若二次函数f(x)=ax2+bx+c(a≠0),满足f(x+2)-f(x)=16x且f(0)=2.(Ⅰ)求函数f(x)的解析式;

2025-01-08 09:20:25
推荐回答(1个)
回答1:

(Ⅰ)由f(0)=2,解得:c=2,
∴f(x)=ax2+bx+2(a≠0),
由f(x+2)-f(x)
=[a(x+2)2+b(x+2)+2]-[ax2+bx+2]
=4ax+4a+2b
=16x,

4a=16
4a+2b=0
,解得:
a=4
b=?8

∴f(x)=4x2-8x+2;
(Ⅱ)∵?x∈[1,2],使不等式f(x)>2x+m,
即?x∈[1,2],使不等式m<4x2-10x+2成立,
令g(x)=4x2-10x+2,x∈[1,2],
故g(x)最大=g(2)=-2,
∴m<-2.