中考數(shù)學(xué)二輪復(fù)習(xí)壓軸題培優(yōu)專練專題15 函數(shù)中的面積問題（解析版）

專題15函數(shù)中的面積問題函數(shù)中面積問題一般包括面積的最大值和最小值或者等于某個(gè)數(shù)值的問題。在解決函數(shù)中的面積問題時(shí)，通常需要過三角形或多邊形的一個(gè)端點(diǎn)，做坐標(biāo)軸的平行線，把三角形或多邊形進(jìn)行割補(bǔ)呈三角形，從而用坐標(biāo)將三角形的底和高表達(dá)出來。如圖，SKIPIF1<0。 （2022·內(nèi)蒙古·中考真題）如圖，拋物線SKIPIF1<0經(jīng)過SKIPIF1<0，SKIPIF1<0兩點(diǎn)，與x軸的另一個(gè)交點(diǎn)為A，與y軸相交于點(diǎn)C．(1)求拋物線的解析式和點(diǎn)C的坐標(biāo)；(2)若點(diǎn)M在直線SKIPIF1<0上方的拋物線上運(yùn)動(dòng)（與點(diǎn)B，C不重合），求使SKIPIF1<0面積最大時(shí)M點(diǎn)的坐標(biāo)，并求最大面積；（請(qǐng)?jiān)趫D1中探索）(3)設(shè)點(diǎn)Q在y軸上，點(diǎn)P在拋物線上，要使以點(diǎn)A，B，P，Q為頂點(diǎn)的四邊形是平行四邊形，求所有滿足條件的點(diǎn)P的坐標(biāo)．（請(qǐng)?jiān)趫D2中探索）（1）用待定系數(shù)法求函數(shù)的解析式即可；（2）作直線BC，過M點(diǎn)作MN∥y軸交BC于點(diǎn)N，求出直線BC的解析式，設(shè)M（m，-SKIPIF1<0+m+SKIPIF1<0），則N（m，-SKIPIF1<0m+SKIPIF1<0），可得S△MBC=SKIPIF1<0?MN?OB=SKIPIF1<0+SKIPIF1<0，再求解即可；（3）設(shè)Q（0，t），P（m，-SKIPIF1<0+m+SKIPIF1<0），分三種情況討論：①當(dāng)AB為平行四邊形的對(duì)角線時(shí)；②當(dāng)AQ為平行四邊形的對(duì)角線時(shí)；③當(dāng)AP為平行四邊形的對(duì)角線時(shí)；根據(jù)平行四邊形的對(duì)角線互相平分，利用中點(diǎn)坐標(biāo)公式求解即可．【答案】(1)SKIPIF1<0，SKIPIF1<0(2)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，S有最大值為SKIPIF1<0(3)滿足條件的點(diǎn)P坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0【詳解】（1）解：把點(diǎn)SKIPIF1<0和SKIPIF1<0分別代入SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0∴拋物線的解析式為SKIPIF1<0把SKIPIF1<0代入SKIPIF1<0可得SKIPIF1<0∴SKIPIF1<0；（2）解：作直線SKIPIF1<0，作SKIPIF1<0軸交直線SKIPIF1<0于點(diǎn)N設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0（SKIPIF1<0）把點(diǎn)SKIPIF1<0和SKIPIF1<0分別代入SKIPIF1<0可得SKIPIF1<0解得SKIPIF1<0∴直線SKIPIF1<0的解析式為SKIPIF1<0設(shè)點(diǎn)M的橫坐標(biāo)為m∴SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0SKIPIF1<0∴SKIPIF1<0SKIPIF1<0（SKIPIF1<0）∴當(dāng)SKIPIF1<0時(shí)，S有最大值為SKIPIF1<0把SKIPIF1<0代入SKIPIF1<0可得SKIPIF1<0∴SKIPIF1<0；（3）解：當(dāng)以SKIPIF1<0為邊時(shí)，只要SKIPIF1<0，且SKIPIF1<0即可∴點(diǎn)P的橫坐標(biāo)為4或-4把SKIPIF1<0代入SKIPIF1<0可得SKIPIF1<0把SKIPIF1<0代入SKIPIF1<0可得SKIPIF1<0∴此時(shí)SKIPIF1<0，SKIPIF1<0當(dāng)以SKIPIF1<0為對(duì)角線時(shí)，作SKIPIF1<0軸于點(diǎn)H∵四邊形SKIPIF1<0是平行四邊形∴SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0和SKIPIF1<0中SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0∴點(diǎn)P的橫坐標(biāo)為2把SKIPIF1<0代入SKIPIF1<0可得SKIPIF1<0∴此時(shí)SKIPIF1<0綜上所述，滿足條件的點(diǎn)P坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0本題考查二次函數(shù)的圖象及性質(zhì)，熟練掌握二次函數(shù)的圖象及性質(zhì)，平行四邊形的性質(zhì)，分類討論是解題的關(guān)鍵。（2022·四川綿陽·統(tǒng)考中考真題）如圖，一次函數(shù)SKIPIF1<0與反比例函數(shù)SKIPIF1<0在第一象限交于SKIPIF1<0、SKIPIF1<0兩點(diǎn)，SKIPIF1<0垂直x軸于點(diǎn)SKIPIF1<0，SKIPIF1<0為坐標(biāo)原點(diǎn)，四邊形SKIPIF1<0的面積為38．(1)求反比例函數(shù)及一次函數(shù)的解析式；(2)點(diǎn)P是反比例函數(shù)第三象限內(nèi)的圖象上一動(dòng)點(diǎn)，請(qǐng)簡要描述使SKIPIF1<0的面積最小時(shí)點(diǎn)P的位置（不需證明），并求出點(diǎn)P的坐標(biāo)和SKIPIF1<0面積的最小值．（1）利用待定系數(shù)法即可求出反比例函數(shù)解析式，再利用四邊形SKIPIF1<0的面積為38．求出SKIPIF1<0，進(jìn)一步利用待定系數(shù)法即可求出一次函數(shù)解析式；（2）平移一次函數(shù)與SKIPIF1<0在第三象限有唯一交點(diǎn)P，此時(shí)P到MN的距離最短，SKIPIF1<0的面積最小，設(shè)平移后的一次函數(shù)解析式為：SKIPIF1<0，聯(lián)立SKIPIF1<0，解得：SKIPIF1<0，進(jìn)一步求出：SKIPIF1<0，即SKIPIF1<0，連接PM，PN，過點(diǎn)P作SKIPIF1<0的延長線交于點(diǎn)B，作SKIPIF1<0交于點(diǎn)C，根據(jù)SKIPIF1<0以及點(diǎn)的坐標(biāo)即可求出SKIPIF1<0的面積．【答案】(1)SKIPIF1<0，SKIPIF1<0；(2)SKIPIF1<0，SKIPIF1<0．【詳解】（1）解：∵SKIPIF1<0在SKIPIF1<0上，∴SKIPIF1<0，即反比例函數(shù)解析式為：SKIPIF1<0，設(shè)SKIPIF1<0，∵四邊形SKIPIF1<0的面積為38．∴SKIPIF1<0，整理得：SKIPIF1<0，解得：SKIPIF1<0（舍去），SKIPIF1<0，∴SKIPIF1<0，將SKIPIF1<0和SKIPIF1<0代入SKIPIF1<0可得：SKIPIF1<0解得：SKIPIF1<0，∴一次函數(shù)解析式為：SKIPIF1<0．（2）解：平移一次函數(shù)SKIPIF1<0到第三象限，與SKIPIF1<0在第三象限有唯一交點(diǎn)P，此時(shí)P到MN的距離最短，SKIPIF1<0的面積最小，設(shè)平移后的一次函數(shù)解析式為：SKIPIF1<0，聯(lián)立SKIPIF1<0可得：SKIPIF1<0，整理得：SKIPIF1<0，∵有唯一交點(diǎn)P，∴SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0（舍去），將SKIPIF1<0代入SKIPIF1<0得：SKIPIF1<0，解得：SKIPIF1<0經(jīng)檢驗(yàn)：SKIPIF1<0是分式方程SKIPIF1<0的根，∴SKIPIF1<0，連接PM，PN，過點(diǎn)P作SKIPIF1<0的延長線交于點(diǎn)B，作SKIPIF1<0交于點(diǎn)C，則：SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0．本題考查一次函數(shù)和反比例函數(shù)的綜合，難度較大，解題的關(guān)鍵是掌握待定系數(shù)法求函數(shù)解析式，掌握平行線之間的距離，解分式方程，解一元二次方程知識(shí)點(diǎn)。（2022·遼寧大連·統(tǒng)考中考真題）在平面直角坐標(biāo)系中，拋物線SKIPIF1<0與x軸相交于點(diǎn)A，B（點(diǎn)A在點(diǎn)B的左側(cè)），與y軸相交于點(diǎn)C，連接SKIPIF1<0．(1)求點(diǎn)B，點(diǎn)C的坐標(biāo)；(2)如圖1，點(diǎn)SKIPIF1<0在線段SKIPIF1<0上（點(diǎn)E不與點(diǎn)B重合），點(diǎn)F在y軸負(fù)半軸上，SKIPIF1<0，連接SKIPIF1<0，設(shè)SKIPIF1<0的面積為SKIPIF1<0，SKIPIF1<0的面積為SKIPIF1<0，SKIPIF1<0，當(dāng)S取最大值時(shí)，求m的值；(3)如圖2，拋物線的頂點(diǎn)為D，連接SKIPIF1<0，點(diǎn)P在第一象限的拋物線上，SKIPIF1<0與SKIPIF1<0相交于點(diǎn)Q，是否存在點(diǎn)P，使SKIPIF1<0，若存在，請(qǐng)求出點(diǎn)P的坐標(biāo)；若不存在，請(qǐng)說明理由．（1）利用拋物線的解析式，令x=0，可得C的坐標(biāo)，令y=0，可得A，C的坐標(biāo)；（2）由SKIPIF1<0可得SKIPIF1<0再分別表示SKIPIF1<0SKIPIF1<0再建立二次函數(shù)關(guān)系式，再利用二次函數(shù)的性質(zhì)可得答案；（3）如圖，延長DC與x軸交于點(diǎn)N，過A作SKIPIF1<0于H，過SKIPIF1<0作SKIPIF1<0軸于K，連接BD，證明SKIPIF1<0證明SKIPIF1<0求解SKIPIF1<0可得SKIPIF1<0再求解SKIPIF1<0及SKIPIF1<0為SKIPIF1<0再聯(lián)立：SKIPIF1<0從而可得答案．【答案】(1)SKIPIF1<0(2)當(dāng)SKIPIF1<0最大時(shí)，SKIPIF1<0(3)SKIPIF1<0【詳解】（1）解：∵SKIPIF1<0，令SKIPIF1<0則SKIPIF1<0SKIPIF1<0令SKIPIF1<0則SKIPIF1<0解得：SKIPIF1<0∴SKIPIF1<0（2）∵SKIPIF1<0∴SKIPIF1<0SKIPIF1<0而SKIPIF1<0∴SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0∴當(dāng)SKIPIF1<0最大時(shí)，則SKIPIF1<0（3）如圖，延長DC與x軸交于點(diǎn)N，過A作SKIPIF1<0于H，過SKIPIF1<0作SKIPIF1<0軸于K，連接BD，SKIPIF1<0SKIPIF1<0，SKIPIF1<0∵拋物線SKIPIF1<0∴頂點(diǎn)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0軸，SKIPIF1<0SKIPIF1<0SKIPIF1<0∴SKIPIF1<0設(shè)SKIPIF1<0為SKIPIF1<0SKIPIF1<0解得SKIPIF1<0∴SKIPIF1<0為SKIPIF1<0聯(lián)立：SKIPIF1<0解得：SKIPIF1<0所以SKIPIF1<0本題考查的是二次函數(shù)與坐標(biāo)軸的交點(diǎn)問題，二次函數(shù)的性質(zhì)，等腰直角三角形的性質(zhì)，銳角三角函數(shù)的應(yīng)用，利用待定系數(shù)法求解一次函數(shù)的解析式，函數(shù)的交點(diǎn)坐標(biāo)問題，求解Q的坐標(biāo)是解本題的關(guān)鍵．1．（2023·廣東佛山·校考一模）如圖，在平面直角坐標(biāo)系中，SKIPIF1<0為坐標(biāo)原點(diǎn)，一次函數(shù)SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，若點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸的對(duì)稱點(diǎn)SKIPIF1<0在一次函數(shù)SKIPIF1<0的圖象上．(1)求SKIPIF1<0的值；(2)若一次函數(shù)SKIPIF1<0與一次函數(shù)SKIPIF1<0交于SKIPIF1<0，且點(diǎn)SKIPIF1<0關(guān)于原點(diǎn)的對(duì)稱點(diǎn)為點(diǎn)SKIPIF1<0．求過SKIPIF1<0，SKIPIF1<0，SKIPIF1<0三點(diǎn)對(duì)應(yīng)的二次函數(shù)表達(dá)式；(3)SKIPIF1<0為拋物線上一點(diǎn)，它關(guān)于原點(diǎn)的對(duì)稱點(diǎn)為點(diǎn)SKIPIF1<0．①當(dāng)四邊形SKIPIF1<0為菱形時(shí)，求點(diǎn)SKIPIF1<0的坐標(biāo)；②若點(diǎn)SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，當(dāng)SKIPIF1<0為何值時(shí)，四邊形SKIPIF1<0的面積最大？請(qǐng)說明理由．【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)①SKIPIF1<0或SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，四邊形SKIPIF1<0的面積最大．理由見解析【詳解】（1）解：SKIPIF1<0一次函數(shù)SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸的對(duì)稱點(diǎn)SKIPIF1<0在一次函數(shù)SKIPIF1<0的圖象上，SKIPIF1<0點(diǎn)SKIPIF1<0坐標(biāo)為SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0坐標(biāo)為SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0在一次函數(shù)SKIPIF1<0的圖象上，SKIPIF1<0，SKIPIF1<0；（2）解：由方程組SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0點(diǎn)坐標(biāo)為SKIPIF1<0，又SKIPIF1<0點(diǎn)為SKIPIF1<0點(diǎn)關(guān)于原點(diǎn)的對(duì)稱點(diǎn)，SKIPIF1<0點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0一次函數(shù)SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，SKIPIF1<0點(diǎn)坐標(biāo)為SKIPIF1<0，設(shè)二次函數(shù)對(duì)應(yīng)的函數(shù)表達(dá)式為SKIPIF1<0，把SKIPIF1<0，SKIPIF1<0，SKIPIF1<0三點(diǎn)的坐標(biāo)分別代入，得SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0二次函數(shù)對(duì)應(yīng)的函數(shù)表達(dá)式為SKIPIF1<0；（3）①當(dāng)四邊形SKIPIF1<0為菱形時(shí)，SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0對(duì)應(yīng)的函數(shù)表達(dá)式為SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0對(duì)應(yīng)的函數(shù)表達(dá)式為SKIPIF1<0．聯(lián)立方程組SKIPIF1<0．解得SKIPIF1<0或SKIPIF1<0，SKIPIF1<0點(diǎn)坐標(biāo)為SKIPIF1<0或SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，四邊形SKIPIF1<0的面積最大．理由如下：如圖，過SKIPIF1<0作SKIPIF1<0，垂足為SKIPIF1<0，過SKIPIF1<0作SKIPIF1<0軸的垂線，交直線SKIPIF1<0于點(diǎn)SKIPIF1<0，易知SKIPIF1<0，SKIPIF1<0線段SKIPIF1<0的長固定不變，SKIPIF1<0當(dāng)SKIPIF1<0最大時(shí)，四邊形SKIPIF1<0的面積最大，易知SKIPIF1<0（固定不變），SKIPIF1<0當(dāng)SKIPIF1<0最大時(shí)，SKIPIF1<0也最大，SKIPIF1<0點(diǎn)在二次函數(shù)圖象上，SKIPIF1<0點(diǎn)在一次函數(shù)SKIPIF1<0的圖象上，SKIPIF1<0點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最大值1，此時(shí)SKIPIF1<0有最大值，即四邊形SKIPIF1<0的面積最大．2．（2022·遼寧盤錦·?？家荒＃┤鐖D：直線SKIPIF1<0交y軸丁點(diǎn)D，交x軸于點(diǎn)SKIPIF1<0，交拋物線SKIPIF1<0于點(diǎn)SKIPIF1<0，點(diǎn)E．點(diǎn)SKIPIF1<0在拋物線上，連接SKIPIF1<0．(1)求拋物線的解析式；(2)點(diǎn)Q從點(diǎn)A出發(fā)，以每秒2個(gè)單位長度的速度沿折線A-B-C做勻速運(yùn)動(dòng)，當(dāng)點(diǎn)Q與點(diǎn)C重合時(shí)停止運(yùn)動(dòng)，設(shè)運(yùn)動(dòng)的時(shí)間為t秒，SKIPIF1<0的面積為S，求S與t的函數(shù)關(guān)系式；(3)在（2）的條件下，若SKIPIF1<0，請(qǐng)直接寫出此時(shí)t的值．【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0或SKIPIF1<0【思路分析】（1）利用待定系數(shù)法即可求出拋物線的解析式；（2）求出直線SKIPIF1<0的解析式，得到SKIPIF1<0，SKIPIF1<0交y軸于點(diǎn)H，分兩種情況：①當(dāng)SKIPIF1<0時(shí)，Q點(diǎn)線段SKIPIF1<0上，②當(dāng)SKIPIF1<0時(shí)，Q點(diǎn)在線段SKIPIF1<0上，分別求出解析式即可；（3）連接SKIPIF1<0，則得到菱形SKIPIF1<0，得到SKIPIF1<0，推出SKIPIF1<0，再分兩種情況：①當(dāng)點(diǎn)Q運(yùn)動(dòng)到SKIPIF1<0邊上時(shí)，②當(dāng)點(diǎn)Q運(yùn)動(dòng)到SKIPIF1<0上時(shí)，分別求出t的值．【詳解】（1）解：將點(diǎn)A、B坐標(biāo)代入SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，∴拋物線的解析式為SKIPIF1<0；（2）將SKIPIF1<0、SKIPIF1<0分別代入SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，∴直線SKIPIF1<0的解析式為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0如圖，SKIPIF1<0交y軸于點(diǎn)H，則SKIPIF1<0，∴SKIPIF1<0，由A,B,C坐標(biāo)知SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，Q點(diǎn)線段SKIPIF1<0上，∴SKIPIF1<0，②當(dāng)SKIPIF1<0時(shí)，Q點(diǎn)在線段SKIPIF1<0上，由SKIPIF1<0、SKIPIF1<0求得SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0，∴SKIPIF1<0，綜上，SKIPIF1<0（3）連接SKIPIF1<0，則得到菱形SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0；①當(dāng)點(diǎn)Q運(yùn)動(dòng)到SKIPIF1<0邊上時(shí)，如答圖2，則SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0；②當(dāng)點(diǎn)Q運(yùn)動(dòng)到SKIPIF1<0上時(shí)，如答圖3，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，綜上，若SKIPIF1<0，此時(shí)t的值為SKIPIF1<0或SKIPIF1<0．3．（2022·重慶璧山·統(tǒng)考一模）如圖，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軸交于點(diǎn)SKIPIF1<0．(1)求拋物線的解析式；(2)如圖1，連接SKIPIF1<0，點(diǎn)SKIPIF1<0為線段SKIPIF1<0下方拋物線上一動(dòng)點(diǎn)，過點(diǎn)SKIPIF1<0作SKIPIF1<0軸交線段SKIPIF1<0于SKIPIF1<0點(diǎn)，連接SKIPIF1<0，記SKIPIF1<0的面積為SKIPIF1<0，SKIPIF1<0的面積為SKIPIF1<0，求SKIPIF1<0的最大值及此時(shí)點(diǎn)SKIPIF1<0的坐標(biāo)；(3)如圖2，在（2）問的條件下，將拋物線沿射線SKIPIF1<0方向平移SKIPIF1<0個(gè)單位長度得到新拋物線，動(dòng)點(diǎn)SKIPIF1<0在原拋物線的對(duì)稱軸上，點(diǎn)SKIPIF1<0為新拋物線上一點(diǎn)，直接寫出所有使得以點(diǎn)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0為頂點(diǎn)的四邊形是平行四邊形的點(diǎn)SKIPIF1<0的坐標(biāo)，并把求其中一個(gè)點(diǎn)SKIPIF1<0的坐標(biāo)的過程寫出來．【答案】(1)SKIPIF1<0(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值，最大值為1，此時(shí)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0(3)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0【思路分析】（1）將SKIPIF1<0，SKIPIF1<0代入拋物線SKIPIF1<0，列方程組求解即可得到答案；（2）延長SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，設(shè)直線SKIPIF1<0的函數(shù)表達(dá)式為SKIPIF1<0，將SKIPIF1<0，SKIPIF1<0代入列方程組求解得出解析式，設(shè)SKIPIF1<0，根據(jù)SKIPIF1<0軸得到SKIPIF1<0，SKIPIF1<0，根據(jù)三角形面積公式用t表示出SKIPIF1<0，利用函數(shù)性質(zhì)即可得到最值；（3）根據(jù)SKIPIF1<0，SKIPIF1<0得到SKIPIF1<0，結(jié)合拋物線沿射線SKIPIF1<0方向平移SKIPIF1<0個(gè)單位長度，得到拋物線向右平移SKIPIF1<0個(gè)單位長度，向上平移3個(gè)單位長度，得到新拋物線解析式，設(shè)點(diǎn)SKIPIF1<0，根據(jù)平行四邊形對(duì)角線互相平分分類討論根據(jù)中點(diǎn)坐標(biāo)公式即可得到答案．【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）解：如圖，延長SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，設(shè)直線SKIPIF1<0的函數(shù)表達(dá)式為SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0，∴直線SKIPIF1<0的函數(shù)表達(dá)式為SKIPIF1<0，設(shè)SKIPIF1<0，其中SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值，最大值為1，此時(shí)點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0；（3）解：∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∵拋物線沿射線SKIPIF1<0方向平移SKIPIF1<0個(gè)單位長度，∴拋物線向右平移SKIPIF1<0個(gè)單位長度，向上平移3個(gè)單位長度，∴平移后的拋物線解析式為SKIPIF1<0，∵點(diǎn)SKIPIF1<0在原拋物線對(duì)稱軸上，∴設(shè)點(diǎn)SKIPIF1<0，①當(dāng)以SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，∵點(diǎn)SKIPIF1<0為新拋物線上一點(diǎn)，∴SKIPIF1<0，②當(dāng)以SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，∵點(diǎn)SKIPIF1<0為新拋物線上一點(diǎn)，∴SKIPIF1<0，③當(dāng)以SKIPIF1<0為對(duì)角線時(shí)，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，∵點(diǎn)SKIPIF1<0為新拋物線上一點(diǎn)，∴SKIPIF1<0，綜上所述，點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．4．（2022·浙江寧波·?？寄M預(yù)測）如圖，直線SKIPIF1<0與雙曲線SKIPIF1<0交于A、B兩點(diǎn)，M是第一象限內(nèi)的雙曲線上任意一點(diǎn)．(1)若點(diǎn)A坐標(biāo)為SKIPIF1<0，求M點(diǎn)坐標(biāo)．(2)若SKIPIF1<0，連接SKIPIF1<0，若SKIPIF1<0的面積是34，求k值．(3)設(shè)直線SKIPIF1<0分別與x軸相交于P、Q兩點(diǎn)，且SKIPIF1<0，求SKIPIF1<0的值．【答案】(1)SKIPIF1<0；(2)SKIPIF1<0；(3)2【思路分析】（1）把點(diǎn)SKIPIF1<0代入SKIPIF1<0可求得反比例函數(shù)解析式，進(jìn)而可得點(diǎn)B的坐標(biāo)，設(shè)SKIPIF1<0，運(yùn)用勾股定理即可求得答案；（2）設(shè)SKIPIF1<0，則SKIPIF1<0，代入代入SKIPIF1<0可求得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，過點(diǎn)O作SKIPIF1<0交SKIPIF1<0于點(diǎn)D，過點(diǎn)B作SKIPIF1<0軸于點(diǎn)E，過點(diǎn)D作SKIPIF1<0軸于點(diǎn)F，可證得SKIPIF1<0，進(jìn)而求得點(diǎn)D的坐標(biāo)，利用待定系數(shù)法求得直線SKIPIF1<0的解析式，聯(lián)立方程組可求得點(diǎn)M的坐標(biāo)，再由SKIPIF1<0的面積是34，建立方程求解即可得出答案；（3）設(shè)SKIPIF1<0，代入SKIPIF1<0得：SKIPIF1<0，聯(lián)立方程組求出A、B兩點(diǎn)的坐標(biāo)，過點(diǎn)A、B、M分別作x軸的垂線SKIPIF1<0，垂足分別為G、K、H，過點(diǎn)M作x軸的平行線交SKIPIF1<0于R，交SKIPIF1<0于L，利用相似三角形性質(zhì)即可得出：SKIPIF1<0，SKIPIF1<0，再由SKIPIF1<0，得出：SKIPIF1<0，從而得出SKIPIF1<0的值．【詳解】（1）解：把點(diǎn)SKIPIF1<0代入SKIPIF1<0得：SKIPIF1<0，∴反比例函數(shù)解析式為SKIPIF1<0，∵SKIPIF1<0，∴由反比例函數(shù)與正比例函數(shù)圖象的對(duì)稱性可得點(diǎn)B坐標(biāo)為SKIPIF1<0，設(shè)SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，整理化簡得SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0（與A重合，舍去）或SKIPIF1<0（舍去）或SKIPIF1<0或SKIPIF1<0（舍去），∴SKIPIF1<0；（2）設(shè)SKIPIF1<0，則SKIPIF1<0，將SKIPIF1<0代入SKIPIF1<0，得：SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，如圖2，過點(diǎn)O作SKIPIF1<0交SKIPIF1<0于點(diǎn)D，過點(diǎn)B作SKIPIF1<0軸于點(diǎn)E，過點(diǎn)D作SKIPIF1<0軸于點(diǎn)F，則SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0是等腰直角三角形，∴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，∴直線SKIPIF1<0的解析式為SKIPIF1<0，聯(lián)立方程組，得：SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，，∵M(jìn)是第一象限內(nèi)的雙曲線上任意一點(diǎn)，∴SKIPIF1<0，∴SKIPIF1<0，過點(diǎn)A作SKIPIF1<0于點(diǎn)H，則SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0的面積是34，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0；（3）設(shè)SKIPIF1<0代入SKIPIF1<0得：SKIPIF1<0，∴SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，過點(diǎn)A、B、M分別作x軸的垂線SKIPIF1<0，垂足分別為G、K、H，過點(diǎn)M作x軸的平行線交SKIPIF1<0于R，交SKIPIF1<0于L，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0的值為2．5．（2022·山東濟(jì)南·模擬預(yù)測）如圖，在平面直角坐標(biāo)系中，點(diǎn)O為坐標(biāo)原點(diǎn)，直線l與拋物線SKIPIF1<0相交于SKIPIF1<0兩點(diǎn)．(1)求出拋物線的解析式；(2)在坐標(biāo)軸上是否存在點(diǎn)D，使得SKIPIF1<0是以線段SKIPIF1<0為斜邊的直角三角形？若存在，求出點(diǎn)D的坐標(biāo)；若不存在，說明理由；(3)點(diǎn)P是線段SKIPIF1<0上一動(dòng)點(diǎn)，（點(diǎn)P不與點(diǎn)A、B重合），過點(diǎn)P作SKIPIF1<0，交第一象限內(nèi)的拋物線于點(diǎn)M，過點(diǎn)M作SKIPIF1<0軸于點(diǎn)C，交SKIPIF1<0于點(diǎn)N，若SKIPIF1<0的面積SKIPIF1<0滿足SKIPIF1<0，求出SKIPIF1<0的值，并求出此時(shí)點(diǎn)M的坐標(biāo)．【答案】(1)SKIPIF1<0(2)存在，D點(diǎn)坐標(biāo)為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0(3)SKIPIF1<0，M點(diǎn)坐標(biāo)為SKIPIF1<0【思路分析】（1）利用待定系數(shù)法來求解；（2）分兩種情況來求解：點(diǎn)D在x軸上和點(diǎn)D在y軸上．當(dāng)點(diǎn)D在x軸上時(shí)，過點(diǎn)A作SKIPIF1<0軸于點(diǎn)D，易求D點(diǎn)的坐標(biāo)；當(dāng)點(diǎn)D在y軸上時(shí)，設(shè)SKIPIF1<0，在SKIPIF1<0中利用勾股定理可求得d的值，可的答案；（3）過P作SKIPIF1<0于點(diǎn)F，易證SKIPIF1<0，從而得到SKIPIF1<0，在SKIPIF1<0中和在SKIPIF1<0中利用三角函數(shù)得出SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，利用SKIPIF1<0和SKIPIF1<0之間的面積關(guān)系，進(jìn)而表示出M的坐標(biāo)，再根據(jù)M點(diǎn)在拋物線上求出a的值，進(jìn)而得到答案．【詳解】（1）解：∵SKIPIF1<0兩點(diǎn)在拋物線SKIPIF1<0的圖像上，∴SKIPIF1<0，解得SKIPIF1<0，∴拋物線解析式為SKIPIF1<0；（2）解：存在三個(gè)點(diǎn)滿足題意，理由如下：當(dāng)點(diǎn)D在x軸上時(shí)，如圖1，過點(diǎn)A作AD⊥x軸于點(diǎn)D，∵SKIPIF1<0，∴D坐標(biāo)為SKIPIF1<0；當(dāng)點(diǎn)D在y軸上時(shí)，設(shè)SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，∵SKIPIF1<0是以SKIPIF1<0為斜邊的直角三角形，∴SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，或SKIPIF1<0∴D點(diǎn)坐標(biāo)為SKIPIF1<0或SKIPIF1<0；綜上可知存在滿足條件的D點(diǎn)，其坐標(biāo)為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0；（3）解：如圖2，過P作SKIPIF1<0于點(diǎn)F，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴M點(diǎn)坐標(biāo)為SKIPIF1<0，又M點(diǎn)在拋物線上，代入可得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），SKIPIF1<0，SKIPIF1<0，∴點(diǎn)M的坐標(biāo)為SKIPIF1<0．6．（2022·甘肅嘉峪關(guān)·校考一模）如圖，已知拋物線SKIPIF1<0與x軸交于A、B兩點(diǎn)，與y軸交于點(diǎn)C，拋物線的對(duì)稱軸交x軸于點(diǎn)D，已知SKIPIF1<0，SKIPIF1<0．(1)求拋物線的表達(dá)式；(2)在拋物線的對(duì)稱軸上是否存在點(diǎn)P，使SKIPIF1<0是以SKIPIF1<0為腰的等腰三角形？如果存在，直接寫出P點(diǎn)的坐標(biāo)；如果不存在，請(qǐng)說明理由；(3)點(diǎn)E是線段SKIPIF1<0上的一個(gè)動(dòng)點(diǎn)，過點(diǎn)E作x軸的垂線與拋物線相交于點(diǎn)F，當(dāng)點(diǎn)E運(yùn)動(dòng)到什么位置時(shí)，四邊形SKIPIF1<0的面積最大？求出四邊形SKIPIF1<0的最大面積及此時(shí)E點(diǎn)的坐標(biāo)．【答案】(1)SKIPIF1<0(2)存在；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0(3)SKIPIF1<0，SKIPIF1<0【思路分析】（1）將點(diǎn)A、C的坐標(biāo)分別代入可得二元一次方程組，解方程組即可得出m、n的值；（2）根據(jù)二次函數(shù)的解析式可得對(duì)稱軸方程，由勾股定理求出SKIPIF1<0的值，以點(diǎn)C為圓心，SKIPIF1<0為半徑作弧，交對(duì)稱軸于SKIPIF1<0；以點(diǎn)D為圓心SKIPIF1<0為半徑作圓交對(duì)稱軸于點(diǎn)SKIPIF1<0，SKIPIF1<0，作SKIPIF1<0垂直于對(duì)稱軸于點(diǎn)H，由等腰三角形的性質(zhì)就可以求出結(jié)論；（3）由二次函數(shù)的解析式可求出B點(diǎn)的坐標(biāo)，從而可求出直線SKIPIF1<0的解析式，從而可設(shè)E點(diǎn)的坐標(biāo)SKIPIF1<0，進(jìn)而可表示出F的坐標(biāo)，由四邊形SKIPIF1<0的面積SKIPIF1<0可求出S與SKIPIF1<0的關(guān)系式，由二次函數(shù)的性質(zhì)就可以求出結(jié)論．【詳解】（1）解：已知拋物線SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0拋物線表達(dá)式為：SKIPIF1<0；（2）解：由（1）可知拋物線對(duì)稱軸為直線SKIPIF1<0，則點(diǎn)SKIPIF1<0坐標(biāo)為SKIPIF1<0，SKIPIF1<0的長為SKIPIF1<0,如圖1所示，使SKIPIF1<0是以SKIPIF1<0為腰的等腰三角形的點(diǎn)SKIPIF1<0有SKIPIF1<0，SKIPIF1<0，SKIPIF1<0三種情況，其中SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0,

垂足為點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，綜上可得，在拋物線的對(duì)稱軸上存在點(diǎn)P，使SKIPIF1<0是以SKIPIF1<0為腰的等腰三角形，P點(diǎn)的坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，；（3）解：根據(jù)題意作圖2，過點(diǎn)SKIPIF1<0作SKIPIF1<0,垂足為點(diǎn)SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故點(diǎn)SKIPIF1<0坐標(biāo)為SKIPIF1<0，SKIPIF1<0，設(shè)直線SKIPIF1<0解析式為SKIPIF1<0，過點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，則直線SKIPIF1<0解析式為SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，故SKIPIF1<0時(shí)，四邊形SKIPIF1<0的面積取得最大值為SKIPIF1<0，此時(shí)點(diǎn)SKIPIF1<0坐標(biāo)為SKIPIF1<0，．7．（2022·山東濟(jì)南·統(tǒng)考一模）如圖，拋物線SKIPIF1<0與x軸交于點(diǎn)A，B，與y軸交于點(diǎn)C，已知A，B兩點(diǎn)坐標(biāo)分別是SKIPIF1<0，SKIPIF1<0，連接SKIPIF1<0．(1)求拋物線的表達(dá)式；(2)將SKIPIF1<0沿SKIPIF1<0所在直線折疊，得到SKIPIF1<0，點(diǎn)A的對(duì)應(yīng)點(diǎn)D是否落在拋物線的對(duì)稱軸上？若點(diǎn)D在對(duì)稱軸上，請(qǐng)求出點(diǎn)D的坐標(biāo)；若點(diǎn)D不在對(duì)稱軸上，請(qǐng)說明理由；(3)若點(diǎn)P是拋物線位于第二象限圖象上的一動(dòng)點(diǎn)，連接SKIPIF1<0交SKIPIF1<0于點(diǎn)Q，連接BP，SKIPIF1<0的面積記為SKIPIF1<0，SKIPIF1<0的面積記為SKIPIF1<0，求SKIPIF1<0的值最大時(shí)點(diǎn)P的坐標(biāo)．【答案】(1)SKIPIF1<0(2)點(diǎn)SKIPIF1<0不在拋物線的對(duì)稱軸上，理由見解析(3)SKIPIF1<0【思路分析】（1）利用待定系數(shù)法可求得函數(shù)的表達(dá)式；（2）拋物線的表達(dá)式為SKIPIF1<0，可證明SKIPIF1<0，繼而可證SKIPIF1<0，則將SKIPIF1<0沿SKIPIF1<0所在直線折疊，點(diǎn)D一定落在直線SKIPIF1<0上，延長SKIPIF1<0至D，使SKIPIF1<0，過點(diǎn)D作SKIPIF1<0軸交y軸于點(diǎn)E，可證SKIPIF1<0，可得點(diǎn)D橫坐標(biāo)．則可判斷D點(diǎn)是否在拋物線對(duì)稱軸上；（3）先求出過點(diǎn)SKIPIF1<0、SKIPIF1<0的直線解析式，分別過A、P作x軸的垂線，利用解析式，用同一個(gè)字母m表示出P，N的坐標(biāo)，再證明SKIPIF1<0，進(jìn)而用m表示出SKIPIF1<0的值，根據(jù)二次函數(shù)的性質(zhì)可以確定出SKIPIF1<0的最大值，進(jìn)而可確定出此時(shí)的P點(diǎn)坐標(biāo)．【詳解】（1）解：∵拋物線SKIPIF1<0過點(diǎn)SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，解得：SKIPIF1<0，∴拋物線的表達(dá)式為SKIPIF1<0．（2）解：點(diǎn)SKIPIF1<0不在拋物線的對(duì)稱軸上，理由是：∵拋物線的表達(dá)式為SKIPIF1<0，∴點(diǎn)SKIPIF1<0坐標(biāo)為SKIPIF1<0．∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0．又∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．∴將SKIPIF1<0沿SKIPIF1<0所在直線折疊，點(diǎn)SKIPIF1<0一定落在直線SKIPIF1<0上，延長SKIPIF1<0至SKIPIF1<0，使SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0軸交SKIPIF1<0軸于點(diǎn)SKIPIF1<0．又∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0