已知二次函数f(x)=ax^2+bx+c满足条件
问题描述：
已知二次函数f(x)=ax^2+bx+c满足条件
1.f(3-x)=f(x)；2.f(1)=0;3.对任意实数f(x)≥1/4a-1/2恒成立,
1.f(3-x)=f(x)；2.f(1)=0;3.对任意实数f(x)≥1/4a-1/2a恒成立,
答
1.f(3-x)=f(x)；
对称轴x=3/2
b/2a=-3/2
b=-3a
2.f(1)=0;
a+b+c=0
c=2a
f(x)=ax^2-3ax+2a
=a(x^2-3x+2)
=a(x-3/2)^2-a/4
3.对任意实数f(x)≥1/4a-1/2恒成立,
a(x-3/2)^2-a/4≥1/4a-1/2
a(x-3/2)^2-a/4-1/4a+1/2≥0
a(x-3/2)^2-(a^2-2a+1)/4≥0
4a(x-3/2)^2-(a-1)^2≥0
a>0
-(a-1)^2≥0
a≤1
01.f(3-x)=f(x)；2.f(1)=0;3.对任意实数f(x)≥1/4a-1/2a恒成立,求解析式前两问会做就是第三问的二分之三哪来的前两个条件得出b=-3a c=2a代入f(x)=ax^2+bx+c得f(x)=ax^2-3ax+2a=a(x^2-3ax+2)=a(x^2-2(3/2)x+(3/2)^2-(3/2)^2+2)=a((x-3/2)^2-9/4+2)=a(x-3/2)^2-a/4 抛物线顶点式