∴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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