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    The curve C has equation y = f(x)
    Given that dy/dx = 3x^2 - 20x + 29

    And that C passes thru the point P(2,6)

    And C goes through the point (4,0)
    Find an equation of the the tangent to C at P (the answer is y = x+4)


    The tangent to C at the point Q is parallel to the tangent at P, so calculate the xcoord of Q.

    So far I have worked out the gradient of Q's tangent is 1..

    what next?

    thanks
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    Find the value of the gradient at the point P, then the tangent has equation
    y = mx + c, where m is the gradient you found.
    Subsitute in the co-ordinates of P for y and x and solve to find c
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    (Original post by koldtoast)
    The curve C has equation y = f(x)
    Given that dy/dx = 3x^2 - 20x + 29

    And that C passes thru the point P(2,6)

    And C goes through the point (4,0)
    Find an equation of the the tangent to C at P (the answer is y = x+4)


    The tangent to C at the point Q is parallel to the tangent at P, so calculate the xcoord of Q.

    So far I have worked out the gradient of Q's tangent is 1..

    what next?

    thanks
    Plug 2 (x-coordinate of P) into your expression for dy/dx to get m. dy/dx=3(2)^2 - 20(2) + 29 = 1. Hence, we have y = x + c. So, 6 = 2 + c, --> c=4. So y=x+4
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    (Original post by koldtoast)
    The curve C has equation y = f(x)
    Given that dy/dx = 3x^2 - 20x + 29

    And that C passes thru the point P(2,6)

    And C goes through the point (4,0)
    Find an equation of the the tangent to C at P (the answer is y = x+4)


    The tangent to C at the point Q is parallel to the tangent at P, so calculate the xcoord of Q.
    dy/dx = 3x^2 - 20x + 29
    ---> y = x^3 - 10x^2 + 29x + k
    When x = 2, y = 6:
    ---> 6 = 8 - 40 + 58 + k
    ---> 6 = 26 + k
    ---> k = - 20

    ---> Curve C: y = x^3 - 10x^2 + 29x - 20

    At P: Gradient = 12 - 40 + 29 = 1
    ---> Tangent at P has gradient = 1 and passes through P(2, 6).
    Equation Of Tangent: y - 6 = 1(x - 2)
    ---> y - 6 = x - 2
    ---> y = x + 4

    As the tangent at P is parallel to the tangent at Q, therefore the tangents at P and Q have the same gradient and thus the curve has the same gradient at these 2 points.

    Hence: Gradient (At P) = Gradient (At Q) = 1.
    At Q when dy/dx = 1:
    ---> 1 = 3x^2 - 20x + 29
    ---> 3x^2 - 20x + 28 = 0
    ---> x = {20(+/-)Sqrt[400 - 4(3)(28)]}/6
    = [20(+/-)8]/6
    ---> x Co-ord Of Q = 28/6 = 28/6 = 14/3 OR:
    ---> x Co-ord Of Q = 12/6 = 2
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    Thanks nima..so x = 14/3 or 2 is a perfectly valid answer? I wouldn't have to work out which of the x values it is?
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    (Original post by koldtoast)
    Thanks nima..so x = 14/3 or 2 is a perfectly valid answer? I wouldn't have to work out which of the x values it is?
    its 14/3

    the 2 relates to the P
 
 
 
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