a-maths~~~~~~

y = x3 - 3x2 + 2x - 1
y` = 3x2 - 6x + 2
slope = 3(1)2 - 6(1) + 2 = -1
切綫 : y - (-1) = (-1)(x - 1)
x - 1 + y + 1 = 0
x + y = 0
好明顯 : O點 係 切綫上
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slope = 3(a)2 - 6(a) + 2 = 3a2 - 6a + 2
切綫 : y - b = (3a2 - 6a + 2)(x - a)
B點的切綫通過O點, 0 - b = (3a2 - 6a + 2)(0 - a)
b = a(3a2 - 6a + 2)
B(a,b)為曲綫上的點, b = a3 - 3a2 + 2a - 1
a(3a2 - 6a + 2) = a3 - 3a2 + 2a - 1
2a3 - 3a2 + 1= 0
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2a3 - 3a2 + 1 = (a - 1)(2a2 - a - 1) = (a - 1)2 (2a + 1)
2a3 - 3a2 + 1= 0
a = 1 or a = -1/2
a = -1/2, b = (-1/2)3 - 3(-1/2)2 + 2(-1/2) - 1 = -23/8
切綫, y - (-23/8) = [ 3(-1/2)2 - 6(-1/2) + 2 ] [ x - (-1/2) ]
y + 23/8 = [ 23/4 ] [ x + 1/2 ]
8y + 23 = [ 23 ] [ 2x + 1 ]
46x - 8y = 0
23x - 4y = 0


2008-06-08 12:29:42 補充：
投我一票哦

y=x^3 -3x^2 +2x -1. dy/dx = 3x^2 -6x + 2 .
For point A(1,-1), dy/dx = 3 -6 +2 = -1.
Therefore, tangent passing through A is y+1 = -1(x -1). y= -x + 1 -1 = -x.
which passes the origin.
For point B(a,b), dy/dx = 3a^2 - 6a + 2 . Therefore, tangent passing through B is
y-b = (3a^2 -6a +2)(x -a). If it passes through (0,0), that is
-b = -a(3a^2 -6a +2)
-b = 6a^2 -3a^3 -2a.......................(1)
Since B is on the curve, therefore, b=a^3 -3a^2 +2a -1 ................(2)
(1) + (2) We get 0=6a^2 -3a^3 -2a + a^3 -3a^2 +2a -1
That is 2a^3-3a^2 +1 = 0.

2008-06-08 12:36:10 補充：
Substitute (1) into equation of tangent, we get y+6a^2 -3a^3 -2a = (3a^2 -6a +2)(x -a)
=(3a^2 -6a +2)x -3a^3 +6a^2 -2a. That is y = (3a^2 -6a +2)x is the tangent from origin to curve.