中考數(shù)學(xué)二輪復(fù)習(xí)壓軸題培優(yōu)專練專題16 函數(shù)的圖像變換問(wèn)題（解析版）

專題16函數(shù)的圖像變換問(wèn)題函數(shù)圖像的變換問(wèn)題的考查一般難度較大，但關(guān)鍵是要理清圖像變換前后的解析式的關(guān)系，圖像變換的規(guī)律以二次函數(shù)為例：1．保持y=ax2的形狀不變，將其頂點(diǎn)平移到（h，k）處，具體平移方法如下：2．二次函數(shù)平移遵循“上加下減，左加右減”的原則，據(jù)此，可以直接由解析式中常數(shù)的加或減求出變化后的解析式；二次函數(shù)圖象的平移可看作頂點(diǎn)間的平移，可根據(jù)頂點(diǎn)之間的平移求出變化后的解析式． （2022·湖北恩施·統(tǒng)考中考真題）在平面直角坐標(biāo)系中，O為坐標(biāo)原點(diǎn)，拋物線SKIPIF1<0與y軸交于點(diǎn)SKIPIF1<0．(1)直接寫(xiě)出拋物線的解析式．(2)如圖，將拋物線SKIPIF1<0向左平移1個(gè)單位長(zhǎng)度，記平移后的拋物線頂點(diǎn)為Q，平移后的拋物線與x軸交于A、B兩點(diǎn)（點(diǎn)A在點(diǎn)B的右側(cè)），與y軸交于點(diǎn)C．判斷以B、C、Q三點(diǎn)為頂點(diǎn)的三角形是否為直角三角形，并說(shuō)明理由．(3)直線BC與拋物線SKIPIF1<0交于M、N兩點(diǎn)（點(diǎn)N在點(diǎn)M的右側(cè)），請(qǐng)?zhí)骄吭趚軸上是否存在點(diǎn)T，使得以B、N、T三點(diǎn)為頂點(diǎn)的三角形與SKIPIF1<0相似，若存在，請(qǐng)求出點(diǎn)T的坐標(biāo)；若不存在，請(qǐng)說(shuō)明理由．(4)若將拋物線SKIPIF1<0進(jìn)行適當(dāng)?shù)钠揭疲?dāng)平移后的拋物線與直線BC最多只有一個(gè)公共點(diǎn)時(shí)，請(qǐng)直接寫(xiě)出拋物線SKIPIF1<0平移的最短距離并求出此時(shí)拋物線的頂點(diǎn)坐標(biāo)．（1）待定系數(shù)法求二次函數(shù)解析式；（2）分別求得B、C、Q的坐標(biāo)，勾股定理的逆定理驗(yàn)證即可求解；（3）由SKIPIF1<0，故分兩種情況討論，根據(jù)相似三角形的性質(zhì)與判定即可求解；（4）如圖，作SKIPIF1<0且與拋物線只有1個(gè)交點(diǎn)，交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0于點(diǎn)SKIPIF1<0，則SKIPIF1<0是等腰直角三角形，作SKIPIF1<0于SKIPIF1<0，進(jìn)而求得直線SKIPIF1<0與SKIPIF1<0的距離，即為所求最短距離，進(jìn)而求得平移方式，將頂點(diǎn)坐標(biāo)平移即可求解．【答案】(1)SKIPIF1<0(2)以B、C、Q三點(diǎn)為頂點(diǎn)的三角形是直角三角形，理由見(jiàn)解析(3)存在，SKIPIF1<0或SKIPIF1<0，(4)最短距離為SKIPIF1<0，平移后的頂點(diǎn)坐標(biāo)為SKIPIF1<0【詳解】（1）解：∵拋物線SKIPIF1<0與y軸交于點(diǎn)SKIPIF1<0∴SKIPIF1<0SKIPIF1<0拋物線解析式為SKIPIF1<0（2）以B、C、Q三點(diǎn)為頂點(diǎn)的三角形是直角三角形，理由如下：SKIPIF1<0SKIPIF1<0的頂點(diǎn)坐標(biāo)為SKIPIF1<0依題意得，SKIPIF1<0SKIPIF1<0平移后的拋物線解析式為SKIPIF1<0令SKIPIF1<0，解SKIPIF1<0得SKIPIF1<0SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0以B、C、Q三點(diǎn)為頂點(diǎn)的三角形是直角三角形（3）存在，SKIPIF1<0或SKIPIF1<0，理由如下，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0是等腰直角三角形設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0的解析式為SKIPIF1<0，聯(lián)立SKIPIF1<0解得SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是等腰直角三角形SKIPIF1<0SKIPIF1<0，SKIPIF1<0設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0直線SKIPIF1<0的解析式為SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0設(shè)SKIPIF1<0的解析式為SKIPIF1<0，由NT過(guò)點(diǎn)SKIPIF1<0則SKIPIF1<0解得SKIPIF1<0SKIPIF1<0SKIPIF1<0的解析式為SKIPIF1<0，令SKIPIF1<0解得SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0②當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0即SKIPIF1<0解得SKIPIF1<0SKIPIF1<0SKIPIF1<0綜上所述，SKIPIF1<0或SKIPIF1<0（4）如圖，作SKIPIF1<0，交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0于點(diǎn)SKIPIF1<0，則SKIPIF1<0是等腰直角三角形，作SKIPIF1<0于SKIPIF1<0SKIPIF1<0直線SKIPIF1<0的解析式為SKIPIF1<0設(shè)與SKIPIF1<0平行的且與SKIPIF1<0只有一個(gè)公共點(diǎn)的直線SKIPIF1<0解析式為SKIPIF1<0則SKIPIF1<0整理得：SKIPIF1<0則SKIPIF1<0解得SKIPIF1<0SKIPIF1<0直線SKIPIF1<0的解析式為SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0即拋物線SKIPIF1<0平移的最短距離為SKIPIF1<0，方向?yàn)镾KIPIF1<0方向SKIPIF1<0∴把點(diǎn)P先向右平移EF的長(zhǎng)度，再向下平移FC的長(zhǎng)度即得到平移后的坐標(biāo)SKIPIF1<0平移后的頂點(diǎn)坐標(biāo)為SKIPIF1<0，即SKIPIF1<0本題是二次函數(shù)綜合，考查了相似三角形的性質(zhì)，求二次函數(shù)與一次函數(shù)解析式，二次函數(shù)圖象的平移，勾股定理的逆定理，正確的添加輔助線以及正確的計(jì)算是解題的關(guān)鍵．（2022·遼寧沈陽(yáng)·統(tǒng)考中考真題）如圖，平面直角坐標(biāo)系中，O是坐標(biāo)原點(diǎn)，拋物線SKIPIF1<0經(jīng)過(guò)點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0與x軸另一個(gè)交點(diǎn)A．拋物線與y軸交于點(diǎn)C，作直線AD．(1)①求拋物線的函數(shù)表達(dá)式②并直接寫(xiě)出直線AD的函數(shù)表達(dá)式．(2)點(diǎn)E是直線AD下方拋物線上一點(diǎn)，連接BE交AD于點(diǎn)F，連接BD，DE，SKIPIF1<0的面積記為SKIPIF1<0，SKIPIF1<0的面積記為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，求點(diǎn)E的坐標(biāo)；(3)點(diǎn)G為拋物線的頂點(diǎn)，將拋物線圖象中x軸下方部分沿x軸向上翻折，與拋物線剩下部分組成新的曲線為SKIPIF1<0，點(diǎn)C的對(duì)應(yīng)點(diǎn)SKIPIF1<0，點(diǎn)G的對(duì)應(yīng)點(diǎn)SKIPIF1<0，將曲線SKIPIF1<0，沿y軸向下平移n個(gè)單位長(zhǎng)度（SKIPIF1<0）．曲線SKIPIF1<0與直線BC的公共點(diǎn)中，選兩個(gè)公共點(diǎn)作點(diǎn)P和點(diǎn)Q，若四邊形SKIPIF1<0是平行四邊形，直接寫(xiě)出P的坐標(biāo)．（1）①利用待定系數(shù)解答，即可求解；②利用待定系數(shù)解答，即可求解；（2）過(guò)點(diǎn)E作EG⊥x軸交AD于點(diǎn)G，過(guò)點(diǎn)B作BH⊥x軸交AD于點(diǎn)H，設(shè)點(diǎn)SKIPIF1<0，則點(diǎn)SKIPIF1<0，可得SKIPIF1<0，然后根據(jù)△EFG∽△BFH，即可求解；（3）先求出向上翻折部分的圖象解析式為SKIPIF1<0，可得向上翻折部分平移后的函數(shù)解析式為SKIPIF1<0，平移后拋物線剩下部分的解析式為SKIPIF1<0，分別求出直線BC和直線SKIPIF1<0的解析式為，可得BC∥C′G′，再根據(jù)平行四邊形的性質(zhì)可得點(diǎn)SKIPIF1<0，然后分三種情況討論：當(dāng)點(diǎn)P，Q均在向上翻折部分平移后的圖象上時(shí)；當(dāng)點(diǎn)P在向上翻折部分平移后的圖象上，點(diǎn)Q在平移后拋物線剩下部分的圖象上時(shí)；當(dāng)點(diǎn)P在平移后拋物線剩下部分的圖象上，點(diǎn)Q在向上翻折部分平移后的圖象上時(shí)，即可求解．【答案】(1)①SKIPIF1<0；②SKIPIF1<0(2)（2，-4）或（0，-3）(3)（1+SKIPIF1<0，SKIPIF1<0）或SKIPIF1<0【詳解】（1）解：①把點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0代入得：SKIPIF1<0，解得：SKIPIF1<0，∴拋物線解析式為SKIPIF1<0；②令y=0，則SKIPIF1<0，解得：SKIPIF1<0，∴點(diǎn)A（-2，0），設(shè)直線AD的解析式為SKIPIF1<0，∴把點(diǎn)SKIPIF1<0和點(diǎn)A（-2，0）代入得：SKIPIF1<0，解得：SKIPIF1<0，∴直線AD的解析式為SKIPIF1<0；（2）解：如圖，過(guò)點(diǎn)E作EG⊥x軸交AD于點(diǎn)G，過(guò)點(diǎn)B作BH⊥x軸交AD于點(diǎn)H，當(dāng)x=6時(shí)，SKIPIF1<0，∴點(diǎn)H（6，-4），即BH=4，設(shè)點(diǎn)SKIPIF1<0，則點(diǎn)SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0的面積記為SKIPIF1<0，SKIPIF1<0的面積記為SKIPIF1<0，且SKIPIF1<0，∴BF=2EF，∵EG⊥x，BH⊥x軸，∴△EFG∽△BFH，∴SKIPIF1<0，∴SKIPIF1<0，解得：SKIPIF1<0或0，∴點(diǎn)E的坐標(biāo)為（2，-4）或（0，-3）；（3）解：SKIPIF1<0，∴點(diǎn)G的坐標(biāo)為（2，-4），當(dāng)x=0時(shí)，y=-3，即點(diǎn)C（0，-3），∴點(diǎn)SKIPIF1<0，∴向上翻折部分的圖象解析式為SKIPIF1<0，∴向上翻折部分平移后的函數(shù)解析式為SKIPIF1<0，平移后拋物線剩下部分的解析式為SKIPIF1<0，設(shè)直線BC的解析式為SKIPIF1<0，把點(diǎn)B（6，0），C（0，-3）代入得：SKIPIF1<0，解得：SKIPIF1<0，∴直線BC的解析式為SKIPIF1<0，同理直線SKIPIF1<0的解析式為SKIPIF1<0，∴BC∥C′G′，設(shè)點(diǎn)P的坐標(biāo)為SKIPIF1<0，∵點(diǎn)SKIPIF1<0，∴點(diǎn)C′向右平移2個(gè)單位，再向上平移1個(gè)單位得到點(diǎn)G′，∵四邊形SKIPIF1<0是平行四邊形，∴點(diǎn)SKIPIF1<0，當(dāng)點(diǎn)P，Q均在向上翻折部分平移后的圖象上時(shí)，SKIPIF1<0，解得：SKIPIF1<0（不合題意，舍去），當(dāng)點(diǎn)P在向上翻折部分平移后的圖象上，點(diǎn)Q在平移后拋物線剩下部分的圖象上時(shí)，SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0（不合題意，舍去），當(dāng)點(diǎn)P在平移后拋物線剩下部分的圖象上，點(diǎn)Q在向上翻折部分平移后的圖象上時(shí)，SKIPIF1<0，解得：SKIPIF1<0（舍去，不合題意）或SKIPIF1<0，綜上所述，點(diǎn)P的坐標(biāo)為綜上所述，點(diǎn)P的坐標(biāo)為（1+SKIPIF1<0，SKIPIF1<0）或（1﹣SKIPIF1<0，SKIPIF1<0）．本題主要考查了二次函數(shù)的綜合題，熟練掌握二次函數(shù)的圖象和性質(zhì)，平行四邊形的性質(zhì)，相似三角形的判定和性質(zhì)，并利用數(shù)形結(jié)合思想解答是解題的關(guān)鍵．（2021·廣西梧州·統(tǒng)考中考真題）如圖，在平面直角坐標(biāo)系中，拋物線y＝x2+bx+c經(jīng)過(guò)點(diǎn)A（﹣1，0），B（0，3），頂點(diǎn)為C．平移此拋物線，得到一條新的拋物線，且新拋物線上的點(diǎn)D（3，﹣1）為原拋物線上點(diǎn)A的對(duì)應(yīng)點(diǎn)，新拋物線頂點(diǎn)為E，它與y軸交于點(diǎn)G，連接CG，EG，CE．（1）求原拋物線對(duì)應(yīng)的函數(shù)表達(dá)式；（2）在原拋物線或新拋物線上找一點(diǎn)F，使以點(diǎn)C，E，F(xiàn)，G為頂點(diǎn)的四邊形是平行四邊形，并求出點(diǎn)F的坐標(biāo)；（3）若點(diǎn)K是y軸上的一個(gè)動(dòng)點(diǎn)，且在點(diǎn)B的上方，過(guò)點(diǎn)K作CE的平行線，分別交兩條拋物線于點(diǎn)M，N，且點(diǎn)M，N分別在y軸的兩側(cè)，當(dāng)MN＝CE時(shí)，請(qǐng)直接寫(xiě)出點(diǎn)K的坐標(biāo)．（1）根據(jù)待定系數(shù)法將點(diǎn)A（﹣1，0），B（0，3）代入拋物線y＝x2+bx+c，即可求出原拋物線解析式；（2）根據(jù)新拋物線上的點(diǎn)D（3，﹣1）為原拋物線上點(diǎn)A的對(duì)應(yīng)點(diǎn)可知拋物線平移方式為右移4個(gè)單位下移1個(gè)單位，從而確定新拋物線解析式，進(jìn)而確定點(diǎn)C、D、G坐標(biāo)，由以點(diǎn)C，E，F(xiàn)，G為頂點(diǎn)的四邊形是平行四邊形即可確定點(diǎn)F坐標(biāo)的可能位置，判斷是否在原拋物線或新拋物線上即可解答；（3）由SKIPIF1<0，MN＝CE，可知M點(diǎn)到N點(diǎn)的平移方式和C點(diǎn)到E點(diǎn)平移方式相同，故可設(shè)點(diǎn)M坐標(biāo)為（a，b），可得點(diǎn)N坐標(biāo)為（a+4，b-1），由圖像可知M在新拋物線、N在原拋物線上，據(jù)此列方程求出點(diǎn)M、N坐標(biāo)，由直線MN解析式即可求出與y軸交點(diǎn)坐標(biāo)即K點(diǎn)坐標(biāo)．【答案】（1）SKIPIF1<0；（2）F（-4，3），（3）SKIPIF1<0．【詳解】解：（1）由拋物線y＝x2+bx+c經(jīng)過(guò)點(diǎn)A（﹣1，0），B（0，3），得：SKIPIF1<0，解得：SKIPIF1<0，∴原拋物線對(duì)應(yīng)的函數(shù)表達(dá)式為：SKIPIF1<0；（2）由（1）得：原拋物線為：SKIPIF1<0，故頂點(diǎn)C坐標(biāo)為SKIPIF1<0∵新拋物線上的點(diǎn)D（3，﹣1）為原拋物線上點(diǎn)A的對(duì)應(yīng)點(diǎn)，∴原拋物線向右移4個(gè)單位，向下移1個(gè)單位得到新拋物線，∴新拋物線對(duì)應(yīng)的函數(shù)表達(dá)式為：SKIPIF1<0，即：SKIPIF1<0故新拋物線頂E點(diǎn)坐標(biāo)為SKIPIF1<0，與y軸交點(diǎn)G坐標(biāo)為SKIPIF1<0，以點(diǎn)C，E，F(xiàn)，G為頂點(diǎn)的四邊形是平行四邊形，點(diǎn)F不可能在CE下方，故如圖所示：當(dāng)平行四邊形為SKIPIF1<0時(shí)，點(diǎn)F坐標(biāo)為SKIPIF1<0，即SKIPIF1<0，根據(jù)平移性質(zhì)可知：SKIPIF1<0一定在原拋物線；當(dāng)平行四邊形為SKIPIF1<0時(shí)，點(diǎn)F坐標(biāo)為SKIPIF1<0，即SKIPIF1<0，此時(shí)SKIPIF1<0；故不在新拋物線上，綜上所述：以點(diǎn)C，E，F(xiàn)，G為頂點(diǎn)的四邊形是SKIPIF1<0時(shí)，F(xiàn)的坐標(biāo)為SKIPIF1<0；（3）∵SKIPIF1<0，MN＝CE，∴M點(diǎn)到N點(diǎn)的平移方式和C點(diǎn)到E點(diǎn)平移方式相同，設(shè)M在左側(cè)，坐標(biāo)為（a，b），則點(diǎn)N坐標(biāo)為（a+4，b-1），由圖可知，點(diǎn)M在新拋物線，點(diǎn)N在原拋物線，SKIPIF1<0，解得：SKIPIF1<0，即M點(diǎn)坐標(biāo)為SKIPIF1<0，∴點(diǎn)N坐標(biāo)為SKIPIF1<0，設(shè)直線MN解析式為SKIPIF1<0，∴SKIPIF1<0，解得：SKIPIF1<0，即：SKIPIF1<0，故直線MN與y軸交點(diǎn)K坐標(biāo)為SKIPIF1<0．本題主要考查了函數(shù)圖像的平移、函數(shù)圖像與幾何圖形結(jié)合的綜合能力的培養(yǎng)，要會(huì)利用數(shù)形結(jié)合的思想把代數(shù)和幾何圖形結(jié)合起來(lái)，掌握?qǐng)D像平移的性質(zhì)確定函數(shù)解析式和點(diǎn)的坐標(biāo)是解題關(guān)鍵．1．（2022·重慶開(kāi)州·校聯(lián)考模擬）如圖1，拋物線SKIPIF1<0與x軸交于A、B兩點(diǎn)（點(diǎn)A在點(diǎn)B的左側(cè)），與y軸交于點(diǎn)C，過(guò)點(diǎn)B作直線SKIPIF1<0直線SKIPIF1<0，交拋物線y于另一點(diǎn)D，點(diǎn)P為直線SKIPIF1<0上方拋物線上一動(dòng)點(diǎn)．(1)求線段SKIPIF1<0的長(zhǎng)．(2)過(guò)點(diǎn)P作SKIPIF1<0軸交SKIPIF1<0于點(diǎn)Q，交直線SKIPIF1<0于點(diǎn)F，過(guò)點(diǎn)P作SKIPIF1<0于點(diǎn)E，求SKIPIF1<0的最大值及此時(shí)點(diǎn)P的坐標(biāo)．(3)如圖2，將拋物線SKIPIF1<0向右平移3個(gè)單位得到新拋物線SKIPIF1<0，點(diǎn)M為新拋物線上一點(diǎn)，點(diǎn)N為原拋物線對(duì)稱軸一點(diǎn)，直接寫(xiě)出所有使得A、B、M、N為頂點(diǎn)的四邊形是平行四邊形時(shí)點(diǎn)N的坐標(biāo)，并寫(xiě)出其中一個(gè)點(diǎn)N的坐標(biāo)的求解過(guò)程．【答案】(1)4(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最大值為SKIPIF1<0，此時(shí)SKIPIF1<0；(3)SKIPIF1<0，SKIPIF1<0和SKIPIF1<0【思路分析】（1）令SKIPIF1<0，求解即可；（2）求直線SKIPIF1<0的解析式，設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，利用SKIPIF1<0，將所求轉(zhuǎn)化為SKIPIF1<0，再求解即可；（3）推出平移后的解析式，設(shè)SKIPIF1<0，SKIPIF1<0，分三種情況討論；再利用平行四邊形的性質(zhì)結(jié)合中點(diǎn)坐標(biāo)求解即可．【詳解】（1）令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0；（2）SKIPIF1<0，SKIPIF1<0，設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，∴直線SKIPIF1<0的解析式為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0的解析式為SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，∵點(diǎn)P為直線SKIPIF1<0上方拋物線上一動(dòng)點(diǎn)，SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最大值為SKIPIF1<0，此時(shí)SKIPIF1<0；（3）SKIPIF1<0，∴拋物線對(duì)稱軸為直線SKIPIF1<0，∵拋物線SKIPIF1<0向右平移3個(gè)單位得到新拋物線SKIPIF1<0，∴新拋物線SKIPIF1<0的解析式為SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，①當(dāng)SKIPIF1<0為平行四邊形的對(duì)角線時(shí)，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0；②當(dāng)SKIPIF1<0為平行四邊形的對(duì)角線時(shí)，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0；③當(dāng)SKIPIF1<0為平行四邊形的對(duì)角線時(shí)，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0；綜上，N點(diǎn)坐標(biāo)分別為SKIPIF1<0，SKIPIF1<0和SKIPIF1<0．2．（2021·山東濱州·模擬）如圖，在平面直角坐標(biāo)系中，拋物線SKIPIF1<0分別交SKIPIF1<0軸，SKIPIF1<0軸于點(diǎn)A，SKIPIF1<0和點(diǎn)SKIPIF1<0，拋物線SKIPIF1<0與拋物線SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，兩條拋物線的交點(diǎn)為SKIPIF1<0，SKIPIF1<0（點(diǎn)SKIPIF1<0在點(diǎn)SKIPIF1<0的左側(cè)）．(1)求拋物線SKIPIF1<0的表達(dá)式；(2)將拋物線SKIPIF1<0沿SKIPIF1<0軸正方向平移，使點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0重合，求平移的距離；(3)在（2）的條件下：規(guī)定拋物線SKIPIF1<0和拋物線SKIPIF1<0在直線SKIPIF1<0下方的圖象所組成的圖象為SKIPIF1<0，點(diǎn)SKIPIF1<0，SKIPIF1<0和SKIPIF1<0，SKIPIF1<0在函數(shù)SKIPIF1<0上（點(diǎn)SKIPIF1<0在點(diǎn)SKIPIF1<0的右側(cè)），在（2）的條件下，若SKIPIF1<0，且SKIPIF1<0，求點(diǎn)SKIPIF1<0坐標(biāo)．【答案】(1)SKIPIF1<0(2)1(3)點(diǎn)SKIPIF1<0坐標(biāo)為：SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0【思路分析】（1）由SKIPIF1<0得拋物線SKIPIF1<0的頂點(diǎn)坐標(biāo)為：SKIPIF1<0，即得拋物線SKIPIF1<0的頂點(diǎn)為SKIPIF1<0，從而拋物線SKIPIF1<0的表達(dá)式為SKIPIF1<0；（2）由SKIPIF1<0得SKIPIF1<0，設(shè)拋物線SKIPIF1<0向右平移SKIPIF1<0個(gè)單位后SKIPIF1<0與SKIPIF1<0重合，即SKIPIF1<0過(guò)SKIPIF1<0，可得平移的距離是1；（3）拋物線SKIPIF1<0向右平移1個(gè)單位得SKIPIF1<0，由SKIPIF1<0，的SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0在SKIPIF1<0左側(cè)圖象上時(shí)，SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0在SKIPIF1<0、SKIPIF1<0之間的圖象上時(shí)，分兩種情況：①SKIPIF1<0在拋物線SKIPIF1<0上，SKIPIF1<0，即得SKIPIF1<0，SKIPIF1<0；②SKIPIF1<0在拋物線SKIPIF1<0上，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0．【詳解】（1）解：SKIPIF1<0，SKIPIF1<0拋物線SKIPIF1<0的頂點(diǎn)坐標(biāo)為：SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱點(diǎn)為SKIPIF1<0，拋物線SKIPIF1<0與拋物線SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，SKIPIF1<0拋物線SKIPIF1<0的頂點(diǎn)為SKIPIF1<0，且拋物線SKIPIF1<0與拋物線SKIPIF1<0的形狀、大小相同，開(kāi)口方向相反，SKIPIF1<0拋物線SKIPIF1<0的表達(dá)式為SKIPIF1<0．（2）解：在SKIPIF1<0中，令SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，設(shè)拋物線SKIPIF1<0向右平移SKIPIF1<0個(gè)單位后SKIPIF1<0與SKIPIF1<0重合，即SKIPIF1<0過(guò)SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），SKIPIF1<0平移的距離是1．（3）解：由（2）知，拋物線SKIPIF1<0向右平移1個(gè)單位，可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0在SKIPIF1<0左側(cè)圖象上時(shí)，如圖：SKIPIF1<0在拋物線SKIPIF1<0上，SKIPIF1<0在拋物線SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0（舍去）或SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0在SKIPIF1<0、SKIPIF1<0之間的圖象上時(shí)，分兩種情況：①SKIPIF1<0在拋物線SKIPIF1<0上，如圖：SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，即得SKIPIF1<0或SKIPIF1<0（舍去），SKIPIF1<0，SKIPIF1<0；②SKIPIF1<0在SKIPIF1<0、SKIPIF1<0之間的圖象上，如圖：SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，綜上所述，點(diǎn)SKIPIF1<0坐標(biāo)為：SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0．3．（2022·陜西渭南·統(tǒng)考三模）在平面直角坐標(biāo)系中，已知拋物線SKIPIF1<0（b、c為常數(shù)）與x軸交于SKIPIF1<0、SKIPIF1<0兩點(diǎn)．(1)求拋物線SKIPIF1<0的函數(shù)表達(dá)式；(2)將該拋物線SKIPIF1<0向右平移4個(gè)單位長(zhǎng)度得到新的拋物線SKIPIF1<0，與原拋物線SKIPIF1<0交于點(diǎn)C，點(diǎn)D是點(diǎn)C關(guān)于x軸的對(duì)稱點(diǎn)，點(diǎn)N在平面直角坐標(biāo)系中，請(qǐng)問(wèn)在拋物線SKIPIF1<0上是否存在點(diǎn)M，使得以點(diǎn)C、D、M、N為頂點(diǎn)的四邊形是以CD為邊的矩形？若存在，求出點(diǎn)M的坐標(biāo)；若不存在，請(qǐng)說(shuō)明理由．【答案】(1)SKIPIF1<0(2)存在，點(diǎn)M的坐標(biāo)為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0【思路分析】（1）利用待定系數(shù)法直接求解即可；（2）存在，根據(jù)題意求得拋物線SKIPIF1<0的表達(dá)式，再與拋物線SKIPIF1<0聯(lián)立，求得點(diǎn)C的坐標(biāo)，進(jìn)而求得點(diǎn)D的坐標(biāo)；要使得以點(diǎn)C、D、M、N為頂點(diǎn)的四邊形是以CD為邊的矩形，分當(dāng)M在x軸上方時(shí)和當(dāng)M在x軸下方時(shí)，兩種情況討論，根據(jù)矩形的性質(zhì)列出方程，求解即可．【詳解】(1)解：把SKIPIF1<0、SKIPIF1<0代入SKIPIF1<0中，得SKIPIF1<0解得SKIPIF1<0∴拋物線SKIPIF1<0的函數(shù)表達(dá)式為SKIPIF1<0．(2)解：存在．理由如下：∵SKIPIF1<0，∴拋物線SKIPIF1<0的函數(shù)表達(dá)式SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴點(diǎn)C的坐標(biāo)為SKIPIF1<0，∵點(diǎn)D是點(diǎn)C關(guān)于x軸的對(duì)稱點(diǎn)，∴點(diǎn)D的坐標(biāo)為SKIPIF1<0．①當(dāng)M在x軸上方時(shí)，要使得以點(diǎn)C、D、M、N為頂點(diǎn)的四邊形是以CD為邊的矩形，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0（舍去），SKIPIF1<0，∴SKIPIF1<0；②當(dāng)M在x軸下方時(shí)，要使得以點(diǎn)C、D、M、N為頂點(diǎn)的四邊形是以CD為邊的矩形，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0．綜上，在拋物線SKIPIF1<0上存在點(diǎn)M，使得以點(diǎn)C、D、M、N為頂點(diǎn)的四邊形是以CD為邊的矩形，點(diǎn)M的坐標(biāo)為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0．4．（2022·廣東深圳·深圳市寶安第一外國(guó)語(yǔ)學(xué)校?？既＃┮阎獟佄锞€SKIPIF1<0：SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0的直線與SKIPIF1<0交于點(diǎn)SKIPIF1<0．(1)求直線SKIPIF1<0的函數(shù)表達(dá)式；(2)如圖SKIPIF1<0，若點(diǎn)SKIPIF1<0為直線SKIPIF1<0下方的SKIPIF1<0上一點(diǎn)，求點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離的最大值；(3)如圖SKIPIF1<0，將直線SKIPIF1<0繞點(diǎn)SKIPIF1<0順時(shí)針旋轉(zhuǎn)SKIPIF1<0后恰好經(jīng)過(guò)SKIPIF1<0的頂點(diǎn)SKIPIF1<0，沿射線SKIPIF1<0的方向平移拋物線SKIPIF1<0得到拋物線SKIPIF1<0，SKIPIF1<0的頂點(diǎn)為SKIPIF1<0，兩拋物線相交于點(diǎn)SKIPIF1<0設(shè)交點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0若SKIPIF1<0，求SKIPIF1<0的值．【答案】(1)y=x+2(2)SKIPIF1<0(3)SKIPIF1<0【思路分析】（1）先根據(jù)拋物線的函數(shù)表達(dá)式求出點(diǎn)A的坐標(biāo)，再將點(diǎn)A的坐標(biāo)和（1，3）代入y=kx+b，即可求出直線AB的函數(shù)表達(dá)式；（2）過(guò)點(diǎn)P作SKIPIF1<0交直線AB于點(diǎn)Q，過(guò)點(diǎn)P作PM⊥AB，垂足為點(diǎn)M，易證△MPQ為等腰直角三角形，分別表示出點(diǎn)P和點(diǎn)Q的坐標(biāo)，求出PQ的最大值，當(dāng)PQ取最大值時(shí)PM也取最大值，（3）過(guò)點(diǎn)E作SKIPIF1<0，交x軸于點(diǎn)P，過(guò)點(diǎn)D作DQ⊥PQ，垂足為Q，易證△APE～△DEQ，將點(diǎn)D的坐標(biāo)用m表示出來(lái)，根據(jù)SKIPIF1<0即可求出m的值．【詳解】（1）解：當(dāng)x=0時(shí)，SKIPIF1<0，∴A（0，2），設(shè)直線AB的函數(shù)表達(dá)式為：y=kx+b，把A（0，2）和（1，3）代入y=kx+b，SKIPIF1<0，解得：SKIPIF1<0，∴直線AB得函數(shù)表達(dá)式為：y=x+2．（2）將拋物線的函數(shù)表達(dá)式整理為一般式為：SKIPIF1<0，如圖，過(guò)點(diǎn)P作SKIPIF1<0交直線AB于點(diǎn)Q，過(guò)點(diǎn)P作PM⊥AB，垂足為點(diǎn)M，設(shè)點(diǎn)P的坐標(biāo)為（a，SKIPIF1<0），∵SKIPIF1<0，∴點(diǎn)Q的橫坐標(biāo)為a，∵點(diǎn)Q在直線AB上，∴點(diǎn)Q的坐標(biāo)為（a，a+2），∴SKIPIF1<0，整理得：SKIPIF1<0，當(dāng)a=SKIPIF1<0時(shí)，PQ有最大值，最大值為SKIPIF1<0，∵直線AB與豎直方向得夾角為45°，∴∠MQP=45°，∴△MPQ為等腰直角三角形，∴PM=SKIPIF1<0，當(dāng)PQ取最大值時(shí)，PM也取最大值，∴PM的最大值為：SKIPIF1<0，（3）∵拋物線的函數(shù)表達(dá)式為：SKIPIF1<0，∴頂點(diǎn)C（1，1），設(shè)直線AC的函數(shù)表達(dá)式為：y=kx+b，將點(diǎn)C和點(diǎn)A的坐標(biāo)代入得：SKIPIF1<0，解得：SKIPIF1<0，∴直線AC的函數(shù)表達(dá)式為：y=-x+2，設(shè)點(diǎn)D的橫坐標(biāo)為b，∵點(diǎn)D在直線AC上，∴點(diǎn)D的縱坐標(biāo)為-b+2，即D（b，-b+2），∴SKIPIF1<0的函數(shù)表達(dá)式為：SKIPIF1<0，E的橫坐標(biāo)為m，∵點(diǎn)E在拋物線SKIPIF1<0上，∴點(diǎn)E的縱坐標(biāo)為：SKIPIF1<0，∵點(diǎn)E也在拋物線SKIPIF1<0上，∴點(diǎn)E的縱坐標(biāo)為：SKIPIF1<0，∴SKIPIF1<0=SKIPIF1<0，整理得：SKIPIF1<0解得：b=2m或b=1（舍），∴D（2m，-2m+2），過(guò)點(diǎn)E作SKIPIF1<0，交x軸于點(diǎn)P，過(guò)點(diǎn)D作DQ⊥PQ，垂足為Q，∵∠AED=90°，∠EPA=90°，∴∠AEP+∠DEQ=90°，∠AEP+∠EAP=90°，∴∠DEQ=∠EAP，在△APE和△DEQ中，∠DEQ=∠EAP，∠APE=∠DQE，∴△APE～△DEQ，∴SKIPIF1<0，∵A（0，2），E（m，SKIPIF1<0），D（2m，-2m+2），∴PE=m，EQ=m，DQ=SKIPIF1<0，AP=SKIPIF1<0，∴SKIPIF1<0，整理得：SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0（舍）．5．（2022·重慶·西南大學(xué)附中校考三模）如圖，在平面直角坐標(biāo)系中，拋物線SKIPIF1<0與x軸交于SKIPIF1<0，B兩點(diǎn)，其對(duì)稱軸SKIPIF1<0與x軸交于點(diǎn)D．圖1

圖2(1)求該拋物線的函數(shù)表達(dá)式；(2)如圖1，點(diǎn)P為第四象限內(nèi)的拋物線上一動(dòng)點(diǎn)，連接PB，PC，CD，求四邊形PBDC面積的最大值和此時(shí)點(diǎn)P的坐標(biāo)；(3)將該拋物線向左平移3個(gè)單位長(zhǎng)度得到拋物線y'，平移后的拋物線與原拋物線的對(duì)稱軸相交于點(diǎn)E，點(diǎn)F為拋物線y'對(duì)稱軸上的一點(diǎn)，M是原拋物線上的動(dòng)點(diǎn)，直接寫(xiě)出所有使得以點(diǎn)A，E，F(xiàn)，M為頂點(diǎn)的四邊形是平行四邊形的點(diǎn)M的坐標(biāo)，并把求其中一個(gè)點(diǎn)M的坐標(biāo)的過(guò)程寫(xiě)出來(lái)．【答案】(1)SKIPIF1<0；(2)SKIPIF1<0的最大值為SKIPIF1<0，此時(shí)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，SKIPIF1<0；(3)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0．【思路分析】（1）運(yùn)用待定系數(shù)法即可求得答案；（2）利用待定系數(shù)法求得直線SKIPIF1<0的解析式為SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸交SKIPIF1<0于點(diǎn)SKIPIF1<0，如圖1，設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，得出SKIPIF1<0，再運(yùn)用二次函數(shù)的性質(zhì)即可得出答案；（3）根據(jù)平移的性質(zhì)可得SKIPIF1<0，新拋物線的對(duì)稱軸為直線SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，又SKIPIF1<0，由以點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為頂點(diǎn)的四邊形是平行四邊形，分三種情況：①當(dāng)SKIPIF1<0、SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0、SKIPIF1<0的中點(diǎn)重合，②當(dāng)SKIPIF1<0、SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0、SKIPIF1<0的中點(diǎn)重合，③當(dāng)SKIPIF1<0、SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0、SKIPIF1<0的中點(diǎn)重合，分別畫(huà)出圖形，建立方程求解即可．【詳解】（1）解：（1）SKIPIF1<0拋物線SKIPIF1<0經(jīng)過(guò)點(diǎn)SKIPIF1<0，其對(duì)稱軸SKIPIF1<0，SKIPIF1<0SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0該拋物線的函數(shù)表達(dá)式為SKIPIF1<0；（2）解：如圖，連接BC，作PH∥y軸，交BC于H，SKIPIF1<0點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0關(guān)于對(duì)稱軸SKIPIF1<0對(duì)稱，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0的解析式為SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最大值為SKIPIF1<0，此時(shí)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，SKIPIF1<0；（3）解：將拋物線SKIPIF1<0向左平移3個(gè)單位長(zhǎng)度得到拋物線SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0新拋物線的對(duì)稱軸為直線SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0；①當(dāng)SKIPIF1<0、SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0、SKIPIF1<0的中點(diǎn)重合，SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0；②當(dāng)SKIPIF1<0、SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0、SKIPIF1<0的中點(diǎn)重合，SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0；③當(dāng)SKIPIF1<0、SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0、SKIPIF1<0的中點(diǎn)重合，SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0；綜上所述，點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0．6．（2022·內(nèi)蒙古呼和浩特·統(tǒng)考三模）拋物線SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，SKIPIF1<0，直線SKIPIF1<0與拋物線交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)．(1)求拋物線的解析式．(2)在此拋物線的對(duì)稱軸上是否存在一點(diǎn)SKIPIF1<0，使得SKIPIF1<0的周長(zhǎng)最??？若存在，請(qǐng)求出點(diǎn)SKIPIF1<0的坐標(biāo)；若不存在，請(qǐng)說(shuō)明理由．(3)若點(diǎn)SKIPIF1<0為直線SKIPIF1<0上方的拋物線上的一個(gè)動(dòng)點(diǎn)（不與點(diǎn)SKIPIF1<0，SKIPIF1<0重合），將直線SKIPIF1<0上方的拋物線部分關(guān)于直線SKIPIF1<0對(duì)稱形成愛(ài)心圖案，動(dòng)點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱的點(diǎn)為SKIPIF1<0，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)存在，SKIPIF1<0，理由見(jiàn)詳解(3)SKIPIF1<0【思路分析】（1）將SKIPIF1<0，SKIPIF1<0代入拋物線SKIPIF1<0求解即可：（2）連接BC，BC與對(duì)稱軸的交點(diǎn)即點(diǎn)P，此時(shí)SKIPIF1<0的周長(zhǎng)最?。唬?）過(guò)點(diǎn)E作SKIPIF1<0軸，進(jìn)而得到SKIPIF1<0，由三角函數(shù)即可求解；【詳解】（1）解：將SKIPIF1<0，SKIPIF1<0代入拋物線SKIPIF1<0得，SKIPIF1<0解得：SKIPIF1<0，∴拋物線的解析式為：SKIPIF1<0．（2）由SKIPIF1<0解得：SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，設(shè)BC的解析式為：SKIPIF1<0，將SKIPIF1<0，SKIPIF1<0代入SKIPIF1<0得，SKIPIF1<0解得：SKIPIF1<0，∴SKIPIF1<0，拋物線的對(duì)稱軸為：SKIPIF1<0，當(dāng)點(diǎn)P在BC上時(shí)，SKIPIF1<0的周長(zhǎng)最小，∴將SKIPIF1<0代入SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0．（3）設(shè)點(diǎn)SKIPIF1<0，由SKIPIF1<0，SKIPIF1<0可求得CD的解析式為：SKIPIF1<0，過(guò)點(diǎn)E作SKIPIF1<0軸，∴SKIPIF1<0，將SKIPIF1<0代入SKIPIF1<0得，SKIPIF1<0，將SKIPIF1<0代入SKIPIF1<0得，SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0軸，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0最大，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0的取值范圍為：SKIPIF1<0．7．（2022·陜西寶雞·統(tǒng)考二模）如圖，在平面直角坐標(biāo)系中，拋物線SKIPIF1<0的圖象經(jīng)過(guò)SKIPIF1<0，SKIPIF1<0兩點(diǎn)，將拋物線SKIPIF1<0向右平移2個(gè)單位得到拋物線SKIPIF1<0，平移后點(diǎn)A的對(duì)應(yīng)點(diǎn)為點(diǎn)B．(1)求拋物線SKIPIF1<0與SKIPIF1<0的函數(shù)表達(dá)式；(2)若點(diǎn)M是拋物線SKIPIF1<0上一動(dòng)點(diǎn)，點(diǎn)N是拋物線SKIPIF1<0上一動(dòng)點(diǎn)，請(qǐng)問(wèn)是否存在這樣的點(diǎn)M、N，使得以A、B、M、N為頂點(diǎn)且以AB為邊的四邊形是面積為8的平行四邊形？若存在，求出點(diǎn)M、N的坐標(biāo)；若不存在，請(qǐng)說(shuō)明理由．【答案】(1)SKIPIF1<0；SKIPIF1<0(2)存在，SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0【思路分析】（1）用待定系數(shù)法求出b與c的值即可；（2）先求點(diǎn)B的坐標(biāo)，再根據(jù)平行四邊形的性質(zhì)進(jìn)行分類討論．【詳解】（1）解：∵SKIPIF1<0的圖象經(jīng)過(guò)SKIPIF1<0，∴SKIPIF1<0，將SKIPIF1<0代入SKIPIF1<0中，得SKIPIF1<0，解得SKIPIF1<0，∴拋物線SKIPIF1<0的函數(shù)表達(dá)式為SKIPIF1<0．∵將拋物線SKIPIF1<0向右平移2個(gè)單位得到拋物線SKIPIF1<0，∴拋物線SKIPIF1<0的函數(shù)表達(dá)式為SKIPIF1<0．（2）解：存在．理由如下：∵點(diǎn)SKIPIF1<0向右平移2個(gè)單位得到點(diǎn)B，∴SKIPIF1<0，∴SKIPIF1<0．由題意知，以AB為邊的平行四邊形的面積為8，則SKIPIF1<0，SKIPIF1<0，AB邊上的高為4．易得拋物線SKIPIF1<0的頂點(diǎn)為SKIPIF1<0，而SKIPIF1<0，∴在x軸下方不存在滿足條件的點(diǎn)M、N．在SKIPIF1<0中，令SKIPIF1<0