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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.
人气:218 ℃ 时间:2019-10-02 07:52:58
解答
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>.
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