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    Finding the tangent and normal to a curve

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    The curve C has equation: y = x^3 - 2x^2 - x + 9, x>0

    The point P has coordinates (2,7).

    (a) Show that P lies on C.
    (b) Find the equation of the tangent to C at P, giving your answer in the form of y = mx+c, where m and c are constants.

    The point Q also lies on C.

    Given that the tangent to C at Q is perpendicular to the tangent to C at P,
    (c) show that the x-coordinate of Q is 1/3(2+sqrt(6)).

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    Solution Preview

    a) When x = 2, y = 8 - 2*4 - 2 + 9 = 7

    Since coordinates of P satisfy the equation of the curve, it lies on this curve.

    b) dy/dx = 3 x^2 - 4 x - 1

    at P, dy/dx = 3* 4 - 4* 2 - 1 = 12 - 8 - 1 = 3

    Slope of the tangent ...

    Solution Summary

    Provided is a very clear and step by step solution to this problem of coordinate geometry. This is a typical problem in this area and one can learn a lot from this solution.