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Math Help - Sketching the graphs of functions using differentiation.

  1. #1
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    Sketching the graphs of functions using differentiation.

    Hi, i have got the horzontal asymptote wrong on the following question but can't work out why.

    The function is f(x) = 8(5+x)/16-x^2

    I have worked out that the vertical asymtotes are x=4 and x=-4 because the denominator is 0 when x = 4 and -4.

    and that the stationary points are f(-8)=1/2 and f(-2) = 2

    and that the x intercept is -5 and the y is 5/2

    But when i divide both the numerator and the denomiator by x^2 to get the horizontal asymtote and look at the behaviour as x tend to infinity i keep getting 0. This doesn't work on the graph because of the values of the stationary point f(-8) and the x intercept.

    I'm really stuck on it, if anyone can see where i'm going wrong i'd be grateful. Cheers.

    Nikki
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  2. #2
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    I don't check the details of your work, but you got it right with the horizontal asymptote. It's y = 0

    FYI, some of the graphs have intersection with horizontal asymptote
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  3. #3
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    Quote Originally Posted by NIKKI020 View Post
    Hi, i have got the horzontal asymptote wrong on the following question but can't work out why.

    The function is f(x) = 8(5+x)/16-x^2

    I have worked out that the vertical asymtotes are x=4 and x=-4 because the denominator is 0 when x = 4 and -4.

    and that the stationary points are f(-8)=1/2 and f(-2) = 2

    and that the x intercept is -5 and the y is 5/2

    But when i divide both the numerator and the denomiator by x^2 to get the horizontal asymtote and look at the behaviour as x tend to infinity i keep getting 0. This doesn't work on the graph because of the values of the stationary point f(-8) and the x intercept.

    I'm really stuck on it, if anyone can see where i'm going wrong i'd be grateful. Cheers.

    Nikki
    Hmmm ...

    everything is OK! Have a look on the graph!
    Attached Thumbnails Attached Thumbnails Sketching the graphs of functions using differentiation.-horiz_asympt.png  
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  4. #4
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    Thanks for that i'll leave it how it is then.
    Cheers

    Nikki

    Quote Originally Posted by songoku View Post
    I don't check the details of your work, but you got it right with the horizontal asymptote. It's y = 0

    FYI, some of the graphs have intersection with horizontal asymptote
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  5. #5
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    Thanks, that's been a big help drawing the graph as well.

    Cheers
    Nikki

    Quote Originally Posted by earboth View Post
    Hmmm ...

    everything is OK! Have a look on the graph!
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