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sclowers Group Title

Find the vertical asymptotes, if any, for the following function. Please show all of your work. f(x) = x+11/x^2-16x

  • 3 years ago
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  1. jamesm Group Title
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    x^2 - 16x = 0

    • 3 years ago
  2. jim_thompson5910 Group Title
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    the denominator cannot be zero if it were, then x^2-16x = 0 x(x-16) = 0 x=0 or x-16=0 x=0 or x=16 So the values x=0 or x=16 cause a division by zero error So the vertical asymptotes are x=0 and x=16

    • 3 years ago
  3. whatevs Group Title
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    james I have a question..after your done with this one..can you come back to my question? thank you

    • 3 years ago
  4. whatevs Group Title
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    never mind..thank you for helping me with those questions

    • 3 years ago
  5. jamesm Group Title
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    i can help you. which question do i go bck to?

    • 3 years ago
  6. sclowers Group Title
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    Thank you Jamesm!!

    • 3 years ago
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