解答:解:(1)∵對(duì)稱(chēng)軸為直線x=4,圖象在x軸上截得的線段長(zhǎng)為6,
∴A(1,0)、B(7,0);
(2)設(shè)二次函數(shù)的解析式為:y=a(x-h)
2+k,
∵頂點(diǎn)C的橫坐標(biāo)為4,且過(guò)點(diǎn)(0,
),
∴y=a(x-4)
2+k
=16a+k①,
又∵對(duì)稱(chēng)軸為直線x=4,圖象在x軸上截得的線段長(zhǎng)為6,
∴A(1,0),B(7,0),
∴0=9a+k②,
由①②解得a=
,k=-
,
∴二次函數(shù)的解析式為:y=
(x-4)
2-
或y=
x
2-
x+
.
(3)解法一:∵點(diǎn)A、B關(guān)于直線x=4對(duì)稱(chēng),
∴PA=PB,
∴PA+PD=PB+PD≥DB,
∴當(dāng)點(diǎn)P在線段DB上時(shí)PA+PD取得最小值,
∴DB與對(duì)稱(chēng)軸的交點(diǎn)即為所求點(diǎn)P,
設(shè)直線x=4與x軸交于點(diǎn)M,
∵PM∥OD,
∴∠BPM=∠BDO,又∠PBM=∠DBO,
∴△BPM∽△BDO,
∴
=,
∴PM=
=
,
∴點(diǎn)P的坐標(biāo)為(4,
).
解法二:利用待定系數(shù)法求一次函數(shù)解析式,即直線DB為y=-
x+
.
(4)由(1)知點(diǎn)C(4,-
),
又∵AM=3,
∴在Rt△AMC中,cos∠ACM=
,
∴∠ACM=60°,
∵AC=BC,
∴∠ACB=120°
①當(dāng)點(diǎn)Q在x軸上方時(shí),過(guò)Q作QN⊥x軸于N,如果AB=BQ,
由△ACB∽△ABQ有BQ=6,∠ABQ=∠ACB=120°,則∠QBN=60°,
∴QN=3
,BN=3,ON=10,此時(shí)點(diǎn)Q(10,3
),如果AB=AQ,由對(duì)稱(chēng)性知Q(-2,3
)
②當(dāng)點(diǎn)Q在x軸下方時(shí),△QAB就是△ACB,此時(shí)點(diǎn)Q的坐標(biāo)是(4,-
),
經(jīng)檢驗(yàn),點(diǎn)(10,3
)與(-2,3
)都在拋物線上,
綜上所述,經(jīng)驗(yàn)證:存在這樣的點(diǎn)Q,使△QAB∽△ABC,
點(diǎn)Q的坐標(biāo)為(10,3
)或(-2,3
)或(4,-
).