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Math Help - Discontinuity question

  1. #1
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    Discontinuity question

    The graph of y=(x^2-9)/(3x-9) has:
    (sorry for not typing out the answer not related to my question)
    a) a vertical asymptote at x=3
    b)....
    c) a removable discontinuity at x=3
    d)...
    e)...

    I know the answer is c but I got confused with answer a.
    Doesn't the graph of y=(x^2-9)/(3x-9) has a vertical asymptote at x=3 if numbers are plugged into the equation? How do we know if (x-3) should be factored out or not?
    Thanks
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  2. #2
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    Re: Discontinuity question

    You need to recognise that for all \displaystyle x \neq 3 that  \displaystyle \frac{x^2 - 9}{3x - 9} \equiv \frac{(x - 3)(x + 3)}{3(x - 3)} \equiv \frac{x + 3}{3}.

    Since the function is identical to a linear function, which does not have asymptotes, that means that x = 3 is not an asymptote, it's just a "hole", or a removable discontinuity.
    Thanks from LLLLLL
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  3. #3
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    Re: Discontinuity question

    Quote Originally Posted by Prove It View Post
    You need to recognise that for all \displaystyle x \neq 3 that  \displaystyle \frac{x^2 - 9}{3x - 9} \equiv \frac{(x - 3)(x + 3)}{3(x - 3)} \equiv \frac{x + 3}{3}.

    Since the function is identical to a linear function, which does not have asymptotes, that means that x = 3 is not an asymptote, it's just a "hole", or a removable discontinuity.
    Thanks a lot!
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