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Math Help - [SOLVED] P=(x,y) on the graph of y=sqrt(x)...keep coming up with wrong answer

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    [SOLVED] P=(x,y) on the graph of y=sqrt(x)...keep coming up with wrong answer

    Let P=(x,y) be a point on the graph of y=sqrt(x). Express the distance d from P to the point (1,0) as a funtion of x.

    so I did:
    f(x)=sqrt((x^2)+(y^2))
    f(x)=sqrt((x^2)+(sqrt(x)^2))
    f(x)=sqrt(x^2+x)
    f(x)=x(x+1)

    my answer options are:
    a) d(x) = x^2 +2x+2
    b) d(x) = sqrt(x^2-x+1)
    c) d(x) = sqrt(x^2-2x+2)
    d) d(x) = x^2 -x+1

    What am I doing wrong?
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  2. #2
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    The distance from any point (a,b) to the point (1,0) is given by \sqrt{(1-a)^2 + (0-b)^2}.
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    Quote Originally Posted by ninobrn99 View Post
    What am I doing wrong?
    You're finding the distance from the origin, not from P.

    You should have

    d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

    = \sqrt{(x - 1)^2+(y-0)^2}

    = \sqrt{(x-1)^2+\left(\sqrt x\right)^2}
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    Quote Originally Posted by ninobrn99 View Post
    Let P=(x,y) be a point on the graph of y=sqrt(x). Express the distance d from P to the point (1,0) as a funtion of x.

    so I did:
    f(x)=sqrt((x^2)+(y^2))
    f(x)=sqrt((x^2)+(sqrt(x)^2))
    f(x)=sqrt(x^2+x)
    f(x)=x(x+1)

    my answer options are:
    a) d(x) = x^2 +2x+2
    b) d(x) = sqrt(x^2-x+1)
    c) d(x) = sqrt(x^2-2x+2)
    d) d(x) = x^2 -x+1

    What am I doing wrong?
    Okay, why did you do that? What happened to the point (1, 0)? What you give (until your last line) is the distance from (x, \sqrt{x}) to (0, 0). But your last line is wrong- x^2+ x= x(x+1), not \sqrt{x^2+ x}. For that you still need the square root: \sqrt{x^2+ x}= \sqrt{x(x+1)}.

    But as I said, that would give the distance to (0, 0), not (1, 0).
    Start with d= \sqrt{(x- 1)^2+ (y- 0)^2}= \sqrt{(x- 1)^2+ y^2} and then replace y^2 with x.
    Last edited by mr fantastic; February 14th 2009 at 06:00 PM. Reason: Fixed a latex tag
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    Quote Originally Posted by Plato View Post
    The distance from any point (a,b) to the point (1,0) is given by \sqrt{(1-a)^2 + (0-b)^2}.
    im coming up with:
    d(x) = sqrt((a^2-2a+1) + b^2)
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    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by ninobrn99 View Post
    im coming up with:
    d(x) = sqrt((a^2-2a+1) + b^2)
    Yes, and in this case (a,\,b)=\left(x,\,\sqrt x\right), so substitute and simplify.
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  7. #7
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    Quote Originally Posted by Reckoner View Post
    You're finding the distance from the origin, not from P.

    You should have

    d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

    = \sqrt{(x - 1)^2+(y-0)^2}

    = \sqrt{(x-1)^2+\left(\sqrt x\right)^2}
    ok, so then :
    =\sqrt{x^2-x+1}

    Quote Originally Posted by HallsofIvy View Post
    Okay, [b]why[b] did you do that? What happened to the point (1, 0)? What you give (until your last line) is the distance from (x, \sqrt{x}) to (0, 0). But your last line is wrong- x^2+ x= x(x+1), not \sqrt{x^2+ x}. For that you still need the square root: \sqrt{x^2+ x}= \sqrt{x(x+1)}.

    But as I said, that would give the distance to (0, 0), not (1, 0).
    Start with d= \sqrt{(x- 1)^2+ (y- 0)^2}= \sqrt{(x- 1)^2+ y^2}[/tex] and then replace y^2 with x.
    I did what I did because I was following the example in the book. That's what was throwing me off. This book doesn't explain thouroughly enough. It just shows a few examples and that's it. I usually use mathway, but I am really struggling with these last 3 word problems!!
    Last edited by mr fantastic; February 14th 2009 at 06:01 PM. Reason: Merged posts
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