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Math Help - Help with graphics.

  1. #1
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    Dec 2011
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    Help with graphics.

    Hello!
    Today my maths teacher gave me a task. I don't know why he thought that I am some sort of computer GURU but I couldn't deny this task because it is very important for me to automatically get to 10th grade in my school without entrance examination.

    The task is to program a graph which automatically solves functions.

    Basically it has to work for this function system:
    • x(square root) + y(square root) = R(square root)

    • x(square root) - y(square root)=number(square root)

    I didn't quit understand the whole thing. Teacher said that there isn't a graph for x(square root) - y(square root)=number(square root) and he proved it so I don't know if I have to include it.

    Any help will be much apprieciated. Maybe there's already an app or program that can solve this or maybe a website. Anything really will help.
    Last edited by mr fantastic; December 20th 2011 at 03:40 PM. Reason: Title.
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  2. #2
    MHF Contributor

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    Re: URGENT! Need help with graphics

    I doubt your teacher said that there was no graph. He/she may well have said that this is not a "function". That is, that there can be two different y values for one x value. As for graphing it, the simplest thing to do is to calculate values and mark those points. For example, if x= 0, then \sqrt{y}= \sqrt{R} so that y= \pm R. Mark the two points (0, R) and (0, -R). If x= R/4, then \sqrt{R}{4}+ \sqrt{y}= \sqrt{R}{2}+ \sqrt{y}=\sqrt{R} so \sqrt{y}= \sqrt{R}{2} so y= \pm 4 also. another two points on the graph are (R/4, R/4) and (R/4, -R/4). If x= R, the equation becomes \sqrt{R}+ \sqrt{y}= \sqrt{R} which leads to \sqrt{y}= 0 and y= 0. The point (R, 0) is on the graph.
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