Question

（9分）已知二次函數的圖象與x軸相交于A、B兩點（A
左B右），與y軸相交于點C，頂點為D．
（1）求m的取值范圍；
（2）當點A的坐標為，求點B的坐標；
（3）當BC⊥CD時，求m的值．

Answer 1

解：（1）∵二次函數的圖象與x軸相交于A、B兩點
∴b²－4ac＞0，∴4+4m＞0，······································································· 2分
解得：m＞－1························································································· 3分
（2）解法一：
∵二次函數的圖象的對稱軸為直線x＝－＝1························· 4分
∴根據拋物線的對稱性得點B的坐標為（5，0）··············································· 6分
解法二：
把x=－3，y=0代入中得m="15···············································" 4分
∴二次函數的表達式為
令y=0得········································································ 5分
解得x₁=－3，x₂=5
∴點B的坐標為（5，0）··········································································· 6分
（3）如圖，過D作DE⊥y軸，垂足為E．

∴∠DEC＝∠COB＝90°，
當BC⊥CD時，∠DCE +∠BCO＝90°，
∵∠DEC＝90°，∴∠DCE +∠EDC＝90°，∴∠EDC＝∠BCO．
∴△DEC∽△COB，∴＝．····························································· 7分
由題意得：OE＝m+1，OC＝m，DE＝1，∴EC＝1．∴＝．
∴OB＝m，∴B的坐標為（m，0）．······························································ 8分
將（m，0）代入得：－m ²+2 m + m＝0．
解得：m₁＝0（舍去）， m₂＝3．·································································· 9分解析:
略