【答案】
分析:(1)把(-4,8)代入y=ax
2可求得a的值,把x=2代入所求的拋物線解析式,可得n的值,那么P的坐標(biāo)為2,縱坐標(biāo)為-n,求得AP與x軸的交點(diǎn)即為Q的坐標(biāo);
(2)A′C+CB′最短,說(shuō)明拋物線向左平移了線段CQ的距離,用頂點(diǎn)式設(shè)出相應(yīng)的函數(shù)解析式,把新頂點(diǎn)坐標(biāo)代入即可;
(3)左右平移時(shí),使A′D+DB′′最短即可,那么作出點(diǎn)A′關(guān)于x軸對(duì)稱點(diǎn)的坐標(biāo)為A′′,得到直線A′′B′′的解析式,讓y=0,求得相應(yīng)的點(diǎn)的坐標(biāo);進(jìn)而得到拋物線頂點(diǎn)平移的規(guī)律,用頂點(diǎn)式設(shè)出相應(yīng)的函數(shù)解析式,把新頂點(diǎn)坐標(biāo)代入即可.
解答:
解:(1)將點(diǎn)A(-4,8)的坐標(biāo)代入y=ax
2,
解得a=

;
將點(diǎn)B(2,n)的坐標(biāo)代入y=

x
2,
求得點(diǎn)B的坐標(biāo)為(2,2),
則點(diǎn)B關(guān)于x軸對(duì)稱點(diǎn)P的坐標(biāo)為(2,-2),
設(shè)直線AP的解析式為y=kx+b,

,
解得:

,
∴直線AP的解析式是y=-

x+

,
令y=0,得x=

.
即所求點(diǎn)Q的坐標(biāo)是(

,0);
(2)①CQ=|-2-

|=

,(1分)
故將拋物線y=

x
2向左平移

個(gè)單位時(shí),A′C+CB′最短,

此時(shí)拋物線的函數(shù)解析式為y=

(x+

)
2;
②左右平移拋物線y=

x
2,因?yàn)榫€段A′B′和CD的長(zhǎng)是定值,
所以要使四邊形A′B′CD的周長(zhǎng)最短,只要使A′D+CB′最短;(1分)

第一種情況:如果將拋物線向右平移,顯然有A′D+CB′>AD+CB,
因此不存在某個(gè)位置,使四邊形A′B′CD的周長(zhǎng)最短;
第二種情況:設(shè)拋物線向左平移了b個(gè)單位,
則點(diǎn)A′和點(diǎn)B′的坐標(biāo)分別為A′(-4-b,8)和B′(2-b,2).
因?yàn)镃D=2,因此將點(diǎn)B′向左平移2個(gè)單位得B′′(-b,2),
要使A′D+CB′最短,只要使A′D+DB′′最短,
點(diǎn)A′關(guān)于x軸對(duì)稱點(diǎn)的坐標(biāo)為A′′(-4-b,-8),
直線A′′B′′的解析式為y=

x+

b+2.
要使A′D+DB′′最短,點(diǎn)D應(yīng)在直線A′′B′′上,
將點(diǎn)D(-4,0)代入直線A′′B′′的解析式,解得b=

.
故將拋物線向左平移時(shí),存在某個(gè)位置,使四邊形A′B′CD的周長(zhǎng)最短,
此時(shí)拋物線的函數(shù)解析式為y=

(x+

)
2.
點(diǎn)評(píng):用到的知識(shí)點(diǎn)為:兩點(diǎn)關(guān)于x軸對(duì)稱,橫坐標(biāo)相同,縱坐標(biāo)互為相反數(shù);拋物線平移,不改變二次項(xiàng)的系數(shù),看頂點(diǎn)是如何平移的即可;涉及距離之和最小問(wèn)題,應(yīng)從作其中一點(diǎn)關(guān)于直線的對(duì)稱點(diǎn)入手思考.