解:(1)根據(jù)已知條件可設(shè)拋物線的解析式為y=a(x-1)(x-5),
把點(diǎn)A(0,4)代入上式得:a=

,
∴y=

(x-1)(x-5)=

x
2-

x+4=

(x-3)
2-

,
∴拋物線的對(duì)稱軸是:x=3;
(2)P點(diǎn)坐標(biāo)為:(6,4),
由題意可知以A、O、M、P為頂點(diǎn)的四邊形有兩條邊AO=4、OM=3,
又∵點(diǎn)P的坐標(biāo)中x>5,
∴MP>2,AP>2;
∴以1、2、3、4為邊或以2、3、4、5為邊都不符合題意,

∴四條邊的長(zhǎng)只能是3、4、5、6的一種情況,
在Rt△AOM中,AM=

=

=5,
∵拋物線對(duì)稱軸過(guò)點(diǎn)M,
∴在拋物線x>5的圖象上有關(guān)于點(diǎn)A的對(duì)稱點(diǎn)與M的距離為5,
即PM=5,此時(shí)點(diǎn)P橫坐標(biāo)為6,即AP=6;
故以A、O、M、P為頂點(diǎn)的四邊形的四條邊長(zhǎng)度分別是四個(gè)連續(xù)的正整數(shù)3、4、5、6成立,
即P(6,4);
(3)在直線AC的下方的拋物線上存在點(diǎn)N,使△NAC面積最大.
設(shè)N點(diǎn)的橫坐標(biāo)為t,此時(shí)點(diǎn)N(t,

t
2-

t+4)(0<t<5),

過(guò)點(diǎn)N作NG∥y軸交AC于G;作AM⊥NG于M,
由點(diǎn)A(0,4)和點(diǎn)C(5,0)可求出直線AC的解析式為:y=-

x+4;
把x=t代入得:y=-

t+4,則G(t,-

t+4),
此時(shí):NG=-

x+4-(

t
2-

t+4)=-

t
2+4t,
∵AM+CF=CO,
∴S
△ACN=S
△ANG+S
△CGN=

AM×NG+

NG×CF=

NG•OC=

(-

t
2+4t)×5=-2t
2+10t=-2(t-

)
2+

,
∴當(dāng)t=

時(shí),△CAN面積的最大值為

,
由t=

,得:y=

t
2-

t+4=-3,
∴N(

,-3).
分析:(1)拋物線經(jīng)過(guò)點(diǎn)A(0,4),B(1,0),C(5,0),可利用兩點(diǎn)式法設(shè)拋物線的解析式為y=a(x-1)(x-5),代入A(0,4)即可求得函數(shù)的解析式,則可求得拋物線的對(duì)稱軸;
(2)由已知,可求得P(6,4),由題意可知以A、O、M、P為頂點(diǎn)的四邊形有兩條邊AO=4、OM=3,又知點(diǎn)P的坐標(biāo)中x>5,所以MP>2,AP>2;因此以1、2、3、4為邊或以2、3、4、5為邊都不符合題意,所以四條邊的長(zhǎng)只能是3、4、5、6的一種情況,則分析求解即可求得答案;
(3)在直線AC的下方的拋物線上存在點(diǎn)N,使△NAC面積最大.設(shè)N點(diǎn)的橫坐標(biāo)為t,此時(shí)點(diǎn)N(t,

t
2-

t+4)(0<t<5),再求得直線AC的解析式,即可求得NG的長(zhǎng)與△ACN的面積,由二次函數(shù)最大值的問(wèn)題即可求得答案.
點(diǎn)評(píng):此題考查了待定系數(shù)法求二次函數(shù)的解析式,勾股定理以及三角形面積的最大值問(wèn)題.此題綜合性很強(qiáng),難度很大,解題的關(guān)鍵是方程思想與數(shù)形結(jié)合思想的應(yīng)用.