(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.
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
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