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.（1）求m的取值范圍；

2.（2）當(dāng)點A的坐標(biāo)為，求點B的坐標(biāo)；

【小題】（3）當(dāng)BC⊥CD時，求m的值．

 

Answer 1

 

1.（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.（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(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)的表達式為

令y=0得·························································································· 5分

解得x₁=－3，x₂=5

∴點B的坐標(biāo)為（5，0）    

3.（3）如圖，過D作DE⊥y軸，垂足為E．∴∠DEC＝∠COB＝90°，

當(dāng)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(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:略