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Thread: graphing a quartic

  1. #1
    Junior Member
    Mar 2009

    Lightbulb graphing a quartic

    The equation of the graph is f(x)=X^4 -4x^3

    The x intercepts are at (0,0) and (4,0)

    The y-intercept is at (0,0)

    I found f'(x) to be 4x^3-12x^2, meaning the critical points are at (0,0) and (3,-27).

    Am I correct in saying that because its a quartic, there are no horizontal or vertical asymptotes?

    If so, it doesnt make sense when I graph it, I get a slanted N which intersects x in three places (-4,0), (0,0), (4,0).
    But when I enter the equation in an online graphing tool, [IMG]file:///C:/Users/NOORMA%7E1/AppData/Local/Temp/moz-screenshot.jpg[/IMG]I get a deep U.

    Am i doing something wrong?
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  2. #2
    Super Member 11rdc11's Avatar
    Jul 2007
    New Orleans
    correct it has no asymptotes. Also the reason why the looks diff is because you have to zoom out and then it will look like a U.
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