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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) = x-4/x(x-4)

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  1. anonymous
    • 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 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 my answer Teacers reply>You find the VA's by setting the denominator to 0. When we get a factor that cancels out, it is not a vertical asymptote. What is it? Graph it to see what it looks like

  2. anonymous
    • one year ago
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    @campbell_st

  3. anonymous
    • one year ago
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    Is this from a test or something?

  4. anonymous
    • one year ago
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    no its from my work

  5. anonymous
    • one year ago
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    i havent taken my midterm yet lol

  6. Jack1
    • one year ago
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    f(x) = x-4/x(x-4) is that: \(\Large f(x) = x- \frac4x \times(x-4) \) or \(\Large f(x) = x- \frac4{x \times(x-4)} \) or \(\Large f(x) = \frac{x-4}{x \times(x-4)} \)

  7. anonymous
    • one year ago
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    |dw:1439383403225:dw|

  8. Jack1
    • one year ago
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    ok... so find when we're dividing by zero so denominator = x(x-4) ... when does this = 0?

  9. anonymous
    • one year ago
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    when we divide bye zero

  10. Jack1
    • one year ago
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    a(b) = 0 so either a = 0 or b = 0 in this case...a = x... so if a = 0, x = 0 a(b) = 0 in this case... b = x-4 if b = 0... x-4 = 0 ... so x = 4 vertical asymtotes at x=0 and x=4... yeah?

  11. Jack1
    • one year ago
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    is that how it works?

  12. Jack1
    • one year ago
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    or do u have to simplify the equation first?

  13. anonymous
    • one year ago
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    idk i gave my answer to the teacher and she said i was wrong so i dont know what to do

  14. Jack1
    • one year ago
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    kk \(\huge f(x) = \frac{x-4}{x \times(x-4)}\) \(\huge f(x) = \frac{\color{red}{x-4}}{x \times\color{red}{(x-4)}}\) \(\huge f(x) = \frac{1}{x}\)

  15. Jack1
    • one year ago
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    so if that's ur eqn.. then simplified it's f(x) = 1/x so when denominator = 0, then u have vertical asymptotoes only 1 v.a in this equation, it's when x = 0

  16. Jack1
    • one year ago
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    right...?

  17. anonymous
    • one year ago
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    right

  18. Jack1
    • one year ago
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    k... k bye

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