已知二次函數(shù)f(x)=ax^2+bx+c
(1)f(-1)=0,試判斷函數(shù)零點(diǎn)個(gè)數(shù)
(2)是否存在a,b,c使函數(shù)同時(shí)滿足一下條件1.對(duì)任意x,f(x-4)=f(2-x)且f(x)>=0 .2對(duì)任意x,都有0=0
f(x-4)=a(x-4)^2+b(x-4)+c=ax^2-(8a-b)x+16a-4b+c
f(2-x)= a(2-x)^2+b(2-x)+c=ax^2-(4a+b)x+4a+2b+c
∴8a-b=4a+b==>2a=b；16a-4b=4a+2b==>2a=b
a>0,a=c,∴f(x)=ax^2+2ax+a=a(x+1)^2
又對(duì)任意x,都有0