Posted by Allison on Wednesday, August 17, 2011 at 9:05pm.
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
second line should have been ----
at x=a , y = x^3 and dy/dx = 3a^2
thank you soooooo much!
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