显然那就是切点
y'=3x²-6x
所以斜率是k=y'(1)=-3
所以是3x+y=0
曲线y=x^3-3x^2-1在点(1,-3),处的切线方程是多少
曲线y=x^3-3x^2-1在点(1,-3),处的切线方程是多少
数学人气:666 ℃时间:2020-01-30 20:22:05
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