Question

已知函数f(x)=ax2+bx+c（a≠0）满足f(0)=-1,对于任意x∈R都有f(x)≥x-1,且f(-1/2+x)

Answer 1

f(0)=-1 c=-1

f(x)≥x-1 即ax^2+(b-1)x≥0 a≠0时,a0,开口向上,△=（b-1)^2≤0 b=1

f(-1/2+x)=f(-1/2-x),对称轴是x=-1/2 -b/2a=-1/2 a=b=1

所以f(x)=x^2+x-1