y'=3x²-2
设切点为(t, t^3-2t)
则切线为y=(3t²-2)(x-t)+t^3-2t=(3t²-2)x-2t^3
代入点(1, -1): (3t²-2)-2t^3=-1
即2t^3-3t²+1=0
2t^3-2t²-t²+1=0
2t²(t-1)-(t-1)(t+1)=0
(t-1)(2t²-t-1)=0
(t-1)(2t+1)(t-1)=0
t=1, -1/2
当t=1时,切线为y=x-2
当t=-1/2时,切线为y=-(5/4)x+1/4
y'=3x²-2
设切点为(t, t^3-2t)
则切线为y=(3t²-2)(x-t)+t^3-2t=(3t²-2)x-2t^3
代入点(1, -1): (3t²-2)-2t^3=-1
即2t^3-3t²+1=0
2t^3-2t²-t²+1=0
2t²(t-1)-(t-1)(t+1)=0
(t-1)(2t²-t-1)=0
(t-1)(2t+1)(t-1)=0
t=1, -1/2
当t=1时,切线为y=x-2
当t=-1/2时,切线为y=-(5/4)x+1/4