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Math Help - rational function

  1. #1
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    rational function

    Hello,

    I'm left with the instructions:

    "A rational function, R(x) has the following characteristics: a vertical asymptote at x=3, a horizontal asymptote at y=2, and a hole at (2,-2). Sketch the function and determine what it could be."

    The function I have created thus far is
    Code:
    2x(x-2) / (x-3)(x-2)
    The problem is even though I have a hole for my function at x=2, the hole does not occur when y=-2.

    My instructions are to "Describe what you must do in order for the hole to appear at (2,-2). Find a."

    I suppose that a is a factor (or rather (x-a) or something of that format) which I need to apply to my function to create a hole at (2,-2), but I have never done this before and I've tried just plowing through math to figure this out, but I have no method!

    Thanks a lot if you can help.

    -Mike
    Last edited by mikedwd; May 14th 2008 at 02:24 PM.
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  2. #2
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    Oh, I've found this:

    http://www.mathhelpforum.com/math-he...functions.html

    Quite sorry, I'll read through this thread and decide if I should delete!

    EDIT: inability to delete threads?
    Last edited by mikedwd; May 14th 2008 at 02:12 PM.
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  3. #3
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    Quote Originally Posted by mikedwd View Post
    Oh, I've found this:

    http://www.mathhelpforum.com/math-he...functions.html

    Quite sorry, I'll read through this thread and decide if I should delete!

    EDIT: inability to delete threads?
    Moderators (and Ads) can delete threads if they feel it is necessary. I think this thread provides some good food for thought for someone who has similar questions, so I see no need to delete it.

    -Dan
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  4. #4
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    Quote Originally Posted by mikedwd View Post
    Hello,

    I'm left with the instructions:

    "A rational function, R(x) has the following characteristics: a vertical asymptote at x=3, a horizontal asymptote at y=2, and a hole at (2,-2). Sketch the function and determine what it could be."

    The function I have created thus far is
    Code:
    2x(x-2) / (x-3)(x-2)
    The problem is even though I have a hole for my function at x=2, the hole does not occur when y=-2.

    My instructions are to "Describe what you must do in order for the hole to appear at (2,-2). Find a."

    I suppose that a is a factor (or rather (x-a) or something of that format) which I need to apply to my function to create a hole at (2,-2), but I have never done this before and I've tried just plowing through math to figure this out, but I have no method!

    Thanks a lot if you can help.

    -Mike
    Well, the thread here: http://www.mathhelpforum.com/math-he...tml#post144408

    goes a long way to solving the problem.

    But doing it your way, you have
    Code:
    2x(x-2) / (x-3)(x-2)
    but you get the wrong value of y for the hole. You get (2, -4) instead of (2, -2) .....

    Anyway, persevering with your way:

    The problem with your function is that you need
    Code:
    y = 2x / (x-3)
    to be modified in some way so that when x = 2, y = -2. At the same time you need to make sure that y = 2 remains a horizontal asymptote .......

    So why not consider finding the value of A such that
    Code:
    y = (2x + A) / (x-3)
    gives y = -2 when x = 2 ......

    Then your final function will be
    Code:
    (2x + A)(x-2) / (x-3)(x-2)
    which can be simplified as necessary.

    You will get the answer I found at the above hyperlink.
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