?y?x2?3x?1;
(2)由题意,一元二次方程ax2?bx?c?0的判别式△?4.
?△?b2?4ac?4,
?4ac?b2?4,
在函数y1?ax2?(b?1)x?c中,1?(b?1)2?4ac?(b?1)2?(b2?4)?2b?5, 5b??,
2?2b?5?0,
即函数图象与x轴没有交点;
4ac?b2(3)因为函数顶点在直线l上,则有??1,
4a即b2?4ac?4a①,
c2?2c?6, AB?c2c2?2c?6?(x2?x1)?,
c2c2?2c?6即(x1?x2)?4x1x2?,
c2b2?4acc2?2c?6?, ?a2c4c2?2c?6由①得:?②,
ac?OAP??DAB,?OPB??DAB,
??OAP??OPB,
?OAP??OBP??APB,?OPB??OPA??APB, ??OBP??OPA,
则?OAP∽?OPB.
?OAOP, ?OPOB?OAOB?OP2,
?x1x2?(?x0)2?(?1)2.
?c?x0?1, ac?1. a?x0?
c2?2c?6由②得:x0??1,
4?x0?11(c?1)2?, 441. 4?当c?1时,(x0)min?
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