曲线y=x3-3x2+1在点(1,-1)处的切线方程为(  ) A.3x-y-4=0 B.3x+y-2=0 C.4x+y-3=0 D.4x-y-5=0

问题描述:

曲线y=x3-3x2+1在点(1,-1)处的切线方程为(  )
           

A. 3x-y-4=0
B. 3x+y-2=0
C. 4x+y-3=0
D. 4x-y-5=0

因为y=x3-3x2+1,
所以y′=3x2-6x,
曲线y=x3-3x2+1在点P(1,-1)处的切线的斜率为:y′|x=1=-3.
此处的切线方程为:y+1=-3(x-1),即3x+y-2=0.
故答案为:3x+y-2=0.