解:(1)∵二次函數(shù)
的圖象與x軸相交于A、B兩點(diǎn)
∴b
2-4ac>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(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x
1=-3,x
2=5
∴點(diǎn)B的坐標(biāo)為(5,0)··········································································· 6分
(3)如圖,過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:m
1=0(舍去), m
2=3.·································································· 9分解析:
略