Question

（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．

 

Answer 1

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