Question

垂直于直线2x-6y+1=0且与曲线y=x³+3x²-1相切的直线方程是

A.3x-y-2=0                          B.3x-y+2=0

C.3x+y-2=0                         D.3x+y+2=0

Answer 1

D?

解析:y′=3x²+6x,2x-6y+1=0的斜率为，垂线的斜率为-3，解3x²+6x=-3得x²+2x+1=0,?x=-1.?

∴切点为(-1，1).∴切线方程为y-1=-3(x+1),

即3x+y+2=0.∴选D.