曲线y=x3-x2在点P(2,4)处的切线方程为______.
由题意得,y′=3x2-2x,
则点P(2,4)处的切线斜率k=12-4=8,
∴点P(2,4)处的切线方程是:y-4=8(x-2),
即8x-y-12=0,
故答案为:8x-y-12=0.
曲线y=x3-x2在点P(2,4)处的切线方程为______.
由题意得,y′=3x2-2x,
则点P(2,4)处的切线斜率k=12-4=8,
∴点P(2,4)处的切线方程是:y-4=8(x-2),
即8x-y-12=0,
故答案为:8x-y-12=0.