關(guān)于x的二次函數(shù)y=-x2+(k2-4)x+2k-2以y軸為對(duì)稱軸,且與y軸的交點(diǎn)在x軸上方.
(1)求此拋物線的解析式,并在下面建立直角坐標(biāo)系畫出函數(shù)的草圖;
(2)設(shè)A是y軸右側(cè)拋物線上的一個(gè)動(dòng)點(diǎn),過點(diǎn)A作AB垂直于x軸于點(diǎn)B,再過點(diǎn)A作x軸的平行線交拋物線于點(diǎn)D,過點(diǎn)D作DC垂直于x軸于點(diǎn)C,得到矩形ABCD.設(shè)矩形ABCD的周長(zhǎng)為l,點(diǎn)A的橫坐標(biāo)為x,試求l關(guān)于x的函數(shù)關(guān)系式;
(3)當(dāng)點(diǎn)A在y軸右側(cè)的拋物線上運(yùn)動(dòng)時(shí),矩形ABCD能否成為正方形?若能,請(qǐng)求出此時(shí)正方形的周長(zhǎng);若不能,請(qǐng)說明理由.
【答案】
分析:(1)因?yàn)槎魏瘮?shù)y=-x
2+(k
2-4)x+2k-2以y軸為對(duì)稱軸,所以k
2-4=0,即可解出k的值,求出拋物線解析式,并利用描點(diǎn)法畫出圖象;
(2)求出拋物線與x軸的交點(diǎn)坐標(biāo),分矩形在x軸上方和矩形在x軸下方兩種情況,根據(jù)矩形周長(zhǎng)公式解答;
(3)假設(shè)能構(gòu)成正方形,根據(jù)正方形邊長(zhǎng)相等,列等式解出x的值,若x>0,則能構(gòu)成正方形,若x<0,則不能構(gòu)成正方形.
解答:
解:
(1)據(jù)題意得:k
2-4=0,
∴k=±2.
當(dāng)k=2時(shí),2k-2=2>0.
當(dāng)k=-2時(shí),2k-2=-6<0(2分)
又∵拋物線與y軸的交點(diǎn)在x軸上方,
∴k=2.
∴拋物線的解析式為:y=-x
2+2.(1分)
(2)解:令-x
2+2=0,得x=±

.
當(dāng)0<x<
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時(shí),A
1D
1=2x,A
1B
1=-x
2+2,
∴l(xiāng)=2(A
1B
1+A
1D
1)=-2x
2+4x+4(2分)
當(dāng)x>
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時(shí),A
2D
2=2x.
A
2B
2=-(-x
2+2)=x
2-2.
∴l(xiāng)=2(A
2D
2+A
2B
2)=2x
2+4x-4(2分)
(3)當(dāng)0<x<
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時(shí),令A(yù)
1B
1=A
1D
1,得x
2+2x-2=0.
解得x=-1-
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(舍去),或x=-1+

.
將x=-1+
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代入l=-2x
2+4x+4,
得l=8

-8(3分)
當(dāng)x>

時(shí),令A(yù)
2B
2=A
2D
2得:x
2-2x-2=0,
解得x=1-

(舍去),或x=1+

.
代入l=2x
2+4x-4,得L=8

+8(3分)
綜上,矩形ABCD能成為正方形,
且當(dāng)x=

-1時(shí)正方形的周長(zhǎng)是8

-8,
當(dāng)x=

+1時(shí),周長(zhǎng)為8

+8(1分).
點(diǎn)評(píng):解答此題的關(guān)鍵是求出二次函數(shù)的解析式,利用解析式求出各點(diǎn)的坐標(biāo)表達(dá)式,根據(jù)矩形或正方形的性質(zhì)來解答.值得關(guān)注,(3)為探索性問題,有一定的開放性.