已知函数f(x)=ax^2+bx+lnx,曲线y=f(x)在点A(1,f(1))处的切线方程为x+y+1=0

f'(x)=2ax+b+1/x,f'(1)=2a+b+1,f(1)=a+b,切线方程是y=a+b+(2a+b+1)(x-1)->y=a+b+(2a+b+1)x-2a-b-1->y=(2a+b+1)x-a-1->-(2a+b+1)x+y+(a+1)=0->
-(2a+b+1)=1
a+1=1->a=0b=-2
f(x)=-2x+lnx