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230问答网 > （2013?包头）已知二次函数y=ax2+bx+c（a≠0）的图象如图所示，下列结论：①b＜0；②4a+2b+c＜0；③a-b+c

（2013?包头）已知二次函数y=ax2+bx+c（a≠0）的图象如图所示，下列结论：①b＜0；②4a+2b+c＜0；③a-b+c

2025-03-16 18:32:45

推荐回答（1个）

回答1：

①图象开口向上，对称轴在y轴右侧，能得到：a＞0，-

b

2a

＞0，则b＜0，正确；
②∵对称轴为直线x=1，∴x=2与x=0时的函数值相等，∴当x=2时，y=4a+2b+c＞0，错误；
③当x=-1时，y=a-b+c＞0，正确；
④∵a-b+c＞0，∴a+c＞b；∵当x=1时，y=a+b+c＜0，∴a+c＜-b；∴b＜a+c＜-b，∴|a+c|＜|b|，∴（a+c）²＜b²，正确．
所以正确的结论是①③④．
故选C．

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