Please check my maths?
posted by holly .
How should i do this question??
it says:
find the equation of the tangent to the curve y=(x2)^3 at the point (3,1). calculate the coordinates of the point where this tangent meets the curve again.
I know how to get the first part, if i'm not wrong, it's y = 3x  8
but no matter how i do, i cant get another point that is meeting the curve again!
This is what i did for the second part:
y=3x8  (1)
y=(x2)^3 (2)
(x2)^3  3x + 8 = 0
simplify:
x^3  6x^2 + 9x = 0
x^2  6x + 9 = 0
factorize:
(x3)^2=0
!?!??!
I'm pretty sure i did something wrong...

your tangent equation is correct.
also correct up to
x^3  6x^2 + 9x = 0
then
x(x^2  6x + 9) = 0
x(x3)(x3) = 0
so x = 0 or x = 3 (we knew that x=3 because we knew there was a tangent there)
You lost the x=0, a very common error
you can "lose" a constant if you factor it from a function f(x) = 0 but if you factor out a variable, that could be zero, thus a solution.
so when x=0, y = 8, so the other point of contact is (0,8) 
coordinates
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