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.（1）求m的取值范围；

2.（2）当点A的坐标为，求点B的坐标；

【小题】（3）当BC⊥CD时，求m的值．

 

Answer 1

【答案】

 

1.（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.（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）    

3.（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