Saturday

August 23, 2014

August 23, 2014

Posted by **holly** on Friday, June 19, 2009 at 6:34am.

it says:

find the equation of the tangent to the curve y=(x-2)^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=3x-8 ---- (1)

y=(x-2)^3 ----(2)

(x-2)^3 - 3x + 8 = 0

simplify:

x^3 - 6x^2 + 9x = 0

x^2 - 6x + 9 = 0

factorize:

(x-3)^2=0

!?!??!

I'm pretty sure i did something wrong...

- Please check my maths? -
**Reiny**, Friday, June 19, 2009 at 11:07amyour tangent equation is correct.

also correct up to

x^3 - 6x^2 + 9x = 0

then

x(x^2 - 6x + 9) = 0

x(x-3)(x-3) = 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)

- my maths -
**samo**, Monday, June 22, 2009 at 5:54amcoordinates

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