calculus!URGENT

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Show that the tangent line to the curve y=x^3 at the point x=a also hits the curve at the point x=-2a.

Any help?! PLEASE!

  • calculus!URGENT -

    dy/dx = 3x^2
    at x=a , y = x^3 and dy/dx = 3a^2

    equation of tangent:
    y = 3a^2x + b
    but (a,a^3) is on it, so
    a^3 = 3a^2(a) + b
    b = -2a^3

    tangent equation: y = (3a^2)x - 2a^3

    intersect that line with y = x^3
    x^3 = 3a^2x - 2a^3
    x^3 - 3a^2x + 2a^3 = 0
    We already know that x-a is a factor, since x=a is a solution.
    so it factors to
    (x-a)(x^2 + ax - 2a^2) = 0
    (x-a)(x-a)(x - 2a) = 0

    so x=a, or x=a, or x = 2a

  • calculus!URGENT -

    second line should have been ----
    at x=a , y = x^3 and dy/dx = 3a^2

  • calculus!URGENT -

    thank you soooooo much!

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