Question

已知二次函数f(x)=ax^2+bx+c(a,b,c∈R,a≠0),f(-2)=f(0)=0,f(x)的最小值为-1.

Answer 1

(1)

f(-2)=f(0)=0

∴可设f(x)=a(x+2)x,对称轴x=-1,顶点纵坐标是f(-1)=-a=-1,得a=1,

∴f(x)=x²+2x,

(2)

g(x)

=x²-2x-mx²-2mx+1

=(1-m)x²-2(1+m)x+1

当m=1时,g(x)=-4x+1,满足题意,

当m>1时,需(1+m)/(1-m)