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Math Help - Asymptotes and Range

  1. #1
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    Asymptotes and Range

    Function = (X+2)\(X+1)

    When X = -1 for a function, it says "ERROR" on the calculator.

    So if this is an asymtote, doesn't that mean the range would be all real numbers as the values between -2 and -1 would reach infinity?

    Can someone explain why this reasoning is wrong? The real answer is supposed to be all real numbers except 1.
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  2. #2
    Member mathemagister's Avatar
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    Quote Originally Posted by Alan306090 View Post
    When X = -1 for a function, it says "ERROR" on the calculator.

    So if this is an asymtote, doesn't that mean the range would be all real numbers as the values between -2 and -1 would reach infinity?

    Can someone explain why this reasoning is wrong? The real answer is supposed to be all real numbers except 1.
    I want to help you, but it's hard to without knowing the function you are talking about. If you post the function, I'll try my best to help you.
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  3. #3
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    (x+2)/(x+1)
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  4. #4
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    Quote Originally Posted by Alan306090 View Post
    (x+2)/(x+1)
    this might help ...

    does the equation \frac{x+2}{x+1} = 1 have a solution ?
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  5. #5
    Member mathemagister's Avatar
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    Quote Originally Posted by Alan306090 View Post
    (x+2)/(x+1)
    Yes, your reasoning is correct, since there is a vertical asymptote, there will be no upper- or lower limit to the range. However, you need to check for horizontal asymptotes.

    \lim_{x\rightarrow \pm \infty} \frac{x+2}{x+1} = \frac {1}{1} = 1

    This means that y=1 is a horizontal asymptote and can't be part of the range.

    Do you understand?
    Hope I helped you
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