流浪的FOX
幼苗
共回答了16个问题采纳率:100% 举报
(1)x-y-4=0
(2)x-y-4=0或y+2=0
(1)∵f′(x)=3x 2 -8x+5,
∴f′(2)=1,又f(2)=-2,
∴曲线f(x)在点(2,f(2))处的切线方程为y-(-2)=x-2,即x-y-4=0.
(2)设切点坐标为(x 0 ,x 0 3 -4x 0 2 +5x 0 -4),
∵f′(x 0 )=3x 0 2 -8x 0 +5,
∴切线方程为y-(-2)=(3x 0 2 -8x 0 +5)(x-2),
又切线过点(x 0 ,x 0 3 -4x 0 2 +5x 0 -4),
∴x 0 3 -4x 0 2 +5x 0 -2=(3x 0 2 -8x 0 +5)(x 0 -2),
整理得(x 0 -2) 2 (x 0 -1)=0,解得x 0 =2或x 0 =1,
∴经过A(2,-2)的曲线f(x)的切线方程为x-y-4=0或y+2=0.
1年前
2