曲线y=x^3-2x^4x+2在（过）点（1,-3）处的切线方程是,这道题怎么求,请写出两种情况完整过程谢!

Question

曲线y=x^3-2x^4x+2在（过）点（1,-3）处的切线方程是,这道题怎么求,请写出两种情况完整过程谢!

Answer 1

函数抄错了吧,应该是：y=x^3-2x^2-4x+2
1)求在点(1,-3)处的切线方程,此时切点为(1,3).
y'=3x^2-4x-4,y'(1)=-5,即切线斜率为-5
由点斜式可求得切线方程为：y+3=-5(x-1),即5x+y-2=0
2)求过点(1,-3)的切线方程,此时切点不一定是(1,-3)
设切点为P(t,t^3-2t^2-4t+2).
则切线斜率为：(t^3-2t^2-4t+2+3)/(t-1)=(t^3-2t^2-4t+5)/(t-1)
y'=3x^2-4x-4,即切线斜率又为y'(t)=3t^2-4t-4
所以,(t^3-2t^2-4t+5)/(t-1)=3t^2-4t-4
t^3-2t^2-4t+5-(3t^2-4t-4)(t-1)=0
t^2(t-1)-(t^2+4t-5)-(3t^2-4t-4)(t-1)=0
t^2(t-1)-(t+5)(t-1)-(3t^2-4t-4)(t-1)=0
(t-1)(-2t^2+3t-1)=0
(t-1)^2(2t-1)=0
则t1=1、t2=1/2.
所以,除点(1,-3)为切点外,还有一个切点为(1/2,-3/8)、该切线斜率为-21/4
该切线方程为：y+3/8=(-21/4)(x-1/2),即21x+4y-9=0
所以,过点(1,-3)的切线方程是5x+y-2=0和21x+4y-9=0