Question

（9分）已知二次函數(shù)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(shù)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(shù)的圖象的對稱軸為直線x＝－＝1························· 4分

∴根據(jù)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(shù)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