曲线y=x3-2x2-4x+2在点(1,-3)处的切线方程是( ) A.5x+y+2=0 B.5x+y-2=0 C.5x-y-8=0 D.5x-y+8=0
问题描述:
曲线y=x3-2x2-4x+2在点(1,-3)处的切线方程是( )
A. 5x+y+2=0
B. 5x+y-2=0
C. 5x-y-8=0
D. 5x-y+8=0
答
∵曲线y=x3-2x2-4x+2,
∴y′=3x2-4x-4,
当x=1时,y′=-5,即切线斜率为-5,
∴切线方程为y+3=-5(x-1),即5x+y-2=0.
故选B.