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Math Help - Differentiation

  1. #1
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    Differentiation

    I am having trouble understanding the method for finding the y-coordinate.

    My question is:
    Find the y-Coordinate and the value of the dradient at the point P with x-coordinate 1 on the curve with wquation y=3+2x-x^2

    Any help much appreciated
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  2. #2
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    Re: Differentiation

    y = 3+2x-x^2
    dy/dx = 2-2x.
    Therefore the gradient of point P with x-coordinate 1 is 2-2(1) = 0

    the coordinate of point P is (1,y) <-- since y coordinate is unknown.
    (1,y) satisfy the equation y = 3+2x-x^2
    Last edited by mr fantastic; September 18th 2011 at 01:04 PM.
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  3. #3
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    Re: Differentiation

    Quote Originally Posted by jacobjacob44 View Post
    I am having trouble understanding the method for finding the y-coordinate.

    My question is:
    Find the y-Coordinate and the value of the dradient at the point P with x-coordinate 1 on the curve with wquation y=3+2x-x^2
    y(1) = 3 + 2(1) - (1)^2 = 4

    the gradient at that point is y'(1), the derivative of y evaluated at the point in question.

    what do you know about finding the derivative of a function?
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  4. #4
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    Re: Differentiation

    I understand how you have got the Gradient but not how you tell which is actually the Y coordinate.
    As for what i know of a derivative of a function - not much, had one lesson with my teacher on it and understood about 20% of what she said.
    Last edited by mr fantastic; September 18th 2011 at 01:05 PM.
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  5. #5
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    Re: Differentiation

    Quote Originally Posted by jacobjacob44 View Post
    Apologies i should have put it in calculus yes. I understand how you have got the Gradient but not how you tell which is actually the Y coordinate.
    As for what i know of a derivative of a function - not much, had one lesson with my teacher on it and understood about 20% of what she said.
    y = f(x) ... if x = 1 , then y = f(1)

    it's that simple.
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