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∴该抛物线与x轴公共点的坐标是??1···················································· 2分 0?. ·,0?和?,(Ⅱ)当a?b?1时,抛物线为y?3x2?2x?c,且与x轴有公共点.
?1
?3??
1对于方程3x2?2x?c?0,判别式??4?12c≥0,有c≤. ··········································· 3分
3①当c?111时,由方程3x2?2x??0,解得x1?x2??. 333此时抛物线为y?3x2?2x??1?1与x轴只有一个公共点??,··································· 4分 0?. ·3?3?②当c?1时, 3x1??1时,y1?3?2?c?1?c, x2?1时,y2?3?2?c?5?c.
1由已知?1?x?1时,该抛物线与x轴有且只有一个公共点,考虑其对称轴为x??,
3应有??y1≤0,?1?c≤0, 即?
?y2?0.?5?c?0.1或?5?c≤?1. ······················································································· 6分 3解得?5?c≤?1. 综上,c?(Ⅲ)对于二次函数y?3ax2?2bx?c,
由已知x1?0时,y1?c?0;x2?1时,y2?3a?2b?c?0, 又a?b?c?0,∴3a?2b?c?(a?b?c)?2a?b?2a?b. 于是2a?b?0.而b??a?c,∴2a?a?c?0,即a?c?0.
∴a?c?0. ······················································································································ 7分 ∵关于x的一元二次方程3ax2?2bx?c?0的判别式
??4b2?12ac?4(a?c)2?12ac?4[(a?c)2?ac]?0,
∴抛物线y?3ax2?2bx?c与x轴有两个公共点,顶点在x轴下方. ································· 8分 又该抛物线的对称轴x??b, 3ay 由a?b?c?0,c?0,2a?b?0, 得?2a?b??a,
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1b2∴???. 33a3又由已知x1?0时,y1?0;x2?1时,y2?0,观察图象,
可知在0?x?1范围内,该抛物线与x轴有两个公共点. ················································ 10分
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