2024年中考數(shù)學(xué)二輪題型突破練習(xí)題型9 二次函數(shù)綜合題 類型10 二次函數(shù)與矩形有關(guān)的問題（專題訓(xùn)練）（教師版）

PAGE類型十二次函數(shù)與矩形有關(guān)的問題（專題訓(xùn)練）1．（2023·山東東營·統(tǒng)考中考真題）如圖，拋物線過點(diǎn)SKIPIF1<0，SKIPIF1<0，矩形SKIPIF1<0的邊SKIPIF1<0在線段SKIPIF1<0上（點(diǎn)B在點(diǎn)A的左側(cè)），點(diǎn)C，D在拋物線上，設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．

(1)求拋物線的函數(shù)表達(dá)式；(2)當(dāng)t為何值時(shí)，矩形SKIPIF1<0的周長有最大值？最大值是多少？(3)保持SKIPIF1<0時(shí)的矩形SKIPIF1<0不動(dòng)，向右平移拋物線，當(dāng)平移后的拋物線與矩形的邊有兩個(gè)交點(diǎn)G，H，且直線SKIPIF1<0平分矩形SKIPIF1<0的面積時(shí)，求拋物線平移的距離．【答案】(1)SKIPIF1<0；(2)當(dāng)SKIPIF1<0時(shí)，矩形SKIPIF1<0的周長有最大值，最大值為SKIPIF1<0；(3)4【分析】（1）設(shè)拋物線的函數(shù)表達(dá)式為SKIPIF1<0，求出點(diǎn)C的坐標(biāo)，將點(diǎn)C的坐標(biāo)代入即可求出該拋物線的函數(shù)表達(dá)式；（2）由拋物線的對(duì)稱性得SKIPIF1<0，則SKIPIF1<0，再得出SKIPIF1<0，根據(jù)矩形的周長公式，列出矩形周長的表達(dá)式，并將其化為頂點(diǎn)式，即可求解；（3）連接ASKIPIF1<0，SKIPIF1<0相交于點(diǎn)P，連接SKIPIF1<0，取SKIPIF1<0的中點(diǎn)Q，連接SKIPIF1<0，根據(jù)矩形的性質(zhì)和平移的性質(zhì)推出四邊形SKIPIF1<0是平行四邊形，則SKIPIF1<0，SKIPIF1<0．求出SKIPIF1<0時(shí)，點(diǎn)A的坐標(biāo)為SKIPIF1<0，則SKIPIF1<0，即可得出結(jié)論．【詳解】（1）解：設(shè)拋物線的函數(shù)表達(dá)式為SKIPIF1<0．∵當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴點(diǎn)C的坐標(biāo)為SKIPIF1<0．將點(diǎn)C坐標(biāo)代入表達(dá)式，得SKIPIF1<0，解得SKIPIF1<0．∴拋物線的函數(shù)表達(dá)式為SKIPIF1<0．（2）解：由拋物線的對(duì)稱性得：SKIPIF1<0，∴SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．∴矩形SKIPIF1<0的周長為SKIPIF1<0SKIPIF1<0SKIPIF1<0．∵SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，矩形SKIPIF1<0的周長有最大值，最大值為SKIPIF1<0．（3）解：連接SKIPIF1<0，SKIPIF1<0相交于點(diǎn)P，連接SKIPIF1<0，取SKIPIF1<0的中點(diǎn)Q，連接SKIPIF1<0．

∵直線SKIPIF1<0平分矩形SKIPIF1<0的面積，∴直線SKIPIF1<0過點(diǎn)P．．由平移的性質(zhì)可知，四邊形SKIPIF1<0是平行四邊形，∴SKIPIF1<0．∵四邊形SKIPIF1<0是矩形，∴P是SKIPIF1<0的中點(diǎn)．∴SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，點(diǎn)A的坐標(biāo)為SKIPIF1<0，∴SKIPIF1<0．∴拋物線平移的距離是4．【點(diǎn)睛】本題主要考查了求二次函數(shù)的解析式，二次函數(shù)的圖象和性質(zhì)，矩形的性質(zhì)，平移的性質(zhì)，解題的關(guān)鍵是掌握用待定系數(shù)法求解二次函數(shù)表達(dá)式的方法和步驟，二次函數(shù)圖象上點(diǎn)的坐標(biāo)特征，矩形的性質(zhì)，以及平移的性質(zhì)．2．（2023·山西·統(tǒng)考中考真題）如圖，二次函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸的正半軸交于點(diǎn)A，經(jīng)過點(diǎn)A的直線與該函數(shù)圖象交于點(diǎn)SKIPIF1<0，與SKIPIF1<0軸交于點(diǎn)C．

(1)求直線SKIPIF1<0的函數(shù)表達(dá)式及點(diǎn)C的坐標(biāo)；(2)點(diǎn)SKIPIF1<0是第一象限內(nèi)二次函數(shù)圖象上的一個(gè)動(dòng)點(diǎn)，過點(diǎn)SKIPIF1<0作直線SKIPIF1<0軸于點(diǎn)SKIPIF1<0，與直線SKIPIF1<0交于點(diǎn)D，設(shè)點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的值；②當(dāng)點(diǎn)SKIPIF1<0在直線SKIPIF1<0上方時(shí)，連接SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0軸于點(diǎn)SKIPIF1<0，SKIPIF1<0與SKIPIF1<0交于點(diǎn)SKIPIF1<0，連接SKIPIF1<0．設(shè)四邊形SKIPIF1<0的面積為SKIPIF1<0，求SKIPIF1<0關(guān)于SKIPIF1<0的函數(shù)表達(dá)式，并求出S的最大值．【答案】(1)SKIPIF1<0，點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0；(2)①2或3或SKIPIF1<0；②SKIPIF1<0，S的最大值為SKIPIF1<0【分析】（1）利用待定系數(shù)法可求得直線SKIPIF1<0的函數(shù)表達(dá)式，再求得點(diǎn)C的坐標(biāo)即可；（2）①分當(dāng)點(diǎn)SKIPIF1<0在直線SKIPIF1<0上方和點(diǎn)SKIPIF1<0在直線SKIPIF1<0下方時(shí)，兩種情況討論，根據(jù)SKIPIF1<0列一元二次方程求解即可；②證明SKIPIF1<0，推出SKIPIF1<0，再證明四邊形SKIPIF1<0為矩形，利用矩形面積公式得到二次函數(shù)的表達(dá)式，再利用二次函數(shù)的性質(zhì)即可求解．【詳解】（1）解：由SKIPIF1<0得，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．解得SKIPIF1<0．∵點(diǎn)A在SKIPIF1<0軸正半軸上．∴點(diǎn)A的坐標(biāo)為SKIPIF1<0．設(shè)直線SKIPIF1<0的函數(shù)表達(dá)式為SKIPIF1<0．將SKIPIF1<0兩點(diǎn)的坐標(biāo)SKIPIF1<0分別代入SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，∴直線SKIPIF1<0的函數(shù)表達(dá)式為SKIPIF1<0．將SKIPIF1<0代入SKIPIF1<0，得SKIPIF1<0．∴點(diǎn)C的坐標(biāo)為SKIPIF1<0；（2）①解：SKIPIF1<0點(diǎn)SKIPIF1<0在第一象限內(nèi)二次函數(shù)SKIPIF1<0的圖象上，且SKIPIF1<0軸于點(diǎn)SKIPIF1<0，與直線SKIPIF1<0交于點(diǎn)SKIPIF1<0，其橫坐標(biāo)為SKIPIF1<0．∴點(diǎn)SKIPIF1<0的坐標(biāo)分別為SKIPIF1<0．∴SKIPIF1<0．∵點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，∴SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0．如圖，當(dāng)點(diǎn)SKIPIF1<0在直線SKIPIF1<0上方時(shí)，SKIPIF1<0．

∵SKIPIF1<0，∴SKIPIF1<0．解得SKIPIF1<0．如圖2，當(dāng)點(diǎn)SKIPIF1<0在直線SKIPIF1<0下方時(shí)，SKIPIF1<0．

∵SKIPIF1<0，∴SKIPIF1<0．解得SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0．綜上所述，SKIPIF1<0的值為2或3或SKIPIF1<0；②解：如圖3，由（1）得，SKIPIF1<0．

∵SKIPIF1<0軸于點(diǎn)SKIPIF1<0，交SKIPIF1<0于點(diǎn)SKIPIF1<0，點(diǎn)B的坐標(biāo)為SKIPIF1<0，∴SKIPIF1<0．∵點(diǎn)SKIPIF1<0在直線SKIPIF1<0上方，∴SKIPIF1<0．∵SKIPIF1<0軸于點(diǎn)SKIPIF1<0，∴SKIPIF1<0．∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0．∴SKIPIF1<0．∴SKIPIF1<0．∴SKIPIF1<0．∴SKIPIF1<0．∴四邊形SKIPIF1<0為平行四邊形．∵SKIPIF1<0軸，∴四邊形SKIPIF1<0為矩形．∴SKIPIF1<0．即SKIPIF1<0．SKIPIF1<0∵SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，S的最大值為SKIPIF1<0．【點(diǎn)睛】本題屬于二次函數(shù)綜合題，考查了二次函數(shù)、一次函數(shù)、等腰三角形、矩形、勾股定理、相似三角形等知識(shí)點(diǎn)，第二問難度較大，需要分情況討論，畫出大致圖形，用含m的代數(shù)式表示出SKIPIF1<0是解題的關(guān)鍵．3．（2023·遼寧大連·統(tǒng)考中考真題）如圖，在平面直角坐標(biāo)系中，拋物線SKIPIF1<0上有兩點(diǎn)SKIPIF1<0，其中點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，拋物線SKIPIF1<0過點(diǎn)SKIPIF1<0．過SKIPIF1<0作SKIPIF1<0軸交拋物線SKIPIF1<0另一點(diǎn)為點(diǎn)SKIPIF1<0．以SKIPIF1<0長為邊向上構(gòu)造矩形SKIPIF1<0．

(1)求拋物線SKIPIF1<0的解析式；(2)將矩形SKIPIF1<0向左平移SKIPIF1<0個(gè)單位，向下平移SKIPIF1<0個(gè)單位得到矩形SKIPIF1<0，點(diǎn)SKIPIF1<0的對(duì)應(yīng)點(diǎn)SKIPIF1<0落在拋物線SKIPIF1<0上．①求SKIPIF1<0關(guān)于SKIPIF1<0的函數(shù)關(guān)系式，并直接寫出自變量SKIPIF1<0的取值范圍；②直線SKIPIF1<0交拋物線SKIPIF1<0于點(diǎn)SKIPIF1<0，交拋物線SKIPIF1<0于點(diǎn)SKIPIF1<0．當(dāng)點(diǎn)SKIPIF1<0為線段SKIPIF1<0的中點(diǎn)時(shí)，求SKIPIF1<0的值；③拋物線SKIPIF1<0與邊SKIPIF1<0分別相交于點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0在拋物線SKIPIF1<0的對(duì)稱軸同側(cè)，當(dāng)SKIPIF1<0時(shí)，求點(diǎn)SKIPIF1<0的坐標(biāo)．【答案】(1)SKIPIF1<0；(2)①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0或SKIPIF1<0【分析】（1）根據(jù)題意得出點(diǎn)SKIPIF1<0，SKIPIF1<0,待定系數(shù)法求解析式即可求解；（2）①根據(jù)平移的性質(zhì)得出SKIPIF1<0，根據(jù)點(diǎn)SKIPIF1<0的對(duì)應(yīng)點(diǎn)SKIPIF1<0落在拋物線SKIPIF1<0上，可得SKIPIF1<0，進(jìn)而即可求解；②根據(jù)題意得出SKIPIF1<0，求得中點(diǎn)坐標(biāo)，根據(jù)題意即可求解；③連接SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0于點(diǎn)SKIPIF1<0，勾股定理求得SKIPIF1<0，設(shè)SKIPIF1<0點(diǎn)的坐標(biāo)為SKIPIF1<0，則SKIPIF1<0，將SKIPIF1<0代入SKIPIF1<0，求得SKIPIF1<0，求得SKIPIF1<0，進(jìn)而根據(jù)SKIPIF1<0落在拋物線SKIPIF1<0上，將SKIPIF1<0代入SKIPIF1<0，即可求解．【詳解】（1）解：依題意，點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，代入拋物線SKIPIF1<0∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0,將點(diǎn)SKIPIF1<0，SKIPIF1<0,代入拋物線SKIPIF1<0，∴SKIPIF1<0解得：SKIPIF1<0∴拋物線SKIPIF1<0的解析式為SKIPIF1<0；（2）①解：∵SKIPIF1<0軸交拋物線SKIPIF1<0另一點(diǎn)為點(diǎn)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0，∵矩形SKIPIF1<0向左平移SKIPIF1<0個(gè)單位，向下平移SKIPIF1<0個(gè)單位得到矩形SKIPIF1<0，點(diǎn)SKIPIF1<0的對(duì)應(yīng)點(diǎn)SKIPIF1<0落在拋物線SKIPIF1<0上∴SKIPIF1<0，SKIPIF1<0整理得SKIPIF1<0∵SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0；②如圖所示，

∵SKIPIF1<0,SKIPIF1<0∴SKIPIF1<0,∵SKIPIF1<0∴SKIPIF1<0，由①可得SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0,SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，分別代入SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0的中點(diǎn)坐標(biāo)為SKIPIF1<0∵點(diǎn)SKIPIF1<0為線段SKIPIF1<0的中點(diǎn)，∴SKIPIF1<0解得：SKIPIF1<0或SKIPIF1<0（大于4，舍去）③如圖所示，連接SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0于點(diǎn)SKIPIF1<0，

則SKIPIF1<0，∵SKIPIF1<0∴SKIPIF1<0，設(shè)SKIPIF1<0點(diǎn)的坐標(biāo)為SKIPIF1<0，則SKIPIF1<0，將SKIPIF1<0代入SKIPIF1<0，SKIPIF1<0，解得：SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，將SKIPIF1<0代入SKIPIF1<0解得：SKIPIF1<0，∴SKIPIF1<0或SKIPIF1<0．【點(diǎn)睛】本題考查了二次函數(shù)綜合運(yùn)用，矩形的性質(zhì)，平移的性質(zhì)，熟練掌握二次函數(shù)的性質(zhì)是解題的關(guān)鍵．4.（2022·安徽）如圖1，隧道截面由拋物線的一部分AED和矩形ABCD構(gòu)成，矩形的一邊BC為12米，另一邊AB為2米．以BC所在的直線為x軸，線段BC的垂直平分線為y軸，建立平面直角坐標(biāo)系xOy，規(guī)定一個(gè)單位長度代表1米．E（0，8）是拋物線的頂點(diǎn)．(1)求此拋物線對(duì)應(yīng)的函數(shù)表達(dá)式；(2)在隧道截面內(nèi)（含邊界）修建“”型或“”型柵欄，如圖2、圖3中粗線段所示，點(diǎn)SKIPIF1<0，SKIPIF1<0在x軸上，MN與矩形SKIPIF1<0的一邊平行且相等．柵欄總長l為圖中粗線段SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，MN長度之和．請解決以下問題：（?。┬藿ㄒ粋€(gè)“”型柵欄，如圖2，點(diǎn)SKIPIF1<0，SKIPIF1<0在拋物線AED上．設(shè)點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，求柵欄總長l與m之間的函數(shù)表達(dá)式和l的最大值；（ⅱ）現(xiàn)修建一個(gè)總長為18的柵欄，有如圖3所示的修建“”型或“”型柵型兩種設(shè)計(jì)方案，請你從中選擇一種，求出該方案下矩形SKIPIF1<0面積的最大值，及取最大值時(shí)點(diǎn)SKIPIF1<0的橫坐標(biāo)的取值范圍（SKIPIF1<0在SKIPIF1<0右側(cè)）．【答案】(1)y＝SKIPIF1<0x2＋8(2)（?。﹍＝SKIPIF1<0m2＋2m＋24，l的最大值為26；（ⅱ）方案一：SKIPIF1<0＋9≤P1橫坐標(biāo)≤SKIPIF1<0；方案二：SKIPIF1<0＋SKIPIF1<0≤P1橫坐標(biāo)≤SKIPIF1<0【分析】（1）通過分析A點(diǎn)坐標(biāo)，利用待定系數(shù)法求函數(shù)解析式；（2）（?。┙Y(jié)合矩形性質(zhì)分析得出P2的坐標(biāo)為（m，－SKIPIF1<0m2＋8），然后列出函數(shù)關(guān)系式，利用二次函數(shù)的性質(zhì)分析最值；（ⅱ）設(shè)P2P1＝n，分別表示出方案一和方案二的矩形面積，利用二次函數(shù)的性質(zhì)分析最值，從而利用數(shù)形結(jié)合思想確定取值范圍．(1)由題意可得：A（－6，2），D（6，2），又∵E（0，8）是拋物線的頂點(diǎn)，設(shè)拋物線對(duì)應(yīng)的函數(shù)表達(dá)式為y＝ax2＋8，將A（－6，2）代入，（－6）2a＋8＝2，解得：a＝SKIPIF1<0，∴拋物線對(duì)應(yīng)的函數(shù)表達(dá)式為y＝SKIPIF1<0x2＋8；(2)（?。唿c(diǎn)P1的橫坐標(biāo)為m（0＜m≤6），且四邊形P1P2P3P4為矩形，點(diǎn)P2，P3在拋物線AED上，∴P2的坐標(biāo)為（m，SKIPIF1<0m2＋8），∴P1P2＝P3P4＝MN＝SKIPIF1<0m2＋8，P2P3＝2m，∴l(xiāng)＝3（SKIPIF1<0m2＋8）＋2m＝SKIPIF1<0m2＋2m＋24＝SKIPIF1<0（m－2）2＋26，∵SKIPIF1<0＜0，∴當(dāng)m＝2時(shí)，l有最大值為26，即柵欄總長l與m之間的函數(shù)表達(dá)式為l＝SKIPIF1<0m2＋2m＋24，l的最大值為26；（ⅱ）方案一：設(shè)P2P1＝n，則P2P3＝18－3n，∴矩形P1P2P3P4面積為（18－3n）n＝－3n2＋18n＝－3（n－3）2＋27，∵－3＜0，∴當(dāng)n＝3時(shí)，矩形面積有最大值為27，此時(shí)P2P1＝3，P2P3＝9，令SKIPIF1<0x2＋8＝3，解得：x＝SKIPIF1<0，∴此時(shí)P1的橫坐標(biāo)的取值范圍為SKIPIF1<0＋9≤P1橫坐標(biāo)≤SKIPIF1<0，方案二：設(shè)P2P1＝n，則P2P3＝9－n，∴矩形P1P2P3P4面積為（9－n）n＝－n2＋9n＝－（n－SKIPIF1<0）2＋SKIPIF1<0，∵－1＜0，∴當(dāng)n＝SKIPIF1<0時(shí)，矩形面積有最大值為SKIPIF1<0，此時(shí)P2P1＝SKIPIF1<0，P2P3＝SKIPIF1<0，令SKIPIF1<0x2＋8＝SKIPIF1<0，解得：x＝SKIPIF1<0，∴此時(shí)P1的橫坐標(biāo)的取值范圍為SKIPIF1<0＋SKIPIF1<0≤P1橫坐標(biāo)≤SKIPIF1<0．【點(diǎn)睛】本題考查二次函數(shù)的應(yīng)用，掌握待定系數(shù)法求函數(shù)解析式，準(zhǔn)確識(shí)圖，確定關(guān)鍵點(diǎn)的坐標(biāo)，利用數(shù)形結(jié)合思想解題是關(guān)鍵．5.（2021·四川中考真題）如圖，在平面直角坐標(biāo)系中，拋物線SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0和SKIPIF1<0，交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，拋物線的對(duì)稱軸交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，交拋物線于點(diǎn)SKIPIF1<0．

（1）求拋物線的解析式；（2）將線段SKIPIF1<0繞著點(diǎn)SKIPIF1<0沿順時(shí)針方向旋轉(zhuǎn)得到線段SKIPIF1<0，旋轉(zhuǎn)角為SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0的最小值．（3）SKIPIF1<0為平面直角坐標(biāo)系中一點(diǎn)，在拋物線上是否存在一點(diǎn)SKIPIF1<0，使得以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為頂點(diǎn)的四邊形為矩形？若存在，請直接寫出點(diǎn)SKIPIF1<0的橫坐標(biāo)；若不存在，請說明理由；【答案】（1）SKIPIF1<0；（2）SKIPIF1<0；（3）存在，SKIPIF1<0點(diǎn)的橫坐標(biāo)分別為：2，SKIPIF1<0，SKIPIF1<0或SKIPIF1<0．【分析】（1）待定系數(shù)法求二次函數(shù)解析式，設(shè)解析式為SKIPIF1<0將SKIPIF1<0，SKIPIF1<0兩點(diǎn)代入求得SKIPIF1<0，c的值即可；（2）胡不歸問題，要求SKIPIF1<0的值，將折線化為直線，構(gòu)造相似三角形將SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，再利用三角形兩邊之和大于第三邊求得SKIPIF1<0最值；（3）分2種情形討論：①AB為矩形的一條邊，利用等腰直角三角形三角形的性質(zhì)可以求得N點(diǎn)的坐標(biāo);②AB為矩形的對(duì)角線，設(shè)R為AB的中點(diǎn),RN=SKIPIF1<0AB,利用兩點(diǎn)距離公式求解方程可得N點(diǎn)的坐標(biāo)．【詳解】解：（1）∵SKIPIF1<0過SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0∴拋物線的解析式為：SKIPIF1<0（2）在SKIPIF1<0上取一點(diǎn)SKIPIF1<0，使得SKIPIF1<0，連接SKIPIF1<0，SKIPIF1<0∵SKIPIF1<0對(duì)稱軸SKIPIF1<0．∴SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0三點(diǎn)在同一點(diǎn)直線上時(shí)，SKIPIF1<0最小為SKIPIF1<0．在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0即SKIPIF1<0最小值為SKIPIF1<0．（3）情形①如圖，AB為矩形的一條邊時(shí)，聯(lián)立SKIPIF1<0得SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0是等腰SKIPIF1<0，SKIPIF1<0分別過SKIPIF1<0兩點(diǎn)作SKIPIF1<0的垂線，交SKIPIF1<0于點(diǎn)SKIPIF1<0，過SKIPIF1<0作SKIPIF1<0軸，SKIPIF1<0SKIPIF1<0軸,SKIPIF1<0SKIPIF1<0SKIPIF1<0,SKIPIF1<0也是等腰直角三角形設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0代入SKIPIF1<0，解得SKIPIF1<0,SKIPIF1<0（不符題意，舍）SKIPIF1<0SKIPIF1<0同理，設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0代入SKIPIF1<0，解得SKIPIF1<0,SKIPIF1<0（不符題意，舍）SKIPIF1<0②AB為矩形的對(duì)角線，設(shè)R為AB的中點(diǎn),則SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0整理得：SKIPIF1<0解得：SKIPIF1<0（不符題意，舍），SKIPIF1<0（不符題意，舍），SKIPIF1<0，SKIPIF1<0SKIPIF1<0綜上所述：SKIPIF1<0點(diǎn)的橫坐標(biāo)分別為：2，SKIPIF1<0，SKIPIF1<0或SKIPIF1<0．【點(diǎn)睛】本題考查了二次函數(shù)的性質(zhì)，待定系數(shù)法求解析式，三角形相似，勾股定理，二次函數(shù)與一次函數(shù)交點(diǎn)，矩形的性質(zhì)，等腰直角三角形性質(zhì)，平面直角坐標(biāo)系中兩點(diǎn)距離計(jì)算等知識(shí)，能正確做出輔助線，找到相似三角形是解題的關(guān)鍵．6.（2021·甘肅中考真題）如圖，在平面直角坐標(biāo)系中，拋物線SKIPIF1<0與坐標(biāo)軸交于SKIPIF1<0兩點(diǎn)，直線SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0．點(diǎn)SKIPIF1<0為直線SKIPIF1<0下方拋物線上一動(dòng)點(diǎn)，過點(diǎn)SKIPIF1<0作SKIPIF1<0軸的垂線，垂足為SKIPIF1<0分別交直線SKIPIF1<0于點(diǎn)SKIPIF1<0．

（1）求拋物線SKIPIF1<0的表達(dá)式；（2）當(dāng)SKIPIF1<0，連接SKIPIF1<0，求SKIPIF1<0的面積；（3）①SKIPIF1<0是SKIPIF1<0軸上一點(diǎn)，當(dāng)四邊形SKIPIF1<0是矩形時(shí)，求點(diǎn)SKIPIF1<0的坐標(biāo)；②在①的條件下，第一象限有一動(dòng)點(diǎn)SKIPIF1<0，滿足SKIPIF1<0，求SKIPIF1<0周長的最小值．【答案】（1）SKIPIF1<0；（2）SKIPIF1<0；（3）①SKIPIF1<0；②SKIPIF1<0【分析】（1）直接利用待定系數(shù)法即可求出答案．（2）由題意可求出SKIPIF1<0，SKIPIF1<0．利用三角函數(shù)可知在SKIPIF1<0和SKIPIF1<0中，SKIPIF1<0，由此即可求出SKIPIF1<0，從而可求出SKIPIF1<0．即可求出D點(diǎn)坐標(biāo)，繼而求出SKIPIF1<0．再根據(jù)SKIPIF1<0，即可求出FD的長，最后利用三角形面積公式即可求出最后答案．（3）①連接SKIPIF1<0，交SKIPIF1<0于點(diǎn)SKIPIF1<0．根據(jù)矩形的性質(zhì)可知SKIPIF1<0，SKIPIF1<0．由SKIPIF1<0可推出SKIPIF1<0．由SKIPIF1<0，可推出SKIPIF1<0．再根據(jù)直線BC的解析式可求出C點(diǎn)坐標(biāo)，即可得出OC的長，由此可求出AC的長，即可求出CH的長，最后即得出OH的長，即可得出H點(diǎn)坐標(biāo)．②在SKIPIF1<0中，利用勾股定理可求出SKIPIF1<0的長，再根據(jù)SKIPIF1<0結(jié)合SKIPIF1<0可推出SKIPIF1<0，即要使SKIPIF1<0最小，就要SKIPIF1<0最小，由題意可知當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)，SKIPIF1<0為最?。辞蟪鯞C長即可．在SKIPIF1<0中，利用勾股定理求出SKIPIF1<0的長，即得出SKIPIF1<0周長的最小值為SKIPIF1<0．【詳解】解：（1）∵拋物線SKIPIF1<0過SKIPIF1<0兩點(diǎn)，SKIPIF1<0，解得，SKIPIF1<0，SKIPIF1<0．（2）SKIPIF1<0SKIPIF1<0．同理，SKIPIF1<0．又SKIPIF1<0軸，SKIPIF1<0軸，∴在SKIPIF1<0和SKIPIF1<0中，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0．SKIPIF1<0，SKIPIF1<0．（3）①如圖，連接SKIPIF1<0，交SKIPIF1<0于點(diǎn)SKIPIF1<0．∵四邊形SKIPIF1<0是矩形，SKIPIF1<0．又SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0．∵四邊形SKIPIF1<0是矩形，SKIPIF1<0．SKIPIF1<0，∵當(dāng)x=0時(shí)，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．②在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0．SKIPIF1<0∴要使SKIPIF1<0最小，就要SKIPIF1<0最?。甋KIPIF1<0，∴當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)，SKIPIF1<0為最?。赟KIPIF1<0中，SKIPIF1<0．SKIPIF1<0周長的最小值是SKIPIF1<0．

【點(diǎn)睛】本題為二次函數(shù)綜合題．考查二次函數(shù)的圖象和性質(zhì)，解直角三角形，一次函數(shù)的圖象和性質(zhì)，矩形的性質(zhì)，平行線分線段成比例，三角形三邊關(guān)系以及勾股定理等知識(shí)，綜合性強(qiáng)，較難．利用數(shù)形結(jié)合的思想是解答本題的關(guān)鍵．7.（2021·山東中考真題）如圖，在平面直角坐標(biāo)系中，已知拋物線SKIPIF1<0交SKIPIF1<0軸于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，交SKIPIF1<0軸于點(diǎn)SKIPIF1<0．（1）求該拋物線的表達(dá)式；（2）點(diǎn)SKIPIF1<0為第四象限內(nèi)拋物線上一點(diǎn)，連接SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，連接SKIPIF1<0，求SKIPIF1<0面積的最大值及此時(shí)點(diǎn)SKIPIF1<0的坐標(biāo)；（3）在（2）的條件下，將拋物線SKIPIF1<0向右平移經(jīng)過點(diǎn)SKIPIF1<0時(shí)，得到新拋物線SKIPIF1<0，點(diǎn)SKIPIF1<0在新拋物線的對(duì)稱軸上，在坐標(biāo)平面內(nèi)是否存在一點(diǎn)SKIPIF1<0，使得以SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0為頂點(diǎn)的四邊形為矩形，若存在，請直接寫出點(diǎn)SKIPIF1<0的坐標(biāo)；若不存在，請說明理由．參考：若點(diǎn)SKIPIF1<0、SKIPIF1<0，則線段SKIPIF1<0的中點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0．【答案】（1）該拋物線的表達(dá)式為：SKIPIF1<0；（2）SKIPIF1<0面積最大值為8，此時(shí)P點(diǎn)的坐標(biāo)為：P（2，-6）；（3）SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0【分析】（1）將兩個(gè)點(diǎn)分別代入拋物線可得關(guān)于a，b的二元一次方程組，可解得a，b；（2）設(shè)出P、Q兩點(diǎn)坐標(biāo)，應(yīng)用三角形相似，及三角形面積公式，代入化簡可得一個(gè)二次函數(shù)，求其最大值即可；（3）拋物線的平移可確定拋物線解析式及對(duì)稱軸，設(shè)出點(diǎn)E、F，應(yīng)用中點(diǎn)坐標(biāo)公式及矩形特點(diǎn)分成的三角形為直角三角形，可得出答案．【詳解】解：（1）將A（-1,0），B（4,0）代入拋物線SKIPIF1<0可得：SKIPIF1<0，解得：SKIPIF1<0，∴該拋物線的表達(dá)式為：SKIPIF1<0；（2）過點(diǎn)P作PN⊥x軸于點(diǎn)N，如圖所示：設(shè)SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵點(diǎn)SKIPIF1<0在拋物線上，∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，根據(jù)拋物線的基本性質(zhì)：對(duì)稱軸為SKIPIF1<0在SKIPIF1<0內(nèi)，∴SKIPIF1<0在SKIPIF1<0取得最大值，代入得：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0面積的最大值為8，此時(shí)點(diǎn)P的坐標(biāo)為：SKIPIF1<0．（3）在（2）的條件下，原拋物線解析式為SKIPIF1<0，將拋物線向右平移經(jīng)過點(diǎn)SKIPIF1<0，可知拋物線向右平移了SKIPIF1<0個(gè)單位長度，∴可得：SKIPIF1<0，化簡得平移后的拋物線：SKIPIF1<0，對(duì)稱軸為：SKIPIF1<0，由（2）得：A（-1，0），SKIPIF1<0，點(diǎn)E在對(duì)稱軸上，∴設(shè)E（3，e），點(diǎn)F（m，n），矩形AEPF，當(dāng)以AP為矩形的對(duì)角線時(shí)，則AP的中點(diǎn)坐標(biāo)為：SKIPIF1<0，EF的中點(diǎn)坐標(biāo)為：SKIPIF1<0，根據(jù)矩形的性質(zhì)可得，兩個(gè)中點(diǎn)坐標(biāo)相同，可得：SKIPIF1<0解得：SKIPIF1<0∵矩形AEPF，∴SKIPIF1<0為直角三角形，∴SKIPIF1<0，③SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，代入③化簡可得：SKIPIF1<0，④∴將②代入④可得：SKIPIF1<0，化簡得：SKIPIF1<0，根據(jù)判別式得：SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0或SKIPIF1<0；當(dāng)以AP為矩形的邊時(shí)，如圖所示：過點(diǎn)P分別作PG⊥x軸于點(diǎn)G，PH∥x軸，過點(diǎn)F作PH的垂線，垂足為H，設(shè)拋物線的對(duì)稱軸與x軸的交點(diǎn)為M，如圖，∴SKIPIF1<0，SKIPIF1<0，AM=4，∴SKIPIF1<0，∵四邊形SKIPIF1<0是矩形，∴SKIPIF1<0，AE=PF，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴FH=2，∵點(diǎn)SKIPIF1<0，∴SKIPIF1<0，當(dāng)以AP為矩形的邊時(shí)，如圖所示：同理可得SKIPIF1<0；綜上所述：以SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0為頂點(diǎn)的四邊形為矩形，SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0【點(diǎn)睛】題目考查確定二次函數(shù)解析式及其基本性質(zhì)、矩形的性質(zhì)、勾股定理等，難點(diǎn)主要是依據(jù)圖像確定各點(diǎn)、線段間的關(guān)系，得出答案．8.（2021·黑龍江中考真題）綜合與探究如圖，在平面直角坐標(biāo)系中，拋物線SKIPIF1<0與x軸交于點(diǎn)A、B，與y軸交于點(diǎn)C，連接BC，SKIPIF1<0，對(duì)稱軸為SKIPIF1<0，點(diǎn)D為此拋物線的頂點(diǎn)．（1）求拋物線的解析式；（2）拋物線上C，D兩點(diǎn)之間的距離是__________；（3）點(diǎn)E是第一象限內(nèi)拋物線上的動(dòng)點(diǎn)，連接BE和CE．求SKIPIF1<0面積的最大值；（4）點(diǎn)P在拋物線對(duì)稱軸上，平面內(nèi)存在點(diǎn)Q，使以點(diǎn)B、C、P、Q為頂點(diǎn)的四邊形為矩形，請直接寫出點(diǎn)Q的坐標(biāo)．【答案】（1）SKIPIF1<0；（2）SKIPIF1<0；（3）SKIPIF1<0；（4）SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0．【分析】（1）先根據(jù)對(duì)稱軸可得SKIPIF1<0的值，再根據(jù)SKIPIF1<0可得點(diǎn)SKIPIF1<0的坐標(biāo)，代入拋物線的解析式即可得；（2）利用拋物線的解析式分別求出點(diǎn)SKIPIF1<0的坐標(biāo)，再利用兩點(diǎn)之間的距離公式即可得；（3）過點(diǎn)SKIPIF1<0作SKIPIF1<0軸的垂線，交SKIPIF1<0于點(diǎn)SKIPIF1<0，先利用待定系數(shù)法求出直線SKIPIF1<0的解析式，再設(shè)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，從而可得SKIPIF1<0和SKIPIF1<0的坐標(biāo)，然后根據(jù)SKIPIF1<0可得SKIPIF1<0關(guān)于SKIPIF1<0的函數(shù)關(guān)系式，利用二次函數(shù)的性質(zhì)求解即可得；（4）設(shè)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，分①當(dāng)SKIPIF1<0為矩形SKIPIF1<0的邊時(shí)，②當(dāng)SKIPIF1<0為矩形SKIPIF1<0的邊時(shí)，③當(dāng)SKIPIF1<0為矩形SKIPIF1<0的對(duì)角線時(shí)三種情況，再分別利用待定系數(shù)法求直線的解析式、矩形的性質(zhì)、點(diǎn)坐標(biāo)的平移變換規(guī)律求解即可得．【詳解】解：（1）SKIPIF1<0拋物線SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且點(diǎn)SKIPIF1<0在SKIPIF1<0軸負(fù)半軸上，SKIPIF1<0，將點(diǎn)SKIPIF1<0代入SKIPIF1<0得：SKIPIF1<0，解得SKIPIF1<0，則拋物線的解析式為SKIPIF1<0；（2）SKIPIF1<0化成頂點(diǎn)式為SKIPIF1<0，則頂點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，則拋物線上SKIPIF1<0兩點(diǎn)之間的距離是SKIPIF1<0，故答案為：SKIPIF1<0；（3）如圖，過點(diǎn)SKIPIF1<0作SKIPIF1<0軸的垂線，交SKIPIF1<0于點(diǎn)SKIPIF1<0，SKIPIF1<0，拋物線的對(duì)稱軸為SKIPIF1<0，SKIPIF1<0，設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0，將點(diǎn)SKIPIF1<0代入得：SKIPIF1<0，解得SKIPIF1<0，則直線SKIPIF1<0的解析式為SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由二次函數(shù)的性質(zhì)得：在SKIPIF1<0內(nèi)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最大值，最大值為SKIPIF1<0，即SKIPIF1<0面積的最大值為SKIPIF1<0；（4）設(shè)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，由題意，分以下三種情況：①當(dāng)SKIPIF1<0為矩形SKIPIF1<0的邊時(shí)，則SKIPIF1<0，設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0，將點(diǎn)SKIPIF1<0代入得：SKIPIF1<0，則直線SKIPIF1<0的解析式為SKIPIF1<0，將點(diǎn)SKIPIF1<0代入得：SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0將點(diǎn)SKIPIF1<0先向右平移2個(gè)單位長度，再向上平移4個(gè)單位長度可得到點(diǎn)SKIPIF1<0，SKIPIF1<0四邊形SKIPIF1<0是矩形，SKIPIF1<0點(diǎn)SKIPIF1<0平移至點(diǎn)SKIPIF1<0的方式與點(diǎn)SKIPIF1<0平移至點(diǎn)SKIPIF1<0的方式相同，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0；②當(dāng)SKIPIF1<0為矩形SKIPIF1<0的邊時(shí)，則SKIPIF1<0，同（4）①的方法可得：點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0；③當(dāng)SKIPIF1<0為矩形SKIPIF1<0的對(duì)角線時(shí)，則SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，當(dāng)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0時(shí)，則將點(diǎn)SKIPIF1<0先向左平移2個(gè)單位長度，再向下平移SKIPIF1<0個(gè)單位長度可得到點(diǎn)SKIPIF1<0，SKIPIF1<0四邊形SKIPIF1<0是矩形，SKIPIF1<0點(diǎn)SKIPIF1<0