Question

已知曲线C:y=x³-3x²+2x,直线l:y=kx,且直线l与曲线C切于点(x₀,y₀)(x₀≠0),求直线l的方程及切点坐标.

Answer 1

解:∵y=kx过原点,∴k=.∵(x₀,y₀)在曲线C上,∴y₀=x₀³-3x₀²+2x₀.∴=x₀²-3x₀+2.

∵y′=3x²-6x+2,∴k=f′(x₀)=3x₀²-6x₀+2.∴3x₀²-6x₀+2=x₀²-3x₀+2,即2x₀²-3x₀=0.

解得x₀=0或x₀=.由于x₀≠0,∴x₀=.∴y₀=()³-3·()²+2×=.

∴k=.∴l的方程为y=x,切点坐标为(,).