# 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.

• 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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