已知函数f(x)=x3+ax2+bx+c的图象过点A(2,1),且在点A处的切线方程2x-y+a=0,则a+b+c=_.
问题描述:
已知函数f(x)=x3+ax2+bx+c的图象过点A(2,1),且在点A处的切线方程2x-y+a=0,则a+b+c=______.
答
∵函数f(x)=x3+ax2+bx+c的图象过点A(2,1),∴8+4a+2b+c=1,且f′(x)=3x2+2ax+b,∵f(x)在点A处的切线方程2x-y+a=0,∴f′(2)=3×4+2a×2+b=12+4a+b=2,f(x)在点A处的切线方程为y-1=2(x-2),即2x-y-3=...