Posted by **Allison** on Wednesday, August 17, 2011 at 9:05pm.

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 -
**Reiny**, Wednesday, August 17, 2011 at 9:23pm
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 -
**Reiny**, Wednesday, August 17, 2011 at 9:25pm
second line should have been ----

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

- calculus!URGENT -
**Allison**, Wednesday, August 17, 2011 at 9:47pm
thank you soooooo much!

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