问题描述: 二次函数y=-x2+2x+3,当x满足______时,y=0;当x满足______时,y>0;当x满足______3时,y<0. 1个回答 分类:数学 2014-10-18 问题解答: 我来补答 当y=0时,-x2+2x+3=0,即-(x+1)(x-3)=0,所以,x+1=0或x-3=0,解得x=-1或x=3.当y>0时,-x2+2x+3>0,即-(x+1)(x-3)>0,解得-1<x<3;当y<0时,-x2+2x+3<0,即-(x+1)(x-3)<0,解得x<-1或x>3;故答案为:x=-1或x=3;-1<x<3;x<-1或x>. 展开全文阅读