2024年中考數(shù)學(xué)壓軸題型（廣東專用）專題09 二次函數(shù)中最值、變換、新定義型問題（含解析）

PAGE專題09二次函數(shù)中最值、變換、新定義型問題通用的解題思路：第一步:先判定函數(shù)的增減性：一次函數(shù)、反比例函數(shù)看SKIPIF1<0，二次函數(shù)看對(duì)稱軸與區(qū)間的位置關(guān)系；第二步：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；所以SKIPIF1<0.二次函數(shù)求取值范圍之動(dòng)軸定區(qū)間或者定軸動(dòng)區(qū)間的分類方法：分對(duì)稱軸在區(qū)間的左邊、右邊、中間三種情況。若自變量SKIPIF1<0的取值范圍為全體實(shí)數(shù)，如圖①，函數(shù)在頂點(diǎn)處SKIPIF1<0時(shí)，取到最值.若SKIPIF1<0，如圖②，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.若SKIPIF1<0，如圖③，當(dāng)SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0.若SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，如圖④，當(dāng)SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0.1．（2023·廣東·中考真題）如圖，拋物線SKIPIF1<0經(jīng)過正方形SKIPIF1<0的三個(gè)頂點(diǎn)A，B，C，點(diǎn)B在SKIPIF1<0軸上，則SKIPIF1<0的值為（

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A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】連接SKIPIF1<0，交y軸于點(diǎn)D，根據(jù)正方形的性質(zhì)可知SKIPIF1<0，然后可得點(diǎn)SKIPIF1<0，進(jìn)而代入求解即可．【詳解】解：連接SKIPIF1<0，交y軸于點(diǎn)D，如圖所示：

當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，即SKIPIF1<0，∵四邊形SKIPIF1<0是正方形，∴SKIPIF1<0，SKIPIF1<0，∴點(diǎn)SKIPIF1<0，∴SKIPIF1<0，解得：SKIPIF1<0，故選B．【點(diǎn)睛】本題主要考查二次函數(shù)的圖象與性質(zhì)及正方形的性質(zhì)，熟練掌握二次函數(shù)的圖象與性質(zhì)及正方形的性質(zhì)是解題的關(guān)鍵．2．（2022·廣東廣州·中考真題）如圖，拋物線SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，下列結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0隨SKIPIF1<0的增大而減小 D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0隨SKIPIF1<0的增大而減小【答案】C【分析】由圖像可知，拋物線開口向上，因此a＞0．由圖像與y軸的交點(diǎn)在y軸負(fù)半軸上得c＜0．根據(jù)圖像可知，在對(duì)稱軸左側(cè)y隨x的增大而減小，在對(duì)稱軸右側(cè)y隨x的增大而增大．【詳解】拋物線開口向上，因此a＞0，故A選項(xiàng)不符合題意．拋物線與y軸的交點(diǎn)在y軸的負(fù)半軸上，因此c＜0，故B選項(xiàng)不符合題意．拋物線開口向上，因此在對(duì)稱軸左側(cè)，y隨x的增大而減小，故C選項(xiàng)符合題意．拋物線開口向上，因此在對(duì)稱軸右側(cè)y隨x的增大而增大，故D選項(xiàng)不符合題意．故選C【點(diǎn)睛】本題考查了二次函數(shù)圖像的性質(zhì)，掌握二次函數(shù)圖像的性質(zhì)是解題的關(guān)鍵．3．（2022·廣東·中考真題）如圖，拋物線SKIPIF1<0（b，c是常數(shù)）的頂點(diǎn)為C，與x軸交于A，B兩點(diǎn)，SKIPIF1<0，SKIPIF1<0，點(diǎn)P為線段SKIPIF1<0上的動(dòng)點(diǎn)，過P作SKIPIF1<0//SKIPIF1<0交SKIPIF1<0于點(diǎn)Q．(1)求該拋物線的解析式；(2)求SKIPIF1<0面積的最大值，并求此時(shí)P點(diǎn)坐標(biāo)．【答案】(1)SKIPIF1<0(2)2；P（-1，0）【分析】（1）用待定系數(shù)法將A，B的坐標(biāo)代入函數(shù)一般式中，即可求出函數(shù)的解析式；（2）分別求出C點(diǎn)坐標(biāo)，直線AC，BC的解析式，PQ的解析式為：y=-2x+n，進(jìn)而求出P，Q的坐標(biāo)以及n的取值范圍，由SKIPIF1<0列出函數(shù)式求解即可．【詳解】（1）解：∵點(diǎn)A（1，0），AB=4，∴點(diǎn)B的坐標(biāo)為（-3，0），將點(diǎn)A（1，0），B（-3，0）代入函數(shù)解析式中得：SKIPIF1<0，解得：b=2，c=-3，∴拋物線的解析式為SKIPIF1<0；（2）解：由（1）得拋物線的解析式為SKIPIF1<0，頂點(diǎn)式為：SKIPIF1<0，則C點(diǎn)坐標(biāo)為：（-1，-4），由B（-3，0），C（-1，-4）可求直線BC的解析式為：y=-2x-6，由A（1，0），C（-1，-4）可求直線AC的解析式為：y=2x-2，∵PQ∥BC，設(shè)直線PQ的解析式為：y=-2x+n，與x軸交點(diǎn)PSKIPIF1<0，由SKIPIF1<0解得：SKIPIF1<0，∵P在線段AB上，∴SKIPIF1<0，∴n的取值范圍為-6＜n＜2，則SKIPIF1<0SKIPIF1<0SKIPIF1<0∴當(dāng)n=-2時(shí)，即P（-1，0）時(shí)，SKIPIF1<0最大，最大值為2．【點(diǎn)睛】本題考查二次函數(shù)的面積最值問題，二次函數(shù)的圖象與解析式間的關(guān)系，一次函數(shù)的解析式與圖象，熟練掌握數(shù)形結(jié)合思想是解決本題的關(guān)鍵．4．（2022·廣東廣州·中考真題）已知直線SKIPIF1<0：SKIPIF1<0經(jīng)過點(diǎn)（0，7）和點(diǎn)（1，6）．(1)求直線SKIPIF1<0的解析式；(2)若點(diǎn)P（SKIPIF1<0，SKIPIF1<0）在直線SKIPIF1<0上，以P為頂點(diǎn)的拋物線G過點(diǎn)（0，-3），且開口向下①求SKIPIF1<0的取值范圍；②設(shè)拋物線G與直線SKIPIF1<0的另一個(gè)交點(diǎn)為Q，當(dāng)點(diǎn)Q向左平移1個(gè)單長度后得到的點(diǎn)Q'也在G上時(shí)，求G在SKIPIF1<0≤SKIPIF1<0≤SKIPIF1<0的圖象的最高點(diǎn)的坐標(biāo)．【答案】(1)直線SKIPIF1<0解析式為：SKIPIF1<0；(2)①m＜10，且m≠0；②最高點(diǎn)的坐標(biāo)為（-2，9）或（2，5）【分析】（1）根據(jù)待定系數(shù)法求出解析式即可；（2）①設(shè)G的頂點(diǎn)式，根據(jù)點(diǎn)P在直線SKIPIF1<0上得出G的關(guān)系式，根據(jù)題意得出點(diǎn)（0，-3）不能成為拋物線G的頂點(diǎn)，進(jìn)而得出點(diǎn)P必須位于直線SKIPIF1<0的上方，可求m的取值范圍，然后結(jié)合點(diǎn)P不能在SKIPIF1<0軸上得出答案；②先根據(jù)點(diǎn)Q，點(diǎn)SKIPIF1<0的對(duì)稱，得QQ'=1，可表示點(diǎn)Q和SKIPIF1<0的坐標(biāo)，再將點(diǎn)SKIPIF1<0的坐標(biāo)的代入關(guān)系式，求出a，再將點(diǎn)（0，-3）代入可求出m的值，然后分兩種情況結(jié)合取值范圍，求出函數(shù)最大值時(shí)，最高點(diǎn)的坐標(biāo)即可．【詳解】（1）解：∵直線SKIPIF1<0經(jīng)過點(diǎn)（0，7）和點(diǎn)（1，6），∴SKIPIF1<0，解得SKIPIF1<0，∴直線SKIPIF1<0解析式為：SKIPIF1<0；（2）解：①設(shè)G：SKIPIF1<0（SKIPIF1<0），∵點(diǎn)P（SKIPIF1<0，SKIPIF1<0）在直線SKIPIF1<0上，∴SKIPIF1<0；∴G：SKIPIF1<0（SKIPIF1<0）∵（0，-3）不在直線SKIPIF1<0上，∴（0，-3）不能成為拋物線G的頂點(diǎn)，而以P為頂點(diǎn)的拋物線G開口向下，且經(jīng)過（0，-3），∴點(diǎn)P必須位于直線SKIPIF1<0的上方，則SKIPIF1<0，SKIPIF1<0，另一方面，點(diǎn)P不能在SKIPIF1<0軸上，∴SKIPIF1<0，∴所求SKIPIF1<0取值范圍為：SKIPIF1<0，且SKIPIF1<0；②如圖，QQ'關(guān)于直線SKIPIF1<0對(duì)稱，且QQ'=1，∴點(diǎn)Q橫坐標(biāo)為SKIPIF1<0，而點(diǎn)Q在SKIPIF1<0上，∴Q（SKIPIF1<0，SKIPIF1<0），Q'（SKIPIF1<0，SKIPIF1<0）；∵Q'（SKIPIF1<0，SKIPIF1<0）在G：SKIPIF1<0上，∴SKIPIF1<0，SKIPIF1<0，∴G：SKIPIF1<0，或SKIPIF1<0．∵拋物線G過點(diǎn)（0，-3），∴SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，拋物線G為SKIPIF1<0，對(duì)稱軸為直線SKIPIF1<0，對(duì)應(yīng)區(qū)間為-2≤SKIPIF1<0≤-1，整個(gè)區(qū)間在對(duì)稱軸SKIPIF1<0的右側(cè)，此時(shí)，函數(shù)值SKIPIF1<0隨著SKIPIF1<0的增大而減小，如圖，∴當(dāng)SKIPIF1<0取區(qū)間左端點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0達(dá)最大值9，最高點(diǎn)坐標(biāo)為（-2，9）；當(dāng)SKIPIF1<0時(shí)，對(duì)應(yīng)區(qū)間為SKIPIF1<0≤SKIPIF1<0≤SKIPIF1<0，最高點(diǎn)為頂點(diǎn)P（2，5），如圖，∴G在指定區(qū)間圖象最高點(diǎn)的坐標(biāo)為（-2，9）或（2，5）．【點(diǎn)睛】本題考查了二次函數(shù)的綜合問題，考查了待定系數(shù)法求二次函數(shù)的關(guān)系式，求二次函數(shù)的極值等．解題的關(guān)鍵是掌握當(dāng)SKIPIF1<0時(shí)，頂點(diǎn)在直線SKIPIF1<0與SKIPIF1<0軸的交點(diǎn)（0，7），此時(shí)拋物線不可能過點(diǎn)（0，-3），因此，SKIPIF1<0可能會(huì)被忽視．題型一二次函數(shù)圖象與系數(shù)a,b,c的關(guān)系1．已知二次函數(shù)SKIPIF1<0圖象的一部分如圖所示，該函數(shù)圖象經(jīng)過點(diǎn)SKIPIF1<0，對(duì)稱軸為直線SKIPIF1<0．對(duì)于下列結(jié)論：SKIPIF1<0；②SKIPIF1<0；③多項(xiàng)式SKIPIF1<0可因式分解為SKIPIF1<0；④無論m為何值時(shí)，SKIPIF1<0．其中正確個(gè)數(shù)有（

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A．1個(gè) B．2個(gè) C．3個(gè) D．4個(gè)【答案】B【分析】本題主要考查了二次函數(shù)圖象與系數(shù)的關(guān)系，二次函數(shù)圖象的性質(zhì)等等：先根據(jù)圖像的開口方向和對(duì)稱軸可判斷①；由拋物線的對(duì)稱軸為SKIPIF1<0可得拋物線與x軸的另一個(gè)交點(diǎn)為SKIPIF1<0，由此可判斷②；根據(jù)拋物線與x軸的兩個(gè)交點(diǎn)坐標(biāo)可判斷③；根據(jù)函數(shù)的對(duì)稱軸為SKIPIF1<0可知SKIPIF1<0時(shí)y有最大值，由此可判斷④．【詳解】解：∵拋物線開口向下，∴SKIPIF1<0，∵對(duì)稱軸為直線SKIPIF1<0，∴SKIPIF1<0，結(jié)論①正確；∵拋物線與x軸的一個(gè)交點(diǎn)為SKIPIF1<0，且對(duì)稱軸為直線SKIPIF1<0，∴拋物線與x軸的另一個(gè)交點(diǎn)為SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，結(jié)論②錯(cuò)誤；∵拋物線SKIPIF1<0與x軸的兩個(gè)交點(diǎn)為SKIPIF1<0，SKIPIF1<0，∴多項(xiàng)式SKIPIF1<0可因式分解為SKIPIF1<0，結(jié)論③錯(cuò)誤；∵對(duì)稱軸為直線SKIPIF1<0，且函數(shù)開口向下，∴當(dāng)SKIPIF1<0時(shí)，y有最大值，由SKIPIF1<0得，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴無論m為何值時(shí)，SKIPIF1<0，∴SKIPIF1<0，結(jié)論④正確；綜上：正確的有①④．故選：B．2．如圖是二次函數(shù)SKIPIF1<0的圖象，對(duì)稱軸是直線SKIPIF1<0．關(guān)于下列結(jié)論：①SKIPIF1<0；②SKIPIF1<0，③SKIPIF1<0；④SKIPIF1<0；⑤方程SKIPIF1<0兩個(gè)根為SKIPIF1<0，SKIPIF1<0，其中正確的結(jié)論有（

）A．①③④ B．②④⑤ C．①②⑤ D．②③⑤【答案】B【分析】本題考查了二次函數(shù)圖像與性質(zhì)，由拋物線的開口方向判斷a與0的關(guān)系，由拋物線與y軸的交點(diǎn)判斷c與0的關(guān)系，然后根據(jù)對(duì)稱軸及拋物線與x軸交點(diǎn)情況進(jìn)行推理，進(jìn)而對(duì)所得結(jié)論進(jìn)行判斷．根據(jù)二次函數(shù)圖像判定代數(shù)式的正負(fù)和數(shù)形結(jié)合思想是解題的關(guān)鍵．【詳解】解：由圖象可得：拋物線開口向下，∴SKIPIF1<0，對(duì)稱軸在y軸左側(cè)，根據(jù)左同右異，∴SKIPIF1<0，∴SKIPIF1<0，故①錯(cuò)；由圖象可得：拋物線與x軸有兩個(gè)交點(diǎn)，∴SKIPIF1<0，故②正確；由圖象可得：SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0，故③錯(cuò)；由圖象可得：SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，故④正確；由圖象可得：SKIPIF1<0的兩根分別為SKIPIF1<0，SKIPIF1<0，∴方程SKIPIF1<0兩個(gè)根為SKIPIF1<0，SKIPIF1<0，故⑤正確；故選：B．3．拋物線SKIPIF1<0上部分點(diǎn)的橫坐標(biāo)x和縱坐標(biāo)y的對(duì)應(yīng)值如下表，下列說法正確的有（

）．x…SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<001…y…SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<033…①當(dāng)SKIPIF1<0時(shí)，y隨x的增大而減?。虎趻佄锞€的對(duì)稱軸為直線SKIPIF1<0；③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；

④方程SKIPIF1<0的一個(gè)正數(shù)解SKIPIF1<0滿足SKIPIF1<0．A．①② B．①②③ C．②③④ D．①②④【答案】D【分析】本題主要考查了二次函數(shù)圖像的性質(zhì)和二次函數(shù)圖像上點(diǎn)的特征，理解二次函數(shù)圖像的性質(zhì)是解題的關(guān)鍵．根據(jù)表格信息，先確定出拋物線的對(duì)稱軸，然后根據(jù)二次函數(shù)的性質(zhì)逐項(xiàng)判斷即可．【詳解】解：①由表格看出，這個(gè)拋物線的對(duì)稱軸為直線SKIPIF1<0且當(dāng)SKIPIF1<0時(shí)，y隨x的增大而增大，根據(jù)二次函數(shù)圖像的對(duì)稱性可得當(dāng)SKIPIF1<0時(shí)，y隨x的增大而減小，故①的說法正確；②由表格看出，這個(gè)拋物線的對(duì)稱軸為直線SKIPIF1<0，故②的說法正確；③當(dāng)SKIPIF1<0時(shí)的函數(shù)值與SKIPIF1<0時(shí)的函數(shù)值相同為SKIPIF1<0，即，故③的說法錯(cuò)誤；④當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，根據(jù)二次函數(shù)的對(duì)稱性可得當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故方程SKIPIF1<0的正數(shù)解滿足SKIPIF1<0，故④的說法正確．故選：D．題型二二次函數(shù)中線段最小值1．如題，在平面直角坐標(biāo)系SKIPIF1<0中，拋物線SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0，與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0．(1)求拋物線的解析式．(2)點(diǎn)SKIPIF1<0為拋物線的對(duì)稱軸上一動(dòng)點(diǎn)，當(dāng)SKIPIF1<0周長最小時(shí)，求點(diǎn)SKIPIF1<0的坐標(biāo)．(3)點(diǎn)SKIPIF1<0是SKIPIF1<0的中點(diǎn)，射線SKIPIF1<0交拋物線于點(diǎn)SKIPIF1<0，SKIPIF1<0是拋物線上一動(dòng)點(diǎn)，過點(diǎn)SKIPIF1<0作SKIPIF1<0軸的平行線，交射線SKIPIF1<0與點(diǎn)SKIPIF1<0，是否存在點(diǎn)SKIPIF1<0使得SKIPIF1<0與SKIPIF1<0相似？若存在，求出點(diǎn)SKIPIF1<0的坐標(biāo)；若不存在，請(qǐng)說明理由．【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)存在，點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0或SKIPIF1<0【分析】（1）待定系數(shù)法求解析式即可求解；（2）點(diǎn)SKIPIF1<0關(guān)于對(duì)稱軸的對(duì)稱點(diǎn)為點(diǎn)SKIPIF1<0，連接SKIPIF1<0交對(duì)稱軸于點(diǎn)SKIPIF1<0，連接SKIPIF1<0，此時(shí)SKIPIF1<0最小，得出直線SKIPIF1<0的解析式為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，得出SKIPIF1<0即可求解；（3）分兩種情況：SKIPIF1<0，SKIPIF1<0，根據(jù)相似三角形的性質(zhì)，即可求解．【詳解】（1）解：把點(diǎn)SKIPIF1<0，SKIPIF1<0分別代入SKIPIF1<0，得SKIPIF1<0解得SKIPIF1<0∴拋物線的解析式為SKIPIF1<0．（2）∵SKIPIF1<0，SKIPIF1<0∴對(duì)稱軸為直線SKIPIF1<0點(diǎn)SKIPIF1<0關(guān)于對(duì)稱軸的對(duì)稱點(diǎn)為點(diǎn)SKIPIF1<0，連接SKIPIF1<0交對(duì)稱軸于點(diǎn)SKIPIF1<0，連接SKIPIF1<0，此時(shí)SKIPIF1<0最小，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴點(diǎn)SKIPIF1<0．設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0，代入SKIPIF1<0得SKIPIF1<0∴SKIPIF1<0∴直線SKIPIF1<0的解析式為SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴點(diǎn)SKIPIF1<0．（3）存在．∵SKIPIF1<0，SKIPIF1<0是SKIPIF1<0的中點(diǎn)，SKIPIF1<0SKIPIF1<0．又SKIPIF1<0，∴直線SKIPIF1<0的解析式為SKIPIF1<0，SKIPIF1<0．聯(lián)立SKIPIF1<0得SKIPIF1<0．解得SKIPIF1<0，SKIPIF1<0（舍）．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．∴SKIPIF1<0．設(shè)SKIPIF1<0，則SKIPIF1<0．∴SKIPIF1<0．分以下兩種情況：①如圖2，若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0．∴SKIPIF1<0軸．∴SKIPIF1<0．∴SKIPIF1<0．解得SKIPIF1<0或SKIPIF1<0（舍）．∴SKIPIF1<0．②如圖3，若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0．過點(diǎn)SKIPIF1<0作SKIPIF1<0于點(diǎn)SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0．解得SKIPIF1<0或SKIPIF1<0（舍）．∴SKIPIF1<0．綜上，點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0或SKIPIF1<0．【點(diǎn)睛】本題考查了二次函數(shù)的綜合運(yùn)用，待定系數(shù)法求二次函數(shù)解析式，線段周長問題以及相似三角形的性質(zhì)，解題的關(guān)鍵是求出二次函數(shù)解析式．2．如圖1，在平面直角坐標(biāo)系中，直線SKIPIF1<0與拋物線SKIPIF1<0（b，c是常數(shù)）交于A、B兩點(diǎn)，點(diǎn)A在x軸上，點(diǎn)B在y軸上．設(shè)拋物線與x軸的另一個(gè)交點(diǎn)為點(diǎn)C．

(1)求該拋物線的解析式；(2)若點(diǎn)M是拋物線對(duì)稱軸上的一個(gè)動(dòng)點(diǎn)，當(dāng)SKIPIF1<0的值最小時(shí)，求點(diǎn)M的坐標(biāo)；(3)P是拋物線上一動(dòng)點(diǎn)（不與點(diǎn)A、B重合），如圖2，若點(diǎn)P在直線SKIPIF1<0上方，連接SKIPIF1<0交SKIPIF1<0于點(diǎn)D，求SKIPIF1<0的最大值；【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0【分析】（1）直線SKIPIF1<0與兩坐標(biāo)的交點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0，將A、B代入拋物線SKIPIF1<0，利用待定系數(shù)法即可求解；（2）根據(jù)拋物線解析式確定與x軸的交點(diǎn)坐標(biāo)，再由對(duì)稱的性質(zhì)及兩點(diǎn)之間線段最短即可確定點(diǎn)M的位置，然后代入一次函數(shù)解析式求解即可；（3）過點(diǎn)P作SKIPIF1<0交直線SKIPIF1<0于點(diǎn)E，則SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0取最大值時(shí)，SKIPIF1<0有最大值．【詳解】（1）解：SKIPIF1<0直線SKIPIF1<0與坐標(biāo)軸交于A、B兩點(diǎn)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，將A、B代入拋物線SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0拋物線的解析式為：SKIPIF1<0．（2）∵拋物線的解析式為：SKIPIF1<0．∴當(dāng)SKIPIF1<0時(shí)，解得SKIPIF1<0，∴SKIPIF1<0，∴拋物線的對(duì)稱軸為SKIPIF1<0，

∵點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱，連接SKIPIF1<0交對(duì)稱軸于點(diǎn)M，∴SKIPIF1<0，此時(shí)SKIPIF1<0取得最小值，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0；（3）過點(diǎn)P作SKIPIF1<0交直線SKIPIF1<0于點(diǎn)E，則SKIPIF1<0，

SKIPIF1<0設(shè)點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0代數(shù)式SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)有最大值，SKIPIF1<0的最大值為SKIPIF1<0．【點(diǎn)睛】本題是二次函數(shù)與一次函數(shù)的交點(diǎn)問題，考查了用待定系數(shù)法求二次函數(shù)的解析式，三角形相似的判定和性質(zhì)，解題的關(guān)鍵是構(gòu)造輔助線證SKIPIF1<0．3．在平面直角坐標(biāo)系SKIPIF1<0中，拋物線SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，與SKIPIF1<0軸交于SKIPIF1<0、SKIPIF1<0兩點(diǎn)SKIPIF1<0點(diǎn)SKIPIF1<0在點(diǎn)SKIPIF1<0的左側(cè)SKIPIF1<0，其中SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．(1)求拋物線的解析式；(2)線段SKIPIF1<0上有一動(dòng)點(diǎn)SKIPIF1<0，連接SKIPIF1<0，當(dāng)SKIPIF1<0SKIPIF1<0SKIPIF1<0的值最小時(shí)，請(qǐng)直接寫出此時(shí)點(diǎn)SKIPIF1<0的坐標(biāo)和SKIPIF1<0SKIPIF1<0SKIPIF1<0的最小值．(3)如圖2，點(diǎn)SKIPIF1<0為直線SKIPIF1<0上方拋物線上一點(diǎn)，連接SKIPIF1<0、SKIPIF1<0交于點(diǎn)SKIPIF1<0，連接SKIPIF1<0，記SKIPIF1<0的面積為SKIPIF1<0，SKIPIF1<0的面積為SKIPIF1<0，求SKIPIF1<0的最大值．【答案】(1)拋物線的解析式為：SKIPIF1<0(2)SKIPIF1<0，CSKIPIF1<0SKIPIF1<0SKIPIF1<0的最小值為SKIPIF1<0(3)最大值為SKIPIF1<0【分析】（1）根據(jù)點(diǎn)SKIPIF1<0的坐標(biāo)和SKIPIF1<0的值可得出點(diǎn)SKIPIF1<0的坐標(biāo)，將點(diǎn)SKIPIF1<0，SKIPIF1<0的坐標(biāo)代入拋物線，組成方程組，解之即可得出結(jié)論；（2）令SKIPIF1<0，可得點(diǎn)SKIPIF1<0的坐標(biāo)，由此可得SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0，則SKIPIF1<0SKIPIF1<0SKIPIF1<0，則SKIPIF1<0SKIPIF1<0SKIPIF1<0，作點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸的對(duì)稱點(diǎn)SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0SKIPIF1<0于點(diǎn)SKIPIF1<0，SKIPIF1<0SKIPIF1<0與SKIPIF1<0軸的交點(diǎn)即為所求點(diǎn)SKIPIF1<0，再根據(jù)直角三角形的三邊關(guān)系可得出結(jié)論；（3）過點(diǎn)SKIPIF1<0作SKIPIF1<0軸于點(diǎn)SKIPIF1<0，交SKIPIF1<0于點(diǎn)SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0軸交SKIPIF1<0的延長線于點(diǎn)SKIPIF1<0，由此可得SKIPIF1<0，則SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0的坐標(biāo)，表達(dá)SKIPIF1<0的長，再根據(jù)二次函數(shù)的性質(zhì)可得結(jié)論．【詳解】（1）解：∵SKIPIF1<0

∴SKIPIF1<0∵SKIPIF1<0

∴SKIPIF1<0，SKIPIF1<0將SKIPIF1<0、SKIPIF1<0的坐標(biāo)代入SKIPIF1<0得：SKIPIF1<0

∴SKIPIF1<0∴拋物線的解析式為：SKIPIF1<0；（2）解：由SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0作點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸的對(duì)稱點(diǎn)SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0SKIPIF1<0于點(diǎn)SKIPIF1<0，SKIPIF1<0SKIPIF1<0與SKIPIF1<0軸的交點(diǎn)即為所求點(diǎn)SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，綜上所述，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0SKIPIF1<0的最小值為SKIPIF1<0；（3）如圖，過SKIPIF1<0作SKIPIF1<0軸于點(diǎn)SKIPIF1<0，交SKIPIF1<0于SKIPIF1<0，過SKIPIF1<0作SKIPIF1<0軸交SKIPIF1<0延長線于SKIPIF1<0，設(shè)直線SKIPIF1<0解析式為：SKIPIF1<0，由(1)得：SKIPIF1<0，將SKIPIF1<0，SKIPIF1<0分別代入SKIPIF1<0得：SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0的表達(dá)式為：SKIPIF1<0，SKIPIF1<0SKIPIF1<0，故SKIPIF1<0的橫坐標(biāo)SKIPIF1<0，代入SKIPIF1<0，得：SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0軸于點(diǎn)SKIPIF1<0，SKIPIF1<0軸，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，將SKIPIF1<0、SKIPIF1<0分別看作SKIPIF1<0、SKIPIF1<0為底邊，則它們的高相同，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0有最大值，最大值為SKIPIF1<0【點(diǎn)睛】本題主要考查了二次函數(shù)綜合，待定系數(shù)法，解直角三角形，相似三角形的性質(zhì)與判定問題，解本題的關(guān)鍵是設(shè)出點(diǎn)SKIPIF1<0的橫坐標(biāo)，并正確表達(dá)面積的比值．題型三二次函數(shù)中面積最值問題1．如圖，拋物線SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0，與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，連接SKIPIF1<0，點(diǎn)SKIPIF1<0在拋物線上．(1)求拋物線的解析式；(2)如圖1，點(diǎn)D在第一象限內(nèi)的拋物線上，連接SKIPIF1<0，SKIPIF1<0，請(qǐng)求出SKIPIF1<0面積的最大值；(3)點(diǎn)SKIPIF1<0在拋物線上移動(dòng)，連接SKIPIF1<0，存在SKIPIF1<0，請(qǐng)直接寫出點(diǎn)SKIPIF1<0的坐標(biāo)．【答案】(1)SKIPIF1<0(2)4(3)點(diǎn)SKIPIF1<0的坐標(biāo)為：SKIPIF1<0或SKIPIF1<0．【分析】（1）由待定系數(shù)法即可求解；（2）由SKIPIF1<0面積SKIPIF1<0，即可求解；（3）當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0軸上方時(shí)，則點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0關(guān)于拋物線對(duì)稱軸對(duì)稱，即可求解；當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0軸下方時(shí)，由SKIPIF1<0，求出點(diǎn)SKIPIF1<0，即可求解．【詳解】（1）解：拋物線的表達(dá)式為：SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0，則拋物線的表達(dá)式為：SKIPIF1<0①；（2）解：過點(diǎn)SKIPIF1<0作SKIPIF1<0軸交SKIPIF1<0于點(diǎn)SKIPIF1<0，由點(diǎn)SKIPIF1<0、SKIPIF1<0的坐標(biāo)得，直線SKIPIF1<0的表達(dá)式為：SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，則點(diǎn)SKIPIF1<0，則SKIPIF1<0面積SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0面積有最大值，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0面積的最大值為4；（3）解：當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0軸上方時(shí)，SKIPIF1<0所以SKIPIF1<0平行于x軸則點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0關(guān)于拋物線對(duì)稱軸對(duì)稱，則點(diǎn)SKIPIF1<0；當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0軸下方時(shí)，設(shè)SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0，即點(diǎn)SKIPIF1<0，由點(diǎn)SKIPIF1<0、SKIPIF1<0的坐標(biāo)得，直線SKIPIF1<0的表達(dá)式為：SKIPIF1<0②，聯(lián)立①②得：SKIPIF1<0，解得：SKIPIF1<0（舍去）或SKIPIF1<0，即點(diǎn)SKIPIF1<0的坐標(biāo)為：SKIPIF1<0；綜上，點(diǎn)SKIPIF1<0的坐標(biāo)為：SKIPIF1<0或SKIPIF1<0．【點(diǎn)睛】本題考查的是二次函數(shù)綜合運(yùn)用，涉及到等腰三角形的性質(zhì)、面積的計(jì)算等，分類求解是解題的關(guān)鍵．2．如圖，在直角坐標(biāo)系中有一直角三角形SKIPIF1<0，SKIPIF1<0為坐標(biāo)原點(diǎn)，SKIPIF1<0，SKIPIF1<0，將此三角形繞原點(diǎn)SKIPIF1<0逆時(shí)針旋轉(zhuǎn)SKIPIF1<0，得到SKIPIF1<0，拋物線SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0．(1)求拋物線的解析式；(2)若點(diǎn)P是第二象限內(nèi)拋物線上的動(dòng)點(diǎn)，其橫坐標(biāo)為t，①是否存在一點(diǎn)P，使SKIPIF1<0的面積最大？若存在，求出SKIPIF1<0的面積的最大值；若不存在，請(qǐng)說明理由．②設(shè)拋物線對(duì)稱軸l與x軸交于一點(diǎn)E，連接SKIPIF1<0，交SKIPIF1<0于SKIPIF1<0，直接寫出當(dāng)SKIPIF1<0與SKIPIF1<0相似時(shí)，點(diǎn)P的坐標(biāo)．【答案】(1)SKIPIF1<0(2)①存在，最大值為SKIPIF1<0，理由見解析；②SKIPIF1<0或SKIPIF1<0【分析】（1）根據(jù)正切函數(shù)，可得SKIPIF1<0，根據(jù)旋轉(zhuǎn)的性質(zhì)可得SKIPIF1<0，據(jù)此求出A、B、C的坐標(biāo)，再利用待定系數(shù)法即可求出函數(shù)解析式；（2）①可求得直線SKIPIF1<0的解析式，過SKIPIF1<0作SKIPIF1<0軸于點(diǎn)SKIPIF1<0，交SKIPIF1<0于點(diǎn)SKIPIF1<0，可用SKIPIF1<0表示出SKIPIF1<0的長，當(dāng)SKIPIF1<0取最大值時(shí)，則SKIPIF1<0的面積最大，可求得其最大值；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0軸于SKIPIF1<0點(diǎn)，證明SKIPIF1<0，得到SKIPIF1<0，進(jìn)而推出SKIPIF1<0，則SKIPIF1<0，解方程即可；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)，SKIPIF1<0軸，則SKIPIF1<0．【詳解】（1）解：在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是由SKIPIF1<0繞點(diǎn)SKIPIF1<0逆時(shí)針旋轉(zhuǎn)SKIPIF1<0而得到的，SKIPIF1<0．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的坐標(biāo)分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，代入解析式得：SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0拋物線的解析式為SKIPIF1<0；（2）解：SKIPIF1<0存在點(diǎn)SKIPIF1<0使SKIPIF1<0的面積最大，SKIPIF1<0的面積有最大值為SKIPIF1<0理由如下：設(shè)直線SKIPIF1<0解析式為SKIPIF1<0，把SKIPIF1<0、SKIPIF1<0兩點(diǎn)坐標(biāo)代入可得：SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0解析式為SKIPIF1<0，如圖SKIPIF1<0，過SKIPIF1<0作SKIPIF1<0軸，交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，交直線SKIPIF1<0于點(diǎn)SKIPIF1<0，SKIPIF1<0點(diǎn)橫坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0點(diǎn)在第二象限，SKIPIF1<0點(diǎn)在SKIPIF1<0點(diǎn)上方，SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最大值，最大值為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0有最大值時(shí)，SKIPIF1<0的面積有最大值，SKIPIF1<0，綜上可知，存在點(diǎn)SKIPIF1<0使SKIPIF1<0的面積最大，SKIPIF1<0的面積有最大值為SKIPIF1<0；SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0軸于SKIPIF1<0點(diǎn)，∴SKIPIF1<0，又∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在第二象限，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，與SKIPIF1<0在二象限，橫坐標(biāo)小于SKIPIF1<0矛盾，舍去，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)，SKIPIF1<0軸，SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0與SKIPIF1<0相似時(shí)，SKIPIF1<0點(diǎn)的坐標(biāo)為SKIPIF1<0或SKIPIF1<0．【點(diǎn)睛】本題考查了二次函數(shù)綜合題，相似三角形的性質(zhì)與判定，一次函數(shù)與幾何綜合，解直角三角形，旋轉(zhuǎn)的性質(zhì)等等，解（1）的關(guān)鍵是利用旋轉(zhuǎn)的性質(zhì)得出SKIPIF1<0，SKIPIF1<0的長，又利用了待定系數(shù)法；解（2）的關(guān)鍵是利用相似三角形的性質(zhì)得出SKIPIF1<0．3．如圖，在平面直角坐標(biāo)系中，已知拋物線SKIPIF1<0與直線SKIPIF1<0相交于A，B兩點(diǎn)，其中SKIPIF1<0．(1)求該拋物線的函數(shù)解析式；(2)點(diǎn)P為直線SKIPIF1<0下方拋物線上的任意一點(diǎn)，連接SKIPIF1<0，求SKIPIF1<0面積的最大值；(3)若點(diǎn)M為拋物線對(duì)稱軸上的點(diǎn)，拋物線上是否存在點(diǎn)N，使得以A、B、M、N為頂點(diǎn)的四邊形是平行四邊形？如果存在，請(qǐng)求出點(diǎn)N的坐標(biāo)；如果不存在，請(qǐng)說明理由．【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)N的坐標(biāo)為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0【分析】（1）用待定系數(shù)法可得拋物線的函數(shù)解析式為SKIPIF1<0；（2）過P作SKIPIF1<0軸交SKIPIF1<0于Q，求出直線SKIPIF1<0解析式為SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0可得SKIPIF1<0，故SKIPIF1<0，根據(jù)二次函數(shù)性質(zhì)可得SKIPIF1<0面積的最大值為SKIPIF1<0；（3）求出拋物線的對(duì)稱軸為直線SKIPIF1<0，設(shè)SKIPIF1<0，分三種情況：①當(dāng)SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0的中點(diǎn)重合，SKIPIF1<0，②當(dāng)SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0，③當(dāng)SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0，分別解方程組可得答案．【詳解】（1）解：把SKIPIF1<0代入SKIPIF1<0得：SKIPIF1<0，解得SKIPIF1<0，∴拋物線的函數(shù)解析式為SKIPIF1<0；（2）解：過P作SKIPIF1<0軸交SKIPIF1<0于Q，如圖：由SKIPIF1<0得直線SKIPIF1<0解析式為SKIPIF1<0，設(shè)SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最大值SKIPIF1<0，SKIPIF1<0面積的最大值為SKIPIF1<0；（3）解：拋物線上存在點(diǎn)N，使得以A、B、M、N為頂點(diǎn)的四邊形是平行四邊形，理由如下：SKIPIF1<0，∴拋物線SKIPIF1<0的對(duì)稱軸為直線SKIPIF1<0，設(shè)SKIPIF1<0，又SKIPIF1<0，①當(dāng)SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0的中點(diǎn)重合，∴SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0；②當(dāng)SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0；③當(dāng)SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0；綜上所述，N的坐標(biāo)為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0．【點(diǎn)睛】本題考查二次函數(shù)的綜合應(yīng)用，涉及待定系數(shù)法，三角形面積，平行四邊形等知識(shí)，解題的關(guān)鍵是分類討論思想和方程思想的應(yīng)用．題型四二次函數(shù)平移、翻折、旋轉(zhuǎn)問題1．如圖1，拋物線SKIPIF1<0與SKIPIF1<0軸相交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，與SK