【答案】
分析:(1)拋物線的解析式中,令x=0即得二次函數(shù)與y軸交點A的縱坐標,令y=0即得二次函數(shù)與x軸交點的橫坐標.
(2)根據(jù)A、C的坐標,易求得直線AC的解析式,由于等腰△EDC的腰和底不確定,因此要分成三種情況討論:
①CD=DE,由于OD=3,OA=4,那么DA=DC=5,此時A點符合E點的要求,即此時A、E重合;
②CE=DE,根據(jù)等腰三角形三線合一的性質(zhì)知:E點橫坐標為點D的橫坐標加上CD的一半,然后將其代入直線AC的解析式中,即可得到點E的坐標;
③CD=CE,此時CE=5,過E作EG⊥x軸于G,已求得CE、CA的長,即可通過相似三角形(△CEG∽△CAO)所得比例線段求得EG、CG的長,從而得到點E的坐標.
(3)過P作x軸的垂線,交AC于Q,交x軸于H;設(shè)出點P的橫坐標(設(shè)為m),根據(jù)拋物線和直線AC的解析式,即可表示出P、Q的縱坐標,從而可得到PQ的長,然后分兩種情況進行討論:
①P點在第一象限時,即0<m<8時,可根據(jù)PQ的長以及A、C的坐標,分別表示出△APQ、△CPQ的面積,它們的面積和即為△APC的面積,由此可得到S的表達式,通過配方即可得到S的取值范圍;
②當P在第二象限時,即-2<m<0時,同①可求得△APQ、△CPQ的面積,此時它們的面積差為△APC的面積,同理可求得S的取值范圍;根據(jù)兩個S的取值范圍,即可判斷出所求的結(jié)論.
解答:解:(1)在二次函數(shù)中令x=0得y=4,
∴點A的坐標為(0,4),
令y=0得:
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,
即:x
2-6x-16=0,
∴x=-2和x=8,
∴點B的坐標為(-2,0),點C的坐標為(8,0).
(2)易得D(3,0),CD=5,
設(shè)直線AC對應(yīng)的函數(shù)關(guān)系式為y=kx+b,則:
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,
解得
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;
∴y=-
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x+4;
①當DE=DC時,
∵OA=4,OD=3,
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∴DA=5,
∴E
1(0,4);
②過E點作EG⊥x軸于G點,
當DE=EC時,由DG=
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=
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,
把x=OD+DG=3+
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=
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代入到y(tǒng)=-
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x+4,求出y=
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,
可得E
2(
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,
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);
③當DC=EC時,如圖,過點E作EG⊥CD,
則△CEG∽△CAO,
∴
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,又OA=4,OC=8,則AC=4
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,DC=EC=5,
∴EG=
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,CG=2
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,
∴E
3(8-2
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,
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);
綜上所述,符合條件的E點共有三個:E
1(0,4)、E
2(
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,
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)、E
3(8-2
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,
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).
(3)如圖,過P作PH⊥OC,垂足為H,交直線AC與點Q;
設(shè)P(m,-
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m
2+
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m+4),則Q(m,-
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m+4).
①當0<m<8時,
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PQ=(-
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m
2+
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m+4)-(-
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m+4)=-
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m
2+2m,
S=S
△APQ+S
△CPQ=
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×8×(-
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m
2+2m)=-(m-4)
2+16,
∴0<S≤16;
②當-2≤m<0時,
PQ=(-
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m+4)-(-
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m
2+
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m+4)=
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m
2-2m,
S=S
△CPQ-S
△APQ=
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×8×(
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m
2-2m)=(m-4)
2-16,
∴0<S<20;
∴當0<S<16時,0<m<8中有m兩個值,-2<m<0中m有一個值,此時有三個;
當16<S<20時,-2<m<0中m只有一個值;
當S=16時,m=4或m=4-4
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這兩個.
故當S=16時,相應(yīng)的點P有且只有兩個.
點評:此題考查了二次函數(shù)圖象與坐標軸交點坐標的求法、等腰三角形的構(gòu)成條件、圖形面積的求法等知識,(3)題的解題過程并不復(fù)雜,關(guān)鍵在于理解題意.