Question

（9分）已知二次函數(shù)的圖象與x軸相交于A、B兩點(diǎn)（A
左B右），與y軸相交于點(diǎn)C，頂點(diǎn)為D．
（1）求m的取值范圍；
（2）當(dāng)點(diǎn)A的坐標(biāo)為，求點(diǎn)B的坐標(biāo)；
（3）當(dāng)BC⊥CD時(shí)，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(diǎn)
∴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(duì)稱軸為直線x＝－＝1························· 4分
∴根據(jù)拋物線的對(duì)稱性得點(diǎn)B的坐標(biāo)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(dá)式為
令y=0得········································································ 5分
解得x₁=－3，x₂=5
∴點(diǎn)B的坐標(biāo)為（5，0）··········································································· 6分
（3）如圖，過(guò)D作DE⊥y軸，垂足為E．

∴∠DEC＝∠COB＝90°，
當(dāng)BC⊥CD時(shí)，∠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(biāo)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