for helping me

  12. anonymous
    • one year ago
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    for large x, the denominator grows to infinity. and 1 / infinity gets close to zero

  13. anonymous
    • one year ago
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    |dw:1439785307400:dw|

  14. anonymous
    • one year ago
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    the x^2 term in the denominator goes to infinity much faster than the other terms

  15. anonymous
    • one year ago
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    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)

  16. anonymous
    • one year ago
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    @jayzdd should i just leave it like that^?

  17. anonymous
    • one year ago
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    yes, you can leave out lines 3 and 4 since they repeat

  18. anonymous
    • one year ago
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    thank you

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