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(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,...

(9分)已知二次函数6ec8aac122bd4f6e的图象与x轴相交于AB两点(AB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6ec8aac122bd4f6e,求点B的坐标;

【小题】(3)当BCCD时,求m的值.

 

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
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