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Math Help - Point on a Curve parallel to a line [Something Missing ?]

  1. #1
    Junior Member StarlitxSunshine's Avatar
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    Point on a Curve parallel to a line [Something Missing ?]

    This is a question from a worksheet my teacher printed out of a book. It seems like something is missing because I don't know what to do with the question...

    The point on the curve  y = \sqrt[2]{4x+2} at which the normal is parallel to the line y=-x+8 is

    1. (A) (\frac{1}{4}, \sqrt[2]{2})
    2. (B) (\frac{-1}{2}, 0)
    3. (C) (0, \sqrt[2]{2})
    4. (D) (\frac{1}{2}, 2)
    5. (E) (1, \sqrt[2]{6})
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  2. #2
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    Quote Originally Posted by StarlitxSunshine View Post
    This is a question from a worksheet my teacher printed out of a book. It seems like something is missing because I don't know what to do with the question...

    The point on the curve  y = \sqrt[2]{4x+2} at which the normal is parallel to the line y=-x+8 is

    1. (A) (\frac{1}{4}, \sqrt[2]{2})
    2. (B) (\frac{-1}{2}, 0)
    3. (C) (0, \sqrt[2]{2})
    4. (D) (\frac{1}{2}, 2)
    5. (E) (1, \sqrt[2]{6})

    Nothing's missing: the normal to a point on the graph of a function is a straightline through that point which perpendicular to the tangent to the function on that point, so you must:

    1) derivate your function

    2) find what the slope of a NORMAL to any point of the graph is

    3) make this slope equal to the given line's slope (so that they'll be parallel)

    4) solve and find out what x is, and then substitute in the function and find out what the y-coordinate is.

    The answer is (D)

    Tonio
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  3. #3
    Junior Member StarlitxSunshine's Avatar
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    Quote Originally Posted by tonio View Post
    Nothing's missing: the normal to a point on the graph of a function is a straightline through that point which perpendicular to the tangent to the function on that point, so you must:

    1) derivate your function

    2) find what the slope of a NORMAL to any point of the graph is

    3) make this slope equal to the given line's slope (so that they'll be parallel)

    4) solve and find out what x is, and then substitute in the function and find out what the y-coordinate is.

    The answer is (D)

    Tonio
    Oh. That's what a normal means ! >_<

    Okay.

    1). The derivative is \frac{2}{\sqrt[2]{4x+2}}

    2). Wait...so I have to find a line perpendicular to the tangent line -- that's the normal ?
    Last edited by StarlitxSunshine; November 7th 2009 at 02:53 PM. Reason: Wait.. working on it >_<
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