CHECK MY ALGEBRA 2

Write an equation that would have a hole at x = 4 and a vertical asympote at x = -2.

My answer was f(x)= (x-4)/(x+2)

If it is not correct can someone please help me solve it and explain it to me? Thanks!

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asked by Kate
  1. a hole is a disruption at a single point
    ... try [(x - 4) / (x - 4)]
    ... the division by zero causes the discontinuity, but the equal numerator means no other point is affected

    f(x) = (x - 4) / [(x - 4) (x + 2)]

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    posted by Scott
  2. Ohhh I see thanks Scott!

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    posted by Kate
  3. Sorry I actually have one more question. After solving for your equation the vertical asymptote would actually be 2 and not -2?? So can you still help me create the correct equation

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    posted by Kate
  4. Would it be f(x) = (x-4) / (x-4)(x+2)???

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    posted by Kate
  5. ?????

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    posted by Scott
  6. Scott was right, the denominator causing the vertical asymptote at x = -2 is
    x+2

    final answer:
    f(x) = (x-4)/( (x-4)(x+2) )

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    posted by Reiny

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