Posted by holly on Friday, June 19, 2009 at 6:34am.
How should i do this question??
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)
(x-2)^3 - 3x + 8 = 0
x^3 - 6x^2 + 9x = 0
x^2 - 6x + 9 = 0
I'm pretty sure i did something wrong...
- Please check my maths? - Reiny, Friday, June 19, 2009 at 11:07am
your tangent equation is correct.
also correct up to
x^3 - 6x^2 + 9x = 0
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 5:54am
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