已知向量m=(ax^2,1),n=(1,bx+1) (a,b为实数）,函数f(x)=m*n,若函数f(x)的最小值是f(

f(x)=m*n=ax^2+bx+1
函数f(x)的最小值在x=-1处,有a>0,且-b/(2a)=-1 ==> b=2a ==>f(x)=ax^2+2ax+1
由f(-1)=0 ==> a-2a+1=0 ==> a=1 ==> f(x)=x^2+2x+1