2024年中考數(shù)學(xué)壓軸題型（江蘇專用）專題06 二次函數(shù)與幾何結(jié)合（解答壓軸題） （含解析）

PAGE專題06二次函數(shù)與幾何結(jié)合（解答壓軸題）通用的解題思路：1.理解題意：首先，仔細(xì)閱讀題目，理解題目中給出的二次函數(shù)表達(dá)式和幾何圖形的性質(zhì)。確定二次函數(shù)的一般形式，并理解a、b、c的值對(duì)函數(shù)圖像的影響。2.分析函數(shù)圖像：根據(jù)二次函數(shù)的系數(shù)，判斷函數(shù)的開(kāi)口方向。計(jì)算對(duì)稱軸。計(jì)算頂點(diǎn)坐標(biāo)。確定函數(shù)與坐標(biāo)軸的交點(diǎn)。3.應(yīng)用幾何知識(shí)：根據(jù)題目要求，利用幾何知識(shí)分析函數(shù)圖像與坐標(biāo)軸、對(duì)稱軸、頂點(diǎn)等的關(guān)系?？赡苄枰?jì)算線段長(zhǎng)度、角度、面積等。4.建立方程或不等式：根據(jù)幾何關(guān)系，建立關(guān)于x或y的方程或不等式。利用二次函數(shù)的性質(zhì)，如對(duì)稱性、最值等，進(jìn)一步簡(jiǎn)化方程或不等式。5.求解方程或不等式：使用代數(shù)方法，如配方法、公式法、因式分解法等，求解方程或不等式。驗(yàn)證解的合理性，確保解符合題目要求和二次函數(shù)的取值范圍。6.得出結(jié)論：根據(jù)求解結(jié)果，結(jié)合幾何知識(shí)，得出最終結(jié)論。檢查答案是否符合題目要求，確保沒(méi)有遺漏或錯(cuò)誤。

(1)SKIPIF1<0_______；(2)D是第三象限拋物線上的一點(diǎn)，連接OD，SKIPIF1<0；將原拋物線向左平移，使得平移后的拋物線經(jīng)過(guò)點(diǎn)D，過(guò)點(diǎn)SKIPIF1<0作x軸的垂線l．已知在l的左側(cè)，平移前后的兩條拋物線都下降，求k的取值范圍；(3)將原拋物線平移，平移后的拋物線與原拋物線的對(duì)稱軸相交于點(diǎn)Q，且其頂點(diǎn)P落在原拋物線上，連接PC、QC、PQ．已知SKIPIF1<0是直角三角形，求點(diǎn)P的坐標(biāo)．【答案】(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0或SKIPIF1<0．【分析】（1）把SKIPIF1<0代入SKIPIF1<0即可求解；（2）過(guò)點(diǎn)D作DM⊥OA于點(diǎn)M，設(shè)SKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0，進(jìn)而求得平移后得拋物線，平移后得拋物線為SKIPIF1<0，根據(jù)二次函數(shù)得性質(zhì)即可得解；（3）先設(shè)出平移后頂點(diǎn)為SKIPIF1<0，根據(jù)原拋物線SKIPIF1<0，求得原拋物線的頂點(diǎn)SKIPIF1<0，對(duì)稱軸為x=1，進(jìn)而得SKIPIF1<0，再根據(jù)勾股定理構(gòu)造方程即可得解．【詳解】（1）解：把SKIPIF1<0代入SKIPIF1<0得，SKIPIF1<0，解得SKIPIF1<0，故答案為SKIPIF1<0；（2）解：過(guò)點(diǎn)D作DM⊥OA于點(diǎn)M，

∵SKIPIF1<0，∴二次函數(shù)的解析式為SKIPIF1<0設(shè)SKIPIF1<0，∵D是第三象限拋物線上的一點(diǎn)，連接OD，SKIPIF1<0，∴SKIPIF1<0，解得m=SKIPIF1<0或m=8（舍去），當(dāng)m=SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴設(shè)將原拋物線向左平移后的拋物線為SKIPIF1<0，把SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0，解得a=3或a=SKIPIF1<0（舍去），∴平移后得拋物線為SKIPIF1<0∵過(guò)點(diǎn)SKIPIF1<0作x軸的垂線l．已知在l的左側(cè)，平移前后的兩條拋物線都下降，在SKIPIF1<0的對(duì)稱軸x=SKIPIF1<0的左側(cè)，y隨x的增大而減小，此時(shí)原拋物線也是y隨x的增大而減小，∴SKIPIF1<0；（3）解：由SKIPIF1<0，設(shè)平移后的拋物線為SKIPIF1<0，則頂點(diǎn)為SKIPIF1<0，∵頂點(diǎn)為SKIPIF1<0在SKIPIF1<0上，∴SKIPIF1<0，∴平移后的拋物線為SKIPIF1<0，頂點(diǎn)為SKIPIF1<0，∵原拋物線SKIPIF1<0，∴原拋物線的頂點(diǎn)SKIPIF1<0，對(duì)稱軸為x=1，∵平移后的拋物線與原拋物線的對(duì)稱軸相交于點(diǎn)Q，∴SKIPIF1<0，∵點(diǎn)Q、C在直線x=1上，平移后的拋物線頂點(diǎn)P在原拋物線頂點(diǎn)C的上方，兩拋物線的交點(diǎn)Q在頂點(diǎn)P的上方，∴∠PCQ與∠CQP都是銳角，∵SKIPIF1<0是直角三角形，∴∠CPQ=90°，∴SKIPIF1<0，∴SKIPIF1<0化簡(jiǎn)得SKIPIF1<0，∴p=1（舍去），或p=3或p=SKIPIF1<0，當(dāng)p=3時(shí)，SKIPIF1<0，當(dāng)p=SKIPIF1<0時(shí)，SKIPIF1<0，∴點(diǎn)P坐標(biāo)為SKIPIF1<0或SKIPIF1<0．【點(diǎn)睛】本題考查了二次函數(shù)的圖像及性質(zhì)，勾股定理，解直角三角形以及待定系數(shù)法求二次函數(shù)的解析式，熟練掌握二次函數(shù)的圖像及性質(zhì)是解題的關(guān)鍵．2．（2023·江蘇徐州·中考真題）如圖，在平而直角坐標(biāo)系中，二次函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸分別交于點(diǎn)SKIPIF1<0，頂點(diǎn)為SKIPIF1<0．連接SKIPIF1<0，將線段SKIPIF1<0繞點(diǎn)SKIPIF1<0按順時(shí)針?lè)较蛐D(zhuǎn)SKIPIF1<0得到線段SKIPIF1<0，連接SKIPIF1<0．點(diǎn)SKIPIF1<0分別在線段SKIPIF1<0上，連接SKIPIF1<0與SKIPIF1<0交于點(diǎn)SKIPIF1<0．

(1)求點(diǎn)SKIPIF1<0的坐標(biāo)；(2)隨著點(diǎn)SKIPIF1<0在線段SKIPIF1<0上運(yùn)動(dòng)．①SKIPIF1<0的大小是否發(fā)生變化？請(qǐng)說(shuō)明理由；②線段SKIPIF1<0的長(zhǎng)度是否存在最大值？若存在，求出最大值；若不存在，請(qǐng)說(shuō)明理由；(3)當(dāng)線段SKIPIF1<0的中點(diǎn)在該二次函數(shù)的圖象的對(duì)稱軸上時(shí)，SKIPIF1<0的面積為．【答案】(1)SKIPIF1<0，SKIPIF1<0；(2)①SKIPIF1<0的大小不變，理由見(jiàn)解析；②線段SKIPIF1<0的長(zhǎng)度存在最大值為SKIPIF1<0；(3)SKIPIF1<0【分析】（1）SKIPIF1<0得SKIPIF1<0，解方程即可求得SKIPIF1<0的坐標(biāo)，把SKIPIF1<0化為頂點(diǎn)式即可求得點(diǎn)SKIPIF1<0的坐標(biāo)；（2）①在SKIPIF1<0上取點(diǎn)SKIPIF1<0，使得SKIPIF1<0，連接SKIPIF1<0，證明SKIPIF1<0是等邊三角形即可得出結(jié)論；②由SKIPIF1<0，得當(dāng)SKIPIF1<0最小時(shí)，SKIPIF1<0的長(zhǎng)最大，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的長(zhǎng)最大，進(jìn)而解直角三角形即可求解；（3）設(shè)SKIPIF1<0的中點(diǎn)為點(diǎn)SKIPIF1<0，連接SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0于點(diǎn)SKIPIF1<0，證四邊形SKIPIF1<0是菱形，得SKIPIF1<0，進(jìn)而證明SKIPIF1<0得SKIPIF1<0，再證SKIPIF1<0，得SKIPIF1<0即SKIPIF1<0，結(jié)合三角形的面積公式即可求解．【詳解】（1）解：∵SKIPIF1<0，∴頂點(diǎn)為SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，∴SKIPIF1<0；（2）解：①SKIPIF1<0的大小不變，理由如下：在SKIPIF1<0上取點(diǎn)SKIPIF1<0，使得SKIPIF1<0，連接SKIPIF1<0，

∵SKIPIF1<0，∴拋物線對(duì)稱軸為SKIPIF1<0，即SKIPIF1<0，∵將線段SKIPIF1<0繞點(diǎn)SKIPIF1<0按順時(shí)針?lè)较蛐D(zhuǎn)SKIPIF1<0得到線段SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0是等邊三角形，∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0，

∴SKIPIF1<0是等邊三角形，SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0是等邊三角形，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，

∴SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0是等邊三角形，∴SKIPIF1<0，即SKIPIF1<0的大小不變；②，∵SKIPIF1<0，∴當(dāng)SKIPIF1<0最小時(shí)，SKIPIF1<0的長(zhǎng)最大，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的長(zhǎng)最大，∵SKIPIF1<0是等邊三角形，∴SKIPIF1<0SKIPIF1<0∴SKIPIF1<0，∴SKIPIF1<0，

∴SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0，即線段SKIPIF1<0的長(zhǎng)度存在最大值為SKIPIF1<0；（3）解：設(shè)SKIPIF1<0的中點(diǎn)為點(diǎn)SKIPIF1<0，連接SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0于點(diǎn)SKIPIF1<0，SKIPIF1<0

∵SKIPIF1<0，∴四邊形SKIPIF1<0是菱形，∴SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0的中點(diǎn)為點(diǎn)SKIPIF1<0，∴SKIPIF1<0，

∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0的中點(diǎn)為點(diǎn)SKIPIF1<0，SKIPIF1<0是等邊三角形，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0即SKIPIF1<0，∴SKIPIF1<0，

∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，故答案為SKIPIF1<0．【點(diǎn)睛】本題主要考查了二次函數(shù)的圖像及性質(zhì)，菱形的判定及性質(zhì)，全等三角形的判定及性質(zhì)，相似三角形的判定及性質(zhì)，等邊三角形的判定及性質(zhì)以及解直角三角形，題目綜合性較強(qiáng)，熟練掌握各知識(shí)點(diǎn)是解題的關(guān)鍵．3．（2023·江蘇蘇州·中考真題）如圖，二次函數(shù)SKIPIF1<0的圖像與SKIPIF1<0軸分別交于點(diǎn)SKIPIF1<0（點(diǎn)A在點(diǎn)SKIPIF1<0的左側(cè)），直線SKIPIF1<0是對(duì)稱軸．點(diǎn)SKIPIF1<0在函數(shù)圖像上，其橫坐標(biāo)大于4，連接SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0，垂足為SKIPIF1<0，以點(diǎn)SKIPIF1<0為圓心，作半徑為SKIPIF1<0的圓，SKIPIF1<0與SKIPIF1<0相切，切點(diǎn)為SKIPIF1<0．

(1)求點(diǎn)SKIPIF1<0的坐標(biāo)；(2)若以SKIPIF1<0的切線長(zhǎng)SKIPIF1<0為邊長(zhǎng)的正方形的面積與SKIPIF1<0的面積相等，且SKIPIF1<0不經(jīng)過(guò)點(diǎn)SKIPIF1<0，求SKIPIF1<0長(zhǎng)的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0或SKIPIF1<0【分析】（1）令SKIPIF1<0求得點(diǎn)SKIPIF1<0的橫坐標(biāo)即可解答；（2）由題意可得拋物線的對(duì)稱軸為SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0；如圖連接SKIPIF1<0，則SKIPIF1<0，進(jìn)而可得切線長(zhǎng)SKIPIF1<0為邊長(zhǎng)的正方形的面積為SKIPIF1<0；過(guò)點(diǎn)P作SKIPIF1<0軸，垂足為H，可得SKIPIF1<0；由題意可得SKIPIF1<0，解得SKIPIF1<0；然后再分當(dāng)點(diǎn)M在點(diǎn)N的上方和下方兩種情況解答即可．【詳解】（1）解：令SKIPIF1<0，則有：SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，∴SKIPIF1<0．（2）解：∵拋物線過(guò)SKIPIF1<0∴拋物線的對(duì)稱軸為SKIPIF1<0，設(shè)SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0,如圖：連接SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，∴切線SKIPIF1<0為邊長(zhǎng)的正方形的面積為SKIPIF1<0，過(guò)點(diǎn)P作SKIPIF1<0軸，垂足為H，則：SKIPIF1<0，∴SKIPIF1<0∵SKIPIF1<0，∴SKIPIF1<0，

假設(shè)SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0,則有以下兩種情況：①如圖1：當(dāng)點(diǎn)M在點(diǎn)N的上方，即SKIPIF1<0

∴SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，∵SKIPIF1<0∴SKIPIF1<0；②如圖2：當(dāng)點(diǎn)M在點(diǎn)N的上方，即SKIPIF1<0

∴SKIPIF1<0，解得：SKIPIF1<0，∵SKIPIF1<0∴SKIPIF1<0；綜上，SKIPIF1<0或SKIPIF1<0．∴當(dāng)SKIPIF1<0不經(jīng)過(guò)點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0或SKIPIF1<0．【點(diǎn)睛】本題主要考查了二次函數(shù)的性質(zhì)、切線的性質(zhì)、勾股定理等知識(shí)點(diǎn)，掌握分類討論思想是解答本題的關(guān)鍵．4．（2023·江蘇連云港·中考真題）如圖，在平面直角坐標(biāo)系SKIPIF1<0中，拋物線SKIPIF1<0的頂點(diǎn)為SKIPIF1<0．直線SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0，且平行于SKIPIF1<0軸，與拋物線SKIPIF1<0交于SKIPIF1<0兩點(diǎn)（SKIPIF1<0在SKIPIF1<0的右側(cè)）．將拋物線SKIPIF1<0沿直線SKIPIF1<0翻折得到拋物線SKIPIF1<0，拋物線SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，頂點(diǎn)為SKIPIF1<0．

(1)當(dāng)SKIPIF1<0時(shí)，求點(diǎn)SKIPIF1<0的坐標(biāo)；(2)連接SKIPIF1<0，若SKIPIF1<0為直角三角形，求此時(shí)SKIPIF1<0所對(duì)應(yīng)的函數(shù)表達(dá)式；(3)在（2）的條件下，若SKIPIF1<0的面積為SKIPIF1<0兩點(diǎn)分別在邊SKIPIF1<0上運(yùn)動(dòng)，且SKIPIF1<0，以SKIPIF1<0為一邊作正方形SKIPIF1<0，連接SKIPIF1<0，寫(xiě)出SKIPIF1<0長(zhǎng)度的最小值，并簡(jiǎn)要說(shuō)明理由．【答案】(1)SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0(3)SKIPIF1<0，見(jiàn)解析【分析】（1）將拋物線解析式化為頂點(diǎn)式，進(jìn)而得出頂點(diǎn)坐標(biāo)SKIPIF1<0，根據(jù)對(duì)稱性，即可求解．（2）由題意得，SKIPIF1<0的頂點(diǎn)SKIPIF1<0與SKIPIF1<0的頂點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，SKIPIF1<0，則拋物線SKIPIF1<0．進(jìn)而得出可得SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，如圖1，過(guò)SKIPIF1<0作SKIPIF1<0軸，垂足為SKIPIF1<0．求得SKIPIF1<0，代入解析式得出SKIPIF1<0，求得SKIPIF1<0．②當(dāng)SKIPIF1<0時(shí)，如圖2，過(guò)SKIPIF1<0作SKIPIF1<0，交SKIPIF1<0的延長(zhǎng)線于點(diǎn)SKIPIF1<0．同理可得SKIPIF1<0，得出SKIPIF1<0，代入解析式得出SKIPIF1<0代入SKIPIF1<0，得SKIPIF1<0；③當(dāng)SKIPIF1<0時(shí)，此情況不存在．（3）由（2）知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0的面積為1，不合題意舍去．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0的面積為3，符合題意．由題意可求得SKIPIF1<0．取SKIPIF1<0的中點(diǎn)SKIPIF1<0，在SKIPIF1<0中可求得SKIPIF1<0．在SKIPIF1<0中可求得SKIPIF1<0．易知當(dāng)SKIPIF1<0三點(diǎn)共線時(shí)，SKIPIF1<0取最小值，最小值為SKIPIF1<0．【詳解】（1）∵SKIPIF1<0，∴拋物線SKIPIF1<0的頂點(diǎn)坐標(biāo)SKIPIF1<0．∵SKIPIF1<0，點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱．∴SKIPIF1<0．（2）由題意得，SKIPIF1<0的頂點(diǎn)SKIPIF1<0與SKIPIF1<0的頂點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，∴SKIPIF1<0，拋物線SKIPIF1<0．∴當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，如圖1，過(guò)SKIPIF1<0作SKIPIF1<0軸，垂足為SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0．∵SKIPIF1<0∴SKIPIF1<0．∴SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0．∵直線SKIPIF1<0軸，∴SKIPIF1<0．∴SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0．∴SKIPIF1<0．又∵點(diǎn)SKIPIF1<0在SKIPIF1<0圖像上，∴SKIPIF1<0．解得SKIPIF1<0或SKIPIF1<0．∵當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0，此時(shí)SKIPIF1<0重合，舍去．當(dāng)SKIPIF1<0時(shí)，符合題意．將SKIPIF1<0代入SKIPIF1<0，得SKIPIF1<0．

②當(dāng)SKIPIF1<0時(shí)，如圖2，過(guò)SKIPIF1<0作SKIPIF1<0，交SKIPIF1<0的延長(zhǎng)線于點(diǎn)SKIPIF1<0．同理可得SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0．∴SKIPIF1<0．又∵點(diǎn)SKIPIF1<0在SKIPIF1<0圖像上，∴SKIPIF1<0．解得SKIPIF1<0或SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0．此時(shí)SKIPIF1<0符合題意．將SKIPIF1<0代入SKIPIF1<0，得SKIPIF1<0．③當(dāng)SKIPIF1<0時(shí)，此情況不存在．綜上，SKIPIF1<0所對(duì)應(yīng)的函數(shù)表達(dá)式為SKIPIF1<0或SKIPIF1<0．（3）如圖3，由（2）知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0則SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的面積為1，不合題意舍去．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，此時(shí)SKIPIF1<0的面積為3，符合題意∴SKIPIF1<0．依題意，四邊形SKIPIF1<0是正方形，∴SKIPIF1<0．取SKIPIF1<0的中點(diǎn)SKIPIF1<0，在SKIPIF1<0中可求得SKIPIF1<0．在SKIPIF1<0中可求得SKIPIF1<0．∴當(dāng)SKIPIF1<0三點(diǎn)共線時(shí)，SKIPIF1<0取最小值，最小值為SKIPIF1<0．【點(diǎn)睛】本題考查了二次函數(shù)的性質(zhì)，特殊三角形問(wèn)題，正方形的性質(zhì)，勾股定理，面積問(wèn)題，分類討論是解題的關(guān)鍵．5．（2022·江蘇蘇州·中考真題）如圖，在二次函數(shù)SKIPIF1<0（m是常數(shù)，且SKIPIF1<0）的圖像與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，頂點(diǎn)為D．其對(duì)稱軸與線段BC交于點(diǎn)E，與x軸交于點(diǎn)F．連接AC，BD．(1)求A，B，C三點(diǎn)的坐標(biāo)（用數(shù)字或含m的式子表示），并求SKIPIF1<0的度數(shù)；(2)若SKIPIF1<0，求m的值；(3)若在第四象限內(nèi)二次函數(shù)SKIPIF1<0（m是常數(shù)，且SKIPIF1<0）的圖像上，始終存在一點(diǎn)P，使得SKIPIF1<0，請(qǐng)結(jié)合函數(shù)的圖像，直接寫(xiě)出m的取值范圍．【答案】(1)A（-1，0）；B（2m+1，0）；C（0，2m+1）；SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0【分析】（1）分別令SKIPIF1<0等于0，即可求得SKIPIF1<0的坐標(biāo)，根據(jù)SKIPIF1<0，即可求得SKIPIF1<0；（2）方法一：如圖1，連接AE．由解析式分別求得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．根據(jù)軸對(duì)稱的性質(zhì)，可得SKIPIF1<0，由SKIPIF1<0，建立方程，解方程即可求解．方法二：如圖2，過(guò)點(diǎn)D作SKIPIF1<0交BC于點(diǎn)H．由方法一，得SKIPIF1<0，SKIPIF1<0．證明SKIPIF1<0，根據(jù)相似三角形的性質(zhì)建立方程，解方程即可求解；（3）設(shè)PC與x軸交于點(diǎn)Q，當(dāng)P在第四象限時(shí)，點(diǎn)Q總在點(diǎn)B的左側(cè)，此時(shí)SKIPIF1<0，即SKIPIF1<0．【詳解】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．解方程，得SKIPIF1<0，SKIPIF1<0．∵點(diǎn)A在點(diǎn)B的左側(cè)，且SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．∴SKIPIF1<0．∴SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0．（2）方法一：如圖1，連接AE．∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0．∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．∵點(diǎn)A，點(diǎn)B關(guān)于對(duì)稱軸對(duì)稱，∴SKIPIF1<0．∴SKIPIF1<0．∴SKIPIF1<0．∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0．∴SKIPIF1<0．∵SKIPIF1<0，∴解方程，得SKIPIF1<0．方法二：如圖2，過(guò)點(diǎn)D作SKIPIF1<0交BC于點(diǎn)H．由方法一，得SKIPIF1<0，SKIPIF1<0．∴SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0．∴SKIPIF1<0．∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0．∴SKIPIF1<0．∴SKIPIF1<0，即SKIPIF1<0．∵SKIPIF1<0，∴解方程，得SKIPIF1<0．（3）SKIPIF1<0．設(shè)PC與x軸交于點(diǎn)Q，當(dāng)P在第四象限時(shí)，點(diǎn)Q總在點(diǎn)B的左側(cè)，此時(shí)SKIPIF1<0，即SKIPIF1<0．∵SKIPIF1<0，∴SKIPIF1<0．SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0．解得SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0．【點(diǎn)睛】本題考查了二次函數(shù)綜合，求二次函數(shù)與坐標(biāo)軸的交點(diǎn)，角度問(wèn)題，解直角三角形，相似三角形的性質(zhì)，三角形內(nèi)角和定理，綜合運(yùn)用以上知識(shí)是解題的關(guān)鍵．1．如圖，在平面直角坐標(biāo)系xOy中，已知拋物線SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，SKIPIF1<0，與y軸交于點(diǎn)C．(1)求拋物線的函數(shù)表達(dá)式；(2)如圖1，若點(diǎn)M是第四象限內(nèi)拋物線上一點(diǎn)，SKIPIF1<0軸交SKIPIF1<0于點(diǎn)N，SKIPIF1<0求SKIPIF1<0的最大值；(3)如圖2，在SKIPIF1<0軸上取一點(diǎn)SKIPIF1<0，拋物線沿SKIPIF1<0方向平移SKIPIF1<0個(gè)單位得新拋物線，新拋物線與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，SKIPIF1<0，交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0在線段SKIPIF1<0上運(yùn)動(dòng)，線段SKIPIF1<0關(guān)于線段SKIPIF1<0的對(duì)稱線段SKIPIF1<0所在直線交新拋物線于點(diǎn)SKIPIF1<0，直線SKIPIF1<0與直線SKIPIF1<0所成夾角為SKIPIF1<0，直接寫(xiě)出點(diǎn)SKIPIF1<0的橫坐標(biāo)．【答案】(1)拋物線的解析式為SKIPIF1<0；(2)SKIPIF1<0有最大值SKIPIF1<0；(3)SKIPIF1<0點(diǎn)的橫坐標(biāo)為SKIPIF1<0或6或SKIPIF1<0或SKIPIF1<0．【分析】（1）用待定系數(shù)法求函數(shù)的解析式即可；（2）過(guò)SKIPIF1<0點(diǎn)作SKIPIF1<0軸交SKIPIF1<0于點(diǎn)SKIPIF1<0，可得四邊形SKIPIF1<0是平行四邊形，再由SKIPIF1<0，SKIPIF1<0，推導(dǎo)出SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最大值SKIPIF1<0；（3）求出平移后的函數(shù)解析式為SKIPIF1<0，直線SKIPIF1<0的解析式為SKIPIF1<0，設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0軸時(shí)，直線SKIPIF1<0與直線SKIPIF1<0所成夾角為SKIPIF1<0，求出SKIPIF1<0，可得直線SKIPIF1<0的解析式為SKIPIF1<0，直線與拋物線的交點(diǎn)即為SKIPIF1<0點(diǎn)；當(dāng)SKIPIF1<0軸時(shí)，直線SKIPIF1<0與直線SKIPIF1<0所成夾角為SKIPIF1<0，求出SKIPIF1<0，可得直線SKIPIF1<0的解析式為SKIPIF1<0，直線與拋物線的交點(diǎn)即為SKIPIF1<0點(diǎn)．【詳解】（1）解：將點(diǎn)SKIPIF1<0，SKIPIF1<0代入SKIPIF1<0，SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0拋物線的解析式為SKIPIF1<0；（2）解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0，將點(diǎn)SKIPIF1<0代入，可得SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0的解析式為SKIPIF1<0，過(guò)SKIPIF1<0點(diǎn)作SKIPIF1<0軸交SKIPIF1<0于點(diǎn)SKIPIF1<0，∵SKIPIF1<0軸，∴SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0四邊形SKIPIF1<0是平行四邊形，SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最大值SKIPIF1<0；（3）解：SKIPIF1<0拋物線沿SKIPIF1<0方向平移SKIPIF1<0個(gè)單位，SKIPIF1<0拋物線沿SKIPIF1<0軸負(fù)半軸平移2個(gè)單位，沿SKIPIF1<0軸正方向平移2個(gè)單位，SKIPIF1<0平移后的函數(shù)解析式為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0，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，SKIPIF1<0，當(dāng)SKIPIF1<0軸時(shí)，直線SKIPIF1<0與直線SKIPIF1<0所成夾角為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍SKIPIF1<0，SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0的解析式為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，解得SKIPIF1<0或SKIPIF1<0，SKIPIF1<0點(diǎn)橫坐標(biāo)為SKIPIF1<0或6；當(dāng)SKIPIF1<0軸時(shí)，直線SKIPIF1<0與直線SKIPIF1<0所成夾角為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0（舍SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0的解析式為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，解得SKIPIF1<0或SKIPIF1<0，SKIPIF1<0點(diǎn)的橫坐標(biāo)為SKIPIF1<0或SKIPIF1<0；綜上所述：SKIPIF1<0點(diǎn)的橫坐標(biāo)為SKIPIF1<0或6或SKIPIF1<0或SKIPIF1<0．【點(diǎn)睛】本題是二次函數(shù)的綜合問(wèn)題，考查二次函數(shù)的圖象及性質(zhì)，解直角三角形，二次函數(shù)的平移，勾股定理，平行四邊形的判定和性質(zhì)．熟練掌握二次函數(shù)的圖象及性質(zhì)，平行線的性質(zhì)，軸對(duì)稱的性質(zhì)，直角三角形的性質(zhì)是解題的關(guān)鍵．2．蔬菜大棚是一種具有出色保溫性能的框架覆膜結(jié)構(gòu)，它的出現(xiàn)使得人們可以吃到反季節(jié)蔬菜．一般蔬菜大棚使用竹結(jié)構(gòu)或者鋼結(jié)構(gòu)的骨架，上面覆上一層或多層保溫塑料膜，這樣就形成了一個(gè)溫室空間．如圖所示，某個(gè)溫室大棚的橫截面可以看作是由矩形SKIPIF1<0和拋物線的一部分SKIPIF1<0構(gòu)成的（以下簡(jiǎn)記為“拋物線SKIPIF1<0”），其中SKIPIF1<0，SKIPIF1<0，現(xiàn)取SKIPIF1<0中點(diǎn)O，過(guò)點(diǎn)O作線段SKIPIF1<0的垂直平分線SKIPIF1<0交拋物線SKIPIF1<0于點(diǎn)E，SKIPIF1<0．若以O(shè)點(diǎn)為原點(diǎn)，SKIPIF1<0所在直線為x軸，SKIPIF1<0所在直線為y軸建立如圖①所示的平面直角坐標(biāo)系．請(qǐng)結(jié)合圖形解答下列問(wèn)題：(1)求拋物線的解析式．(2)如圖②所示，為了保證蔬菜大棚的通風(fēng)性，該大棚要安裝兩個(gè)正方形孔的排氣裝置SKIPIF1<0，其中L，R在拋物線SKIPIF1<0上，若SKIPIF1<0，求兩個(gè)正方形裝置的間距SKIPIF1<0的長(zhǎng)．【答案】(1)拋物線的解析式為SKIPIF1<0(2)SKIPIF1<0【分析】本題考查二次函數(shù)的實(shí)際應(yīng)用，矩形的性質(zhì)，讀懂題意，正確的求出二次函數(shù)解析式，利用數(shù)形結(jié)合的思想，進(jìn)行求解，是解題的關(guān)鍵．（1）由題意得SKIPIF1<0，SKIPIF1<0，設(shè)函數(shù)解析式為SKIPIF1<0，再利用待定系數(shù)法求出函數(shù)解析式即可；（2）求出SKIPIF1<0時(shí)對(duì)應(yīng)的自變量的值，得到SKIPIF1<0的長(zhǎng)，再減去兩個(gè)正方形的邊長(zhǎng)即可得解；【詳解】（1）由題意，知四邊形SKIPIF1<0為矩形，SKIPIF1<0為SKIPIF1<0的垂直平分線，SKIPIF1<0，SKIPIF1<0．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．設(shè)拋物線的解析式為SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0拋物線的解析式為SKIPIF1<0．（2）SKIPIF1<0四邊形SKIPIF1<0，四邊形SKIPIF1<0均為正方形，SKIPIF1<0，SKIPIF1<0．如圖所示，延長(zhǎng)SKIPIF1<0交SKIPIF1<0于點(diǎn)H，延長(zhǎng)SKIPIF1<0交SKIPIF1<0于點(diǎn)J，則四邊形SKIPIF1<0，四邊形SKIPIF1<0均為矩形，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．3．已知拋物線SKIPIF1<0與x軸負(fù)半軸交于點(diǎn)A．與x軸正半軸交于點(diǎn)B，與y軸交于點(diǎn)C，連接SKIPIF1<0、SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．(1)求拋物線的解析式；(2)點(diǎn)P為第一象限拋物線上一點(diǎn)，連接SKIPIF1<0、SKIPIF1<0，點(diǎn)P的橫坐標(biāo)是t，SKIPIF1<0的面積為S，求S與t之間的函數(shù)關(guān)系式（不用寫(xiě)出t的取值范圍）；(3)在（2）的條件下，SKIPIF1<0交SKIPIF1<0于點(diǎn)D，連接SKIPIF1<0，以SKIPIF1<0為斜邊在SKIPIF1<0的右側(cè)作等腰SKIPIF1<0，連接SKIPIF1<0、SKIPIF1<0，SKIPIF1<0，點(diǎn)F為線段SKIPIF1<0上一動(dòng)點(diǎn)，當(dāng)SKIPIF1<0時(shí)，求點(diǎn)F的坐標(biāo)．【答案】(1)拋物線解析式為SKIPIF1<0；(2)SKIPIF1<0；(3)點(diǎn)F的坐標(biāo)為SKIPIF1<0．【分析】（1）先求出SKIPIF1<0，SKIPIF1<0，然后用待定系數(shù)法求解即可；（2）過(guò)點(diǎn)P作SKIPIF1<0軸于點(diǎn)H，作SKIPIF1<0軸于點(diǎn)T，令SKIPIF1<0交y軸于點(diǎn)SKIPIF1<0，P點(diǎn)在拋物線上，P點(diǎn)橫坐標(biāo)為t．根據(jù)SKIPIF1<0求出SKIPIF1<0，然后根據(jù)三角形面積公式求解即可；（3）過(guò)點(diǎn)E作SKIPIF1<0軸于點(diǎn)N，SKIPIF1<0延長(zhǎng)線交SKIPIF1<0于點(diǎn)K，過(guò)點(diǎn)D作SKIPIF1<0，交SKIPIF1<0延長(zhǎng)線于點(diǎn)M，過(guò)點(diǎn)E作SKIPIF1<0軸于點(diǎn)L．根據(jù)SKIPIF1<0證明SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，進(jìn)而可推出SKIPIF1<0，可求出SKIPIF1<0，結(jié)合SKIPIF1<0求出SKIPIF1<0．作SKIPIF1<0軸于點(diǎn)R，求出直線SKIPIF1<0解析式和直線SKIPIF1<0解析式．由SKIPIF1<0求出SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，求出直線SKIPIF1<0解析式，然后聯(lián)立直線SKIPIF1<0與SKIPIF1<0即可求解．【詳解】（1）令SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0∵SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0∵拋物線經(jīng)過(guò)A、B兩點(diǎn)，∴SKIPIF1<0，解得SKIPIF1<0∴拋物線解析式為SKIPIF1<0（2）過(guò)點(diǎn)P作SKIPIF1<0軸于點(diǎn)H，作SKIPIF1<0軸于點(diǎn)T，令SKIPIF1<0交y軸于點(diǎn)SKIPIF1<0，P點(diǎn)在拋物線上，P點(diǎn)橫坐標(biāo)為t．∴SKIPIF1<0∵SKIPIF1<0∴四邊形SKIPIF1<0為矩形∴SKIPIF1<0，SKIPIF1<0∵SKIPIF1<0，即SKIPIF1<0∴SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0即SKIPIF1<0（3）過(guò)點(diǎn)E作SKIPIF1<0軸于點(diǎn)N，SKIPIF1<0延長(zhǎng)線交SKIPIF1<0于點(diǎn)K，過(guò)點(diǎn)D作SKIPIF1<0，交SKIPIF1<0延長(zhǎng)線于點(diǎn)M，過(guò)點(diǎn)E作SKIPIF1<0軸于點(diǎn)L．∵SKIPIF1<0是等腰直角三角形∴SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0且SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0∴SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，即SKIPIF1<0∴SKIPIF1<0∵SKIPIF1<0∴四邊形SKIPIF1<0為矩形∴SKIPIF1<0且SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0作SKIPIF1<0軸于點(diǎn)R∴四邊形SKIPIF1<0為矩形∴SKIPIF1<0SKIPIF1<0∴SKIPIF1<0且SKIPIF1<0設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0∴直線SKIPIF1<0解析式為SKIPIF1<0∵SKIPIF1<0∴設(shè)直線SKIPIF1<0解析式為SKIPIF1<0∵SKIPIF1