Question

一道高二数学题（请进！请详细说明！谢谢！)

已知曲线 f(x)=x^3+2x^2-4x+1，若斜率为-5的直线是曲线y=f(x)的切线，求此直线方程。

Answer 1

f'(x)=3x^2+4x-4
3x^2+4x-4=-5
3x^2+4x+1=0
(3x+1)(x+1)=0
x=-1/3, x=-1
f(-1/3)=(-1/3)^3+2/9+4/3+1=-1/27+2/9+4/3+1=41/27+27/27=68/27
f(-1)=-1+2+4+1=6
y=kx+b
x1=-1/3,y1=68/27 68/27=-5*(-1/3)+b b= (68-45)/27=23/27
l1: y=-5x+23/27
x2=-1,y2=6 6=-5*(-1)+b b=1
l2: y=-5x+1

Answer 2

求导f'(x)=3x^2+4x-4=-5
x=-1/3 or -1
......