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anonymous

  • one year ago

Find the vertical asymptotes, if any, of the graph of the rational function. Show your work. f(x) =

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  1. anonymous
    • one year ago
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    @dan815 @abb0t

  2. anonymous
    • one year ago
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    @DullJackel09 @TheRaggedyDoctor

  3. anonymous
    • one year ago
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    @Luigi0210

  4. anonymous
    • one year ago
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    @sammixboo

  5. anonymous
    • one year ago
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    @poopsiedoodle @pooja195

  6. anonymous
    • one year ago
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    plz help me id understand this trickery at all

  7. anonymous
    • one year ago
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    @dan815

  8. dan815
    • one year ago
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    okay pick a rational function and tell me

  9. anonymous
    • one year ago
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    what is a rational function lol

  10. anonymous
    • one year ago
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  11. anonymous
    • one year ago
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    haha

  12. freckles
    • one year ago
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    Consider where the fraction is not defined. When you have a fraction, what is that famous number you cannot divide by?

  13. anonymous
    • one year ago
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    0 lol i think or im stupid lol

  14. freckles
    • one year ago
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    You aren't stupid. And that is an awesome answer because it is true. So when is your fraction's bottom zero? For what value of x, do you have x-1 is zero?

  15. anonymous
    • one year ago
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    good question idk how to answer them

  16. freckles
    • one year ago
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    what number can you take away 1 from and have zero?

  17. freckles
    • one year ago
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    is 5-1 zero? is 2-1 zero?

  18. anonymous
    • one year ago
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    no 1-1 is 0

  19. freckles
    • one year ago
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    yes x-1=0 has solution x=1 you could have also just added 1 on both sides

  20. anonymous
    • one year ago
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    true ok i get it somewhat now

  21. anonymous
    • one year ago
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    ok so if i was to answer this question then how would i phrase this

  22. freckles
    • one year ago
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    x=1 is your vertical asymptote though if there was a factor of x-1 on top then it could possibly be a hole (but you do not have that here) that is f(x)=(x-1)/(x-1) has a hole at x=1 (not a vertical asymptote) that is f(x)=x(x-1)/(x-1) also has a hole at x=1 but g(x)=(x-1)/(x-1)^2 has a vertical asympote at x=1 since it has more (x-1)'s on bottom then on top

  23. anonymous
    • one year ago
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    i hate math so bad lol good thing this is my lsast math class lol thnx for your help

  24. freckles
    • one year ago
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    np

  25. anonymous
    • one year ago
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    thank you to dan

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