liangzi041
幼苗
共回答了19个问题采纳率:89.5% 举报
(1)∵
![](https://img.yulucn.com/upload/2/56/2568ca819aa02374e22a388e3fc3a215_thumb.jpg)
经过点(﹣3,0),
∴0=
![](https://img.yulucn.com/upload/1/dc/1dc95156b451aeb310ae8cec1133b786_thumb.jpg)
+m,解得m=
![](https://img.yulucn.com/upload/7/3e/73ed9cfd4c043b49134a6d5eda9c43bf_thumb.jpg)
,
∴直线解析式为
![](https://img.yulucn.com/upload/4/79/479ab463791fda913da8b29e64591e8e_thumb.jpg)
,C(0,
![](https://img.yulucn.com/upload/b/31/b31ca35129460ff1f69b8c37e77c66f0_thumb.jpg)
).
∵抛物线y=ax
2 +bx+c对称轴为x=1,且与x轴交于A(﹣3,0),∴另一交点为B(5,0),
设抛物线解析式为y=a(x+3)(x﹣5),
∵抛物线经过C(0,
![](https://img.yulucn.com/upload/f/65/f657ed7fcd06ab44a6d18c255d0cc8ad_thumb.jpg)
),
∴
![](https://img.yulucn.com/upload/2/60/26017c3978f1b4cd9d5896f3eca27e90_thumb.jpg)
=a
![](https://img.yulucn.com/upload/9/f0/9f0701724083814bd77e205cdc231132_thumb.jpg)
3(﹣5),解得a=
![](https://img.yulucn.com/upload/6/24/6249c362ad577470869f94f7b243ea38_thumb.jpg)
,
∴抛物线解析式为y=
![](https://img.yulucn.com/upload/a/8d/a8d047506dac81efe53c8185107c6c69_thumb.jpg)
x
2 +
![](https://img.yulucn.com/upload/0/56/056d15c573dec7593036e6565fa854d7_thumb.jpg)
x+
![](https://img.yulucn.com/upload/d/12/d12dcaeaa9c575ccac0612df725c6bcd_thumb.jpg)
;
(2)假设存在点E使得以A、C、E、F为顶点的四边形是平行四边形,
则AC∥EF且AC=EF.
如答图1,
(i)当点E在点E位置时,过点E作EG⊥x轴于点G,
∵AC∥EF,∴∠CAO=∠EFG,
又∵
![](https://img.yulucn.com/upload/0/1e/01eedc23f2fcc9a8f936e74bade13e43_thumb.jpg)
,∴△CAO≌△EFG,
∴EG=CO=
![](https://img.yulucn.com/upload/e/28/e2864a4716140a3457950fb4bbd10425_thumb.jpg)
,即y
E =
![](https://img.yulucn.com/upload/a/92/a92da17b21d1cc2d8ce3f6bd1585ef9b_thumb.jpg)
,
∴
![](https://img.yulucn.com/upload/2/e2/2e24cf2dc9698baf988c47616f3a2777_thumb.jpg)
=
![](https://img.yulucn.com/upload/9/01/901d5b8b0867bd4866ebac8236d4562f_thumb.jpg)
xE
2 +
![](https://img.yulucn.com/upload/3/95/395f9884ac62cfeb9803e375a553727f_thumb.jpg)
xE+
![](https://img.yulucn.com/upload/9/a4/9a4fd514e54e1db1cd5422bc902dce11_thumb.jpg)
,解得xE=2(xE=0与C点重合,舍去),
∴E(2,
![](https://img.yulucn.com/upload/9/34/9345813bd63e560b9f250d5e067c5aac_thumb.jpg)
),S
![](https://img.yulucn.com/upload/1/88/1880a436cac2fd3add441404338c4887_thumb.jpg)
ACEF=
![](https://img.yulucn.com/upload/8/44/844d2fecb46c5b6db20db7e3af75c289_thumb.jpg)
;
(ii)当点E在点E′位置时,过点E′作E′G′⊥x轴于点G′,
同理可求得E′(
![](https://img.yulucn.com/upload/2/f5/2f5d826c19b5a62d07c1343ffd178768_thumb.jpg)
+1,
![](https://img.yulucn.com/upload/a/8a/a8a8ba88527914e247a264f21ddd9209_thumb.jpg)
),
S
ACE′F′ =
![](https://img.yulucn.com/upload/7/99/799e4cdc897b80a335d92f6b170a9d55_thumb.jpg)
.
(3)要使△ACP的周长最小,只需AP+CP最小即可.
如答图2,连接BC交x=1于P点,因为点A、B关于x=1对称,根据轴对称性质以及两点之间线段最短,
可知此时AP+CP最小(AP+CP最小值为线段BC的长度).
∵B(5,0),C(0,
![](https://img.yulucn.com/upload/5/bb/5bbfcdd5d01a18713f04197ab05970a8_thumb.jpg)
),∴直线BC解析式为y=
1年前
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