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     Post subject:Summer was long, need a refresher
    PostPosted:Wed Sep 03, 2008 4:26 pm 
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    Joined:Sun Feb 12, 2006 8:56 pm
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    Okay, I let 25% of what I learned in Algebra last year slip away over the Summer like all the experts say, and now coming into Geometry the teacher gave us a simple assignment.

    I just need to be retaught the methods to doing this assignment.

    Quite simply, I need to write three functions each containing the data points from three graphs I made.


    Yeah, my brain isn't exactly a secure fortress, stuff slips out a lot and goes on cruises in the Caribbean and junk like that, never coming back. Also, I have no idea why I added that random and useless statement just now.

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     Post subject:Re: Summer was long, need a refresher
    PostPosted:Wed Sep 03, 2008 5:50 pm 
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    Uh, what are the graphs, exactly? Lines, parabolas, hyperbolas, ellipses, circles? Every graph has its own function ( = equation).

    -- Griffinhart

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     Post subject:Re: Summer was long, need a refresher
    PostPosted:Wed Sep 03, 2008 5:54 pm 
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    Joined:Sun Feb 12, 2006 8:56 pm
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    Well, it curves upward, so it's not linear. Just a slight curve though.
    Attachment:
    curve.PNG
    curve.PNG [2.22KiB |Viewed 11006 times ]

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     Post subject:Re: Summer was long, need a refresher
    PostPosted:Sat Sep 06, 2008 10:31 am 
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    Looks like a semiparabola. y = a(x-h)^2 + k, x > SomeValue, where a stretches or squishes the graph, h moves the graph left or right, and k moves the graph up or down.

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     Post subject:Re: Summer was long, need a refresher
    PostPosted:Sat Sep 06, 2008 5:32 pm 
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    Joined:Sun Feb 12, 2006 8:56 pm
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    Thanks for that, went to class and all the other kids thought it was half a parabola too.

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     Post subject:Re: Summer was long, need a refresher
    PostPosted:Sat Sep 06, 2008 8:17 pm 
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    You're just lucky that I'm reviewing functions (and logarithms/exponentials)in my AP Calc class right now... or I'd be about as stuck as you were.

    Speaking of which, I should probably study for the test... meh, oh well.

    -- Griffinhart

    PS. By the way, can anyone show me how to solve for x in the following equation?

    e^x + e^(-x) = 3

    I know logarithms are involved, and I know what the answer is (it's something like x = (5 * sqrt(3)) / 2), I just don't know how to actually solve the equation. I mean, if you took the log of both sides of the equation, you'd end up with

    x + -x = ln 3

    And that makes no sense.

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     Post subject:Re: Summer was long, need a refresher
    PostPosted:Sun Sep 07, 2008 5:13 am 
     
    First: ln (e^x + e^(-x)) != x + -x

    To solve this you have to solve a quadratic expression
    e^x + e^(-x) = 3
    <=> e^(x²) + 1 = 3e^x
    <=> e^(x²) - 3e^x + 1 = 0
    e^x = y
    <=> y² - 3y + 1 = 0

    Now you can find the answer.


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     Post subject:Re: Summer was long, need a refresher
    PostPosted:Sun Sep 07, 2008 5:30 am 
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    Joined:Sat Jun 03, 2006 3:51 am
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    Quote:
    First: ln (e^x + e^(-x)) != x + -x
    Oh, I see... I thought you had to take the log of each individual part, not the entire side of the equation, because of the addition.

    Okay, how did you go from
    Quote:
    e^x + e^(-x) = 3
    to
    Quote:
    e^(x²) + 1 = 3e^x
    ? That's the only part I don't understand.

    -- Griffinhart

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     Post subject:Re: Summer was long, need a refresher
    PostPosted:Sun Sep 07, 2008 6:01 am 
     
    I made a mistake... It's not
    e^(x²) + 1 = 3e^x
    but
    e^(2x) + 1 = 3e^x

    e^x + e^(-x) = 3
    <=> e^x + 1/e^x = 3
    <=> e^x ( e^x + 1/e^x) = 3e^x
    <=> e^(2x) + 1 = 3e^x
    ...


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     Post subject:Re: Summer was long, need a refresher
    PostPosted:Sun Sep 07, 2008 6:17 am 
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    Joined:Sat Jun 03, 2006 3:51 am
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    Ah, okay. That makes more sense. Thank you so very much for the help! :D

    -- Griffinhart

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