（9分）已知二次函数的图象与x轴相交于A、B两点（A左B右），与y轴相交于点C，顶点为D．（1）求m的取值范围；（2）当点A的坐标为，求点B的坐标；（3）当BC

题型：不详难度：来源：

（9分）已知二次函数的图象与x轴相交于A、B两点（A
左B右），与y轴相交于点C，顶点为D．
（1）求m的取值范围；
（2）当点A的坐标为，求点B的坐标；
（3）当BC⊥CD时，求m的值．

答案

解：（1）∵二次函数的图象与x轴相交于A、B两点
∴b²－4ac＞0，∴4+4m＞0，······································································· 2分
解得：m＞－1························································································· 3分
（2）解法一：
∵二次函数的图象的对称轴为直线x＝－＝1························· 4分
∴根据抛物线的对称性得点B的坐标为（5，0）··············································· 6分
解法二：
把x=－3，y=0代入中得m="15···············································" 4分
∴二次函数的表达式为
令y=0得········································································ 5分
解得x₁=－3，x₂=5
∴点B的坐标为（5，0）··········································································· 6分
（3）如图，过D作DE⊥y轴，垂足为E．

∴∠DEC＝∠COB＝90°，
当BC⊥CD时，∠DCE +∠BCO＝90°，
∵∠DEC＝90°，∴∠DCE +∠EDC＝90°，∴∠EDC＝∠BCO．
∴△DEC∽△COB，∴＝．····························································· 7分
由题意得：OE＝m+1，OC＝m，DE＝1，∴EC＝1．∴＝．
∴OB＝m，∴B的坐标为（m，0）．······························································ 8分
将（m，0）代入得：－m ²+2 m + m＝0．
解得：m₁＝0（舍去）， m₂＝3．·································································· 9分

解析

略