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Math Help - graph of y=e^2x

  1. #1
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    graph of y=e^2x

    how does the graph of y=e^(2x)

    I have tried it using a math software but it shows a straight line with 0 as the y - intercept which seems wrong to me as x=0 , y = e^(2(0)) , y = e^0  , y = 1

    I would be delighted if you can show me the graph of it and is there any asymptote? is the asymptote y = 0? , let x = -n , where n is an integer , y = e^(2(-n)) y = 1 / e^(2n) , the y will be getting closer and closer to 0 but will never reach it
    Last edited by ose90; September 26th 2008 at 05:29 PM. Reason: changed
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  2. #2
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    did you graph

    y = e^2x

    or

    y = e^(2x)

    ?
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  3. #3
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    You probably graphed ax where a is e^2 instead of e^(2x). I attached the graph of y = e^{2x}.

    And e^{2x} approaches 0 as x approaches negative infinity.
    Attached Thumbnails Attached Thumbnails graph of y=e^2x-picture-2.png  
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  4. #4
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    Quote Originally Posted by skeeter View Post
    did you graph

    y = e^2x

    or

    y = e^(2x)

    ?
    sry it is y=e^(2x)
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  5. #5
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    Quote Originally Posted by Chop Suey View Post
    You probably graphed ax where a is e^2 instead of e^(2x). I attached the graph of y = e^{2x}.

    And e^{2x} approaches 0 as x approaches negative infinity.

    Thanks a lot!
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