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Math Help - Distance??

  1. #1
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    Exclamation Distance??

    Let P = (x,y) be a point on the graph of .
    Express the distance d from P to the point (2, 0) as a function of x.
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  2. #2
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    Hey - I can't see the graph that you may have posted?
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  3. #3
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    Oh there is no graph. It just says the graph of y=sqrt x
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  4. #4
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    OK, thanks, that was the bit that was missing!

    To find the distance between (2,0) and P you need to use Pythagorus rule:

    a^2+b^2=c^2

    Where c is the length between (2,0) and P.

    To find a (the bottom of the triangle) you need to find the difference between (2,0) and and the value of x.

    To find b (the height of the triangle) you need to find the value of y in terms of x.

    Both of these distances can be expressed in terms of x. Then you can plug them into the pythagorus formula to get your answer.

    Hope this gets you on the way.

    Oh, and drawing a picture will help!
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  5. #5
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    Ok but I dont understand where I can get my x
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  6. #6
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    Or the y
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  7. #7
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    x can take any value from 0 to positive infinity. (x can't be negative as you can't square root a negative number which can be plotted on this graph).

    It may help if you put some real values of x in, and work through the equations, you then might see how to generalise this to all values of x.

    Try putting x=4. Find the difference between (2,0) and (4,0) - then find the value of y when x=4, then you can find the distance.

    Then try putting x=5, to see if you can see a pattern.
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