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    The gradient of a curve at any point is directly proportional to the x-coordinate of that point. The curve passes through point A with coordinates 3,2. The gradient of the curve at A is 12. Find the equation of the curve
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    Y = 4x +2 maybe
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    if the gradient is proportional to the x coordinate you know the dy/dx = kx where k is a constant value
    then sub in the values you know (gradient at a certain x coordinate) to find k then you can integrate to find the original equation (then sub back in the coordinate you know to find c)
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    (Original post by slaybae)
    The gradient of a curve at any point is directly proportional to the x-coordinate of that point. The curve passes through point A with coordinates 3,2. The gradient of the curve at A is 12. Find the equation of the curve
    Gradient=k(x)
    Solve for k (using given values).

    Then using the equation for the gradient, and the point you are given, you can find the equation of the curve.
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    Is this GCSE?
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    (Original post by Y11_Maths)
    Is this GCSE?
    No - AS, as it requires polynomial calculus.
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    Substitute y & x into the formula. 2=3x12 + c (y=mx+c)
    Now find c, and then you have formulated the equation.
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    (Original post by Pokefan123!)
    Substitute y & x into the formula. 2=3x12 + c (y=mx+c)
    Now find c, and then you have formulated the equation.
    this assumes that it is a linear equation - if the gradient is proportional to the x coordinate, the gradient won't be constant so unless i'm being really thick, i think you need to use calculus
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    (Original post by mupsman2312)
    No - AS, as it requires polynomial calculus.
    Phew ok thanks
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    (Original post by AliG_)
    Y = 4x +2 maybe
    not really curvy enough

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    gradient is proportional to kx
    therefore 12=3k
    k=4
    dy/dx=4x
    intergrate 4x = 2x^2 +c
    y = 2x^2+c
    sub in y and x
    2 = 2(3)^2+c
    c = -16
    final equation is y = 2x^2-16

    someone tell me if i made a stupid mistake(i often do)


    EDIT: just checked on calculator and it looks right
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    (Original post by jb791)
    gradient is proportional to kx
    therefore 12=3k
    k=4
    dy/dx=4x
    intergrate 4x = 2x^2 +c
    y = 2x^2+c
    sub in y and x
    2 = 2(3)^2+c
    c = -16
    final equation is y = 2x^2-16

    someone tell me if i made a stupid mistake(i often do)


    EDIT: just checked on calculator and it looks right
    That's what I got
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    @thekidwhogames
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