中考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)專(zhuān)題43幾何中的最值問(wèn)題之和長(zhǎng)度有關(guān)的最值之函數(shù)法求最值 (含答案)_第1頁(yè)
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43第8章幾何中的最值問(wèn)題之和長(zhǎng)度有關(guān)的最值之函數(shù)法求最值一、選擇題1.如圖,將一張面積為20的大三角形紙片沿著虛線剪成三張小三角形紙片與一張平行四邊形紙片.根據(jù)圖中標(biāo)示的長(zhǎng)度,則平行四邊形紙片的最大面積為()A.5 B.10 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【分析】根據(jù)題意可知△AMN∽△ABC,可知相似比為SKIPIF1<0,再根據(jù)高之比也為相似比,表示出平行四邊形的高,再利用面積公式求得即可.【解答】由題意可知:MN∥BC,∴△AMN∽△ABC,SKIPIF1<0,而S△ABC=SKIPIF1<0,即:SKIPIF1<0,解得:AE=4,SKIPIF1<0,SKIPIF1<0平行四邊形=SKIPIF1<0,SKIPIF1<0,因此平行四邊形紙片的最大面積為10,故選B.【點(diǎn)評(píng)】本題考查相似的性質(zhì),平行四邊形的性質(zhì)與面積計(jì)算,計(jì)算量較大.2.已知x=m是一元二次方程x2+2x+n-3=0的一個(gè)根,則m+n的最大值等于()A.SKIPIF1<0 B.4 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【分析】先利用一元二次方程的根的判別式、根的定義求出m的取值范圍和SKIPIF1<0,再利用二次函數(shù)的性質(zhì)求最值即可得.【解答】由題意得:此方程的根的判別式SKIPIF1<0,解得SKIPIF1<0,SKIPIF1<0是一元二次方程SKIPIF1<0的一個(gè)根,SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0對(duì)于任意實(shí)數(shù)m,SKIPIF1<0均成立,令SKIPIF1<0,整理得:SKIPIF1<0,由二次函數(shù)的性質(zhì)可知,當(dāng)SKIPIF1<0時(shí),y取得最大值,最大值為SKIPIF1<0,即SKIPIF1<0的最大值等于SKIPIF1<0,故選:A.【點(diǎn)評(píng)】本題考查了一元二次方程的根的判別式、根的定義、二次函數(shù)的最值,熟練掌握二次函數(shù)的性質(zhì)是解題關(guān)鍵.3.在平面直角坐標(biāo)系中,已知A(2,4),P(1,0),B為y軸上的動(dòng)點(diǎn),以AB為邊構(gòu)造△ABC,使點(diǎn)C在x軸上,∠BAC=90°,M為BC的中點(diǎn),則PM的最小值為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【分析】作AH⊥y軸,CE⊥AH,證明△AHB∽△CEA,根據(jù)相似三角形的性質(zhì)得到AE=2BH,求出點(diǎn)M的坐標(biāo),根據(jù)兩點(diǎn)間的距離公式用x表示出PM,根據(jù)二次函數(shù)的性質(zhì)解答即可.【解答】解:如圖,過(guò)點(diǎn)A作AH⊥y軸于H,過(guò)點(diǎn)C作CE⊥AH于E,則四邊形CEHO是矩形,∴OH=CE=4,∵∠BAC=∠AHB=∠AEC=90°,∴∠ABH+∠HAB=90°,∠HAB+∠EAC=90°,∴∠ABH=∠EAC,∴△AHB∽△CEA,∴SKIPIF1<0,即SKIPIF1<0∴AE=2BH,設(shè)BH=x,則AE=2x,∵A(2,4),∴OC=HE=2+2x,OB=4﹣x,∴B(0,4﹣x),C(2+2x,0),∵M(jìn)為BC的中點(diǎn),∴BM=CM,∴M(1+x,SKIPIF1<0),∵P(1,0),∴PM=SKIPIF1<0,∴當(dāng)SKIPIF1<0時(shí),PM有最小值為SKIPIF1<0=SKIPIF1<0,故選:C.【點(diǎn)評(píng)】本題考查了坐標(biāo)與圖形性質(zhì)、矩形的判定與性質(zhì)、同角的余角相等、相似三角形的判定與性質(zhì)、兩點(diǎn)間的距離公式、二次函數(shù)的性質(zhì)等知識(shí),認(rèn)真分析圖形,借助作輔助線,利用相似三角形的性質(zhì)及二次函數(shù)的最值求解是解答的關(guān)鍵.4.一塊矩形木板ABCD,長(zhǎng)AD=3cm,寬AB=2cm,小虎將一塊等腰直角三角板的一條直角邊靠在頂點(diǎn)C上,另一條直角邊與AB邊交于點(diǎn)E,三角板的直角頂點(diǎn)P在AD邊上移動(dòng)(不含端點(diǎn)A、D),當(dāng)線段BE最短時(shí),AP的長(zhǎng)為()A.SKIPIF1<0cm B.1cm C.SKIPIF1<0cm D.2cm【答案】C【分析】設(shè)SKIPIF1<0,SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0,構(gòu)建二次函數(shù)即可解決問(wèn)題;【解答】設(shè)BE=y,AP=x,∵四邊形ABCD是矩形,∵SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,∵SKIPIF1<0,∴SKIPIF1<0時(shí),y有最小值.故答案選C.【點(diǎn)評(píng)】本題主要考查了求解二次函數(shù)的最值問(wèn)題,相似三角形的判定與性質(zhì)是解題的關(guān)鍵.5.如圖,線段AB的長(zhǎng)為2,C為AB上一動(dòng)點(diǎn),分別以AC、BC為斜邊在AB的同側(cè)作兩個(gè)直角SKIPIF1<0、SKIPIF1<0,其中∠A=30°,∠B=60°,則DE的最小值為()A.1 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【分析】設(shè)SKIPIF1<0,則SKIPIF1<0,再根據(jù)直角三角形的性質(zhì)、勾股定理分別求出CD、CE的長(zhǎng),然后根據(jù)角的和差可得SKIPIF1<0,從而利用勾股定理可得SKIPIF1<0,最后利用二次函數(shù)的性質(zhì)求解即可得.【解答】如圖,連接DE設(shè)SKIPIF1<0,則SKIPIF1<0,且SKIPIF1<0,即SKIPIF1<0SKIPIF1<0在SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0在SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0在SKIPIF1<0中,SKIPIF1<0整理得:SKIPIF1<0由二次函數(shù)的性質(zhì)可知,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0隨x的增大而減??;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0隨x的增大而增大則當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得最小值,最小值為SKIPIF1<0因此,DE的最小值為SKIPIF1<0故選:C.【點(diǎn)評(píng)】本題考查了直角三角形的性質(zhì)、勾股定理、二次函數(shù)的性質(zhì)等知識(shí)點(diǎn),利用勾股定理得出SKIPIF1<0關(guān)于x的表達(dá)式是解題關(guān)鍵.二、填空題6.已知,四邊形ABCD的兩條對(duì)角線AC、BD互相垂直,且AC+BD=10,當(dāng)AC=_______時(shí),四邊形ABCD的面積最大,最大值為_(kāi)_________.【答案】512.5【分析】根據(jù)已知設(shè)四邊形ABCD面積為S,AC為SKIPIF1<0,則SKIPIF1<0,進(jìn)而求出SKIPIF1<0,再求出最值即可.【解答】解:設(shè)SKIPIF1<0,四邊形ABCD面積為S,則SKIPIF1<0,

則:SKIPIF1<0,

∵SKIPIF1<0,∴S有最大值,

當(dāng)SKIPIF1<0時(shí),四邊形ABCD的面積最大,

即當(dāng)SKIPIF1<0時(shí),四邊形ABCD面積最大,SKIPIF1<0,

故答案為:5,12.5.【點(diǎn)評(píng)】本題主要考查了二次函數(shù)的應(yīng)用,根據(jù)已知正確得出二次函數(shù)關(guān)系是解題關(guān)鍵.7.小敏用一根長(zhǎng)為8cm的細(xì)鐵絲圍成矩形,則矩形的最大面積是_________.【答案】4cm2【分析】設(shè)圍成的矩形的一邊長(zhǎng)為xcm,圍成的矩形面積為Scm2,根據(jù)矩形面積公式可得S與x的關(guān)系式,再根據(jù)二次函數(shù)的性質(zhì)解答即可.【解答】解:設(shè)圍成的矩形的一邊長(zhǎng)為xcm,圍成的矩形面積為Scm2,則另一邊長(zhǎng)為(4-x)cm,根據(jù)題意,得:SKIPIF1<0,∴當(dāng)x=2時(shí),S最大=4cm2.故答案為:4cm2.【點(diǎn)評(píng)】本題考查了二次函數(shù)的應(yīng)用,屬于??碱}型,正確列出函數(shù)關(guān)系式、熟練掌握二次函數(shù)的性質(zhì)是解題的關(guān)鍵.8.如圖,已知SKIPIF1<0,SKIPIF1<0為線段SKIPIF1<0上一個(gè)動(dòng)點(diǎn),分別以SKIPIF1<0、SKIPIF1<0為邊在SKIPIF1<0同側(cè)作菱形SKIPIF1<0和菱形SKIPIF1<0,點(diǎn)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0在同一條直線上,SKIPIF1<0.SKIPIF1<0、SKIPIF1<0別是對(duì)角線SKIPIF1<0、SKIPIF1<0的中點(diǎn),當(dāng)點(diǎn)SKIPIF1<0在線段SKIPIF1<0上移動(dòng)時(shí),則點(diǎn)SKIPIF1<0、SKIPIF1<0之間的最短距離為_(kāi)_____.【答案】SKIPIF1<0【分析】連接PM、PN.首先證明∠MPN=90°,再根據(jù)含30°的直角三角形的特點(diǎn)求出SKIPIF1<0,SKIPIF1<0,再利用勾股定理求得MN2與AP2的關(guān)系,根據(jù)二次函數(shù)的性質(zhì)即可解決問(wèn)題.【解答】如解圖,連接SKIPIF1<0,SKIPIF1<0.∵四邊形APCD,四邊形PBFE是菱形,∠DAP=60°,

∴∠APC=120°,∠EPB=60°,∵SKIPIF1<0、SKIPIF1<0分別是SKIPIF1<0,SKIPIF1<0的中點(diǎn),SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.∴SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.∴SKIPIF1<0,即SKIPIF1<0SKIPIF1<0.∴當(dāng)SKIPIF1<0時(shí),SKIPIF1<0有最小值SKIPIF1<0,∴點(diǎn)SKIPIF1<0、SKIPIF1<0之間的最短距離為SKIPIF1<0.故答案為:SKIPIF1<0.【點(diǎn)評(píng)】本題考查菱形的性質(zhì),含30°的直角三角形,勾股定理,二次函數(shù)與幾何問(wèn)題.解題的關(guān)鍵是學(xué)會(huì)添加常用輔助線,構(gòu)建二次函數(shù)解決最值問(wèn)題.9.如圖,P是拋物線y=x2﹣x﹣4在第四象限的一點(diǎn),過(guò)點(diǎn)P分別向x軸和y軸作垂線,垂足分別為A、B,則四邊形OAPB周長(zhǎng)的最大值為_(kāi)____.【答案】10【分析】設(shè)P(x,x2﹣x﹣4)根據(jù)矩形的周長(zhǎng)公式得到C=﹣2(x﹣1)2+10.根據(jù)二次函數(shù)的性質(zhì)來(lái)求最值即可.【解答】解:設(shè)P(x,x2﹣x﹣4),四邊形OAPB周長(zhǎng)=2PA+2OA=﹣2(x2﹣x﹣4)+2x=﹣2x2+4x+8=﹣2(x﹣1)2+10,當(dāng)x=1時(shí),四邊形OAPB周長(zhǎng)有最大值,最大值為10.故答案為10.【點(diǎn)評(píng)】本題考查了二次函數(shù)圖象上點(diǎn)的坐標(biāo)特征:二次函數(shù)圖象上點(diǎn)的坐標(biāo)滿(mǎn)足其解析式.也考查了二次函數(shù)的性質(zhì).10.在平面直角坐標(biāo)系中,點(diǎn)A的坐標(biāo)為(1,0),點(diǎn)B的坐標(biāo)為(m,5﹣m),當(dāng)AB的長(zhǎng)最小時(shí),m的值為_(kāi)____【答案】3【分析】先根據(jù)兩點(diǎn)間的距離公式求出AB2=2m2﹣12m+26,利用配方法得到AB2=2(m﹣3)2+8,根據(jù)二次函數(shù)的性質(zhì)即可求解.【解答】解:∵點(diǎn)A的坐標(biāo)為(1,0),點(diǎn)B的坐標(biāo)為(m,5﹣m),∴AB2=(m﹣1)2+(5﹣m﹣0)2=m2﹣2m+1+25﹣10m+m2=2m2﹣12m+26=2(m﹣3)2+8,∵2>0,∴當(dāng)m=3時(shí),AB2最小,∵當(dāng)AB2最小時(shí),AB的長(zhǎng)最?。蚀鸢笧椋?.【點(diǎn)評(píng)】本題主要考查的是二次函數(shù)求最值,利用配方法求最值是解題的關(guān)鍵.三、解答題11.已知拋物線SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0,且SKIPIF1<0.(1)求拋物線的解析式及頂點(diǎn)SKIPIF1<0的坐標(biāo);(2)若SKIPIF1<0,SKIPIF1<0均在該拋物線上,且SKIPIF1<0,求SKIPIF1<0點(diǎn)橫坐標(biāo)SKIPIF1<0的取值范圍;(3)點(diǎn)SKIPIF1<0為拋物線在直線SKIPIF1<0下方圖象上的一動(dòng)點(diǎn),當(dāng)SKIPIF1<0面積最大時(shí),求點(diǎn)SKIPIF1<0的坐標(biāo).【答案】(1)SKIPIF1<0,頂點(diǎn)SKIPIF1<0;(2)SKIPIF1<0;(3)SKIPIF1<0.【分析】(1)把SKIPIF1<0代入SKIPIF1<0即可求出a,化為頂點(diǎn)式即可得到頂點(diǎn);(2)根據(jù)函數(shù)圖像及對(duì)稱(chēng)軸SKIPIF1<0即可求解;(3)先求出直線SKIPIF1<0的表達(dá)式,過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸的平行線交SKIPIF1<0于點(diǎn)SKIPIF1<0,設(shè)點(diǎn)SKIPIF1<0,得到點(diǎn)SKIPIF1<0,表示出PH,再根據(jù)SKIPIF1<0列出函數(shù),根據(jù)二次函數(shù)最值即可求出P點(diǎn)坐標(biāo).【解答】解:(1)把SKIPIF1<0代入SKIPIF1<0,即SKIPIF1<0,解得:SKIPIF1<0,故拋物線的表達(dá)式為:SKIPIF1<0,SKIPIF1<0=SKIPIF1<0則頂點(diǎn)SKIPIF1<0.(2)由(1)知拋物線的對(duì)稱(chēng)軸SKIPIF1<0,所以點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱(chēng)點(diǎn)SKIPIF1<0在拋物線上∵SKIPIF1<0∴SKIPIF1<0的取值范圍為SKIPIF1<0(3)令y=0,即SKIPIF1<0=0,解得x1=1,x2=3,∴C(3,0)將點(diǎn)SKIPIF1<0、SKIPIF1<0的坐標(biāo)代入一次函數(shù)表達(dá)式:SKIPIF1<0得SKIPIF1<0解得:SKIPIF1<0∴直線SKIPIF1<0的表達(dá)式為:SKIPIF1<0,過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸的平行線交SKIPIF1<0于點(diǎn)SKIPIF1<0,

設(shè)點(diǎn)SKIPIF1<0,則點(diǎn)SKIPIF1<0,∴SKIPIF1<0則SKIPIF1<0,∵SKIPIF1<0,故SKIPIF1<0有最大值,此時(shí)SKIPIF1<0,故點(diǎn)SKIPIF1<0.【點(diǎn)評(píng)】此題主要考查二次函數(shù)綜合,解題的關(guān)鍵是熟知二次函數(shù)的圖象與性質(zhì)、最值的求解方法.12.如圖,四邊形ABDC為矩形,AB=4,AC=3,點(diǎn)M為邊AB上一點(diǎn)(點(diǎn)M不與點(diǎn)A、B重合),連接CM,過(guò)點(diǎn)M作MN⊥MC,MN與邊BD交于點(diǎn)N.(1)當(dāng)點(diǎn)M為邊AB的中點(diǎn)時(shí),求線段BN的長(zhǎng);(2)直接寫(xiě)出:當(dāng)DN最小時(shí)△MNB的面積為_(kāi)__________.【答案】(1)BN=SKIPIF1<0;(2)SKIPIF1<0【分析】(1)由矩形的性質(zhì)及“一線三等角“推得∠ACM=∠BMN,利用有兩個(gè)角相等的三角形相似,可證得△ACM∽△BMN,利用相似三角形的性質(zhì)可得比例式,將相關(guān)數(shù)據(jù)代入即可求得BN的值;(2)設(shè)BM=x,DN=y(tǒng),根據(jù)SKIPIF1<0,得出y關(guān)于x的二次函數(shù),將其寫(xiě)成頂點(diǎn)式,根據(jù)二次函數(shù)的性質(zhì)可得DN最小時(shí)相應(yīng)的x值及y值,再利用三角形的面積公式求得答案即可.【解答】解:(1)∵AB=4,∴當(dāng)點(diǎn)M為邊AB的中點(diǎn)時(shí),AM=BM=2,∵四邊形ABDC為矩形,∴∠A=∠B=90°,∵M(jìn)N⊥MC,∴∠CMN=90°,∵∠ACM+∠AMC=90°,∠BMN+∠AMC=180°﹣∠CMN=90°,∴∠ACM=∠BMN,又∵∠A=∠B,∴△ACM∽△BMN,∴SKIPIF1<0,∵AC=3,AM=BM=2,∴SKIPIF1<0,∴BN=SKIPIF1<0;(2)設(shè)BM=x,DN=y(tǒng),∵四邊形ABDC為矩形,AB=4,AC=3,∴AM=AB﹣BM=4﹣x,BN=BD﹣DN=3﹣y,由(1)知,SKIPIF1<0,∴SKIPIF1<0,∴(4﹣x)x=3(3﹣y),∴﹣x2+4x=9﹣3y,∴y=SKIPIF1<0x2﹣SKIPIF1<0x+3=SKIPIF1<0(x﹣2)2+SKIPIF1<0,∴當(dāng)x=2時(shí),y取得最小值,即DN最小,此時(shí)DN=y(tǒng)=SKIPIF1<0,∴BM=2,BN=3﹣SKIPIF1<0=SKIPIF1<0,∴△MNB的面積為:SKIPIF1<0×2×SKIPIF1<0=SKIPIF1<0.故答案為:SKIPIF1<0.【點(diǎn)評(píng)】本題考查相似三角形的性質(zhì)和判定,矩形的性質(zhì),二次函數(shù)與幾何問(wèn)題.(1)能證明相似,并通過(guò)相似的性質(zhì)列出比例式是解題關(guān)鍵;(2)能借助相似列出y與x的表達(dá)式是解題關(guān)鍵.13.如圖,在平面直角坐標(biāo)系中,菱形SKIPIF1<0的頂點(diǎn)SKIPIF1<0在SKIPIF1<0軸正半軸上,頂點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,SKIPIF1<0是拋物線SKIPIF1<0上一點(diǎn),且在SKIPIF1<0軸上方,則SKIPIF1<0面積的最大值是多少?【答案】15【分析】SKIPIF1<0是拋物線SKIPIF1<0上一點(diǎn),設(shè)SKIPIF1<0;根據(jù)頂點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0以及菱形SKIPIF1<0的頂點(diǎn)SKIPIF1<0在SKIPIF1<0軸正半軸上,得SKIPIF1<0且SKIPIF1<0軸,從而得到SKIPIF1<0關(guān)系式,再根據(jù)二次函數(shù)的性質(zhì),即可得到SKIPIF1<0最大值.【解答】∵SKIPIF1<0是拋物線SKIPIF1<0上一點(diǎn)∴設(shè)SKIPIF1<0∵頂點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0∴SKIPIF1<0∵四邊形SKIPIF1<0是菱形,且頂點(diǎn)SKIPIF1<0在SKIPIF1<0軸正半軸上∴SKIPIF1<0,SKIPIF1<0軸∴SKIPIF1<0∵SKIPIF1<0∴SKIPIF1<0有最大值當(dāng)SKIPIF1<0時(shí),SKIPIF1<0最大值為15.【點(diǎn)評(píng)】本題考查了菱形、二次函數(shù)、勾股定理的知識(shí);解題的關(guān)鍵是熟練掌握菱形、二次函數(shù)的性質(zhì),從而完成求解.14.如圖,點(diǎn)E,F(xiàn),G,H分別在菱形ABCD的四條邊上,BE=BF=DG=DH,連接EF,F(xiàn)G,GH,HE,得到四邊形EFGH.(1)求證:四邊形EFGH是矩形;(2)若AB=2,∠A=60°,當(dāng)BE為何值時(shí),矩形EFGH的面積最大?【答案】(1)見(jiàn)解析;(2)當(dāng)BE=1時(shí),矩形EFGH的面積最大.【分析】(1)利用等腰三角形的性質(zhì):等邊對(duì)等角,以及平行線的性質(zhì)可以證得∠DGH+∠CGH=90°,則∠HGF=90°,根據(jù)三個(gè)角是直角的四邊形是矩形,即可證得;(2)設(shè)BE的長(zhǎng)是x,則利用x表示出矩形EFGH的面積,根據(jù)函數(shù)的性質(zhì)即可求解.【解答】(1)證明:∵DG=DH,∴∠DHG=∠DGH=SKIPIF1<0,同理,∠CGF=SKIPIF1<0,∴∠DGH+∠CGF=SKIPIF1<0,又∵在菱形ABCD中,AD∥BC,∴∠D+∠C=180°,∴∠DGH+∠CGF=90°,∴∠HGF=90°,同理,∠GHE=∠EFG=90°,∴四邊形EFGH是矩形;(2)如圖,連接BD,∵四邊形ABCD是菱形,∠A=60°,∴AD=AB,∴△ABD和△BCD是等邊三角形,∴BD=AB=2,∴S△BCD=S△ABD=SKIPIF1<0AB2=SKIPIF1<0,則菱形ABCD的面積是2SKIPIF1<0,設(shè)BE=x,則AE=2-x,∵BE=DH,AB=AD,∴AH=AE,∵∠A=60°,∴△AEH是等邊三角形,∴EH=AE=2-x在Rt△BME中,∠ABD=60°,BE=x,∴EM=SKIPIF1<0x∴EF=2EM=SKIPIF1<0x則矩形EFGH的面積y=HE×EF=(2-x)×SKIPIF1<0x=-SKIPIF1<0(x2-2x)=-SKIPIF1<0(x-1)2+SKIPIF1<0,∴當(dāng)x=1時(shí),矩形EFGH的面積最大,即當(dāng)BE=1時(shí),矩形EFGH的面積最大.【點(diǎn)評(píng)】本題考查了菱形的性質(zhì),矩形的判定以及二次函數(shù)的性質(zhì),正確利用x表示出矩形EFGH的面積是關(guān)鍵.15.如圖,直線l:SKIPIF1<0與x軸、y軸分別相交于A、B兩點(diǎn),拋物線SKIPIF1<0經(jīng)過(guò)點(diǎn)B.(1)求該拋物線的函數(shù)表達(dá)式:(2)已知點(diǎn)M是拋物線上的一個(gè)動(dòng)點(diǎn),并且點(diǎn)M在第一象限內(nèi),連接AM、BM,設(shè)點(diǎn)M的橫坐標(biāo)為m,△ABM的面積為S,求S與m的函數(shù)表達(dá)式,并求出S的最大值.【答案】(1)SKIPIF1<0;(2)SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得最大值SKIPIF1<0.【分析】(1)根據(jù)題意先求出點(diǎn)B的坐標(biāo),然后代入二次函數(shù)解析式求解即可;(2)由題意可求點(diǎn)A坐標(biāo),連接SKIPIF1<0,由題意知,點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,則有SKIPIF1<0,然后根據(jù)割補(bǔ)法求面積即可.【解答】解:(1)把SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0,∴SKIPIF1<0.把SKIPIF1<0代入SKIPIF1<0,得SKIPIF1<0,∴SKIPIF1<0.∴拋物線的解析式為SKIPIF1<0;(2)令SKIPIF1<0,則SKIPIF1<0,解得SKIPIF1<0或3,∴拋物線與SKIPIF1<0軸的交點(diǎn)橫坐標(biāo)分別為SKIPIF1<0和3.∵點(diǎn)SKIPIF1<0在拋物線上,且在第一象限內(nèi),∴SKIPIF1<0.將SKIPIF1<0代入SKIPIF1<0,得SKIPIF1<0,解得SKIPIF1<0,∴SKIPIF1<0.如解圖,連接SKIPIF1<0,由題意知,點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0,則SKIPIF1<0SKIPIF1<0,∵SKIPIF1<0,且SKIPIF1<0,∴當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得最大值SKIPIF1<0.【點(diǎn)評(píng)】本題主要考查二次函數(shù)的綜合,熟練掌握二次函數(shù)的性質(zhì)是解題的關(guān)鍵.16.已知:如圖,在△ABC中,∠B=90°,AB=5cm,BC=7cm.點(diǎn)P從點(diǎn)A開(kāi)始沿AB邊向點(diǎn)B以1cm/s的速度移動(dòng),同時(shí)點(diǎn)Q從點(diǎn)B開(kāi)始沿BC邊向點(diǎn)C以2cm/s的速度移動(dòng).當(dāng)一個(gè)點(diǎn)到達(dá)終點(diǎn)時(shí)另一點(diǎn)也隨之停止運(yùn)動(dòng),設(shè)運(yùn)動(dòng)時(shí)間為t秒,(1)求幾秒后,△PBQ的面積等于6cm2?(2)P、Q在運(yùn)動(dòng)過(guò)程中,是否存在時(shí)間t,使得△PBQ的面積最大,若存在求出時(shí)間t和最大面積,若不存在,說(shuō)明理由.【答案】(1)2或3秒后△PBQ的面積等于6cm2;(2)存在,當(dāng)t=SKIPIF1<0時(shí),△PBQ面積最大為SKIPIF1<0cm2【分析】(1)先用t的代數(shù)式表示出BP和BQ,再根據(jù)三角形的面積公式即得關(guān)于t的方程,解方程即得結(jié)果;(2)根據(jù)三角形的面積公式可用t的代數(shù)式表示出S△PBQ,再利用配方法和非負(fù)數(shù)的性質(zhì)解答即可.【解答】解:(1)設(shè)經(jīng)過(guò)t秒以后△PBQ面積為6,根據(jù)題意,得:SKIPIF1<0×(5﹣t)×2t=6,整理得:t2﹣5t+6=0,解得:t=2或t=3;答:2或3秒后△PBQ的面積等于6cm2;(2)存在;由題意得:S△PBQ=SKIPIF1<0BP?PQ=SKIPIF1<0×(5﹣t)×2t=﹣t2+5t=﹣(t﹣SKIPIF1<0)2+SKIPIF1<0,因?yàn)?t﹣SKIPIF1<0)2≥0,所以﹣(t﹣SKIPIF1<0)2≤0,因此,當(dāng)t=SKIPIF1<0時(shí),△PBQ面積最大為SKIPIF1<0cm2.【點(diǎn)評(píng)】本題考查了一元二次方程的應(yīng)用,屬于常見(jiàn)題型,正確理解題意、熟練掌握一元二次方程的相關(guān)知識(shí)是解題的關(guān)鍵.17.已知如圖,二次函數(shù)y=-x2+2x+m的圖象過(guò)點(diǎn)B(0,3),與x軸正半軸交于點(diǎn)A(1)求二次函數(shù)的解析式;(2)求點(diǎn)A的坐標(biāo);(3)若點(diǎn)C為拋物線上位于直線BA上方的一動(dòng)點(diǎn)(不與點(diǎn)A和點(diǎn)B重合),過(guò)點(diǎn)C作CD⊥x軸交直線BA于點(diǎn)D.請(qǐng)問(wèn):是否存在一點(diǎn)C,使線段CD的長(zhǎng)度最大?若不存在,請(qǐng)說(shuō)明理由;若存在,請(qǐng)求點(diǎn)C的坐標(biāo)和線段CD長(zhǎng)度的最大值.【答案】(1)y=-x2+2x+3;(2)(3,0);(3)存在,點(diǎn)C的坐標(biāo)(SKIPIF1<0,SKIPIF1<0),線段CD長(zhǎng)度的最大值為SKIPIF1<0【分析】(1)利用待定系數(shù)法即可求解;(2)令y=0,再解方程-x2+2x+3=0即可求得點(diǎn)A坐標(biāo);(3)先求得直線AB的解析式,點(diǎn)C坐標(biāo)為(n,-n2+2n+3),則點(diǎn)D坐標(biāo)為(n,-n+3),利用二次函數(shù)的性質(zhì)即可求解.【解答】(1)∵二次函數(shù)y=-x2+2x+m的圖象過(guò)點(diǎn)B(0,3),∴m=3,∴二次函數(shù)的解析式為y=-x2+2x+3;(2)把y=0代入解析式y(tǒng)=-x2+2x+3中,得:-x2+2x+3=0,解得:x1=3,x2=-1(舍負(fù)),∴點(diǎn)A坐標(biāo)為(3,0);(3)存在一點(diǎn)C,使線段CD的長(zhǎng)度最大.設(shè)過(guò)點(diǎn)A(3,0)和點(diǎn)B(0,3)的一次函數(shù)的解析式為:SKIPIF1<0,∴SKIPIF1<0,解得:SKIPIF1<0∴直線AB的解析式為SKIPIF1<0,∵點(diǎn)C在二次函數(shù)y=-x2+2x+3的圖象上,∴設(shè)點(diǎn)C坐標(biāo)為(n,-n2+2n+3),則點(diǎn)D坐標(biāo)為(n,-n+3),∴CD=(-n2+2n+3)-(-n+3)=-n2+3n=SKIPIF1<0,∴當(dāng)n=SKIPIF1<0時(shí),線段CD的長(zhǎng)度最大,線段CD長(zhǎng)度的最大值為SKIPIF1<0,此時(shí)點(diǎn)C坐標(biāo)為(SKIPIF1<0,SKIPIF1<0).【點(diǎn)評(píng)】本題考查了待定系數(shù)法求函數(shù)解析式,二次函數(shù)的性質(zhì),熟練掌握二次函數(shù)的性質(zhì)是解題的關(guān)鍵.18.如圖,二次函數(shù)SKIPIF1<0的圖象過(guò)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0三點(diǎn)(1)求二次函數(shù)的解析式;(2)若線段OB的垂直平分線與y軸交于點(diǎn)C,與二次函數(shù)的圖象在x軸上方的部分相交于點(diǎn)D,求直線CD的解析式;(3)在直線CD下方的二次函數(shù)的圖象上有一動(dòng)點(diǎn)P,過(guò)點(diǎn)P作SKIPIF1<0軸,交直線CD于Q,當(dāng)線段PQ的長(zhǎng)最大時(shí),求點(diǎn)P的坐標(biāo).【答案】(1)SKIPIF1<0;(2)y=-SKIPIF1<0x+SKIPIF1<0;(3)(-SKIPIF1<0,SKIPIF1<0).【分析】(1)根據(jù)待定系數(shù)法即可求解;(2)先求出直線OB的解析式為y=SKIPIF1<0x與線段OB的中點(diǎn)E的坐標(biāo),可設(shè)直線CD的解析式為y=SKIPIF1<0x+m,再把E點(diǎn)代入即可求出直線CD的解析式;(3)設(shè)P的橫坐標(biāo)為t,先聯(lián)立直線CD與拋物線得到D點(diǎn)的橫坐標(biāo),得到t的取值,再得到線段PQ關(guān)于t的關(guān)系式,利用二次函數(shù)的性質(zhì)即可求解.【解答】(1)把SKIPIF1<0、SKIPIF1<0、SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0解得SKIPIF1<0∴二次函數(shù)的解析式為SKIPIF1<0;(2)如圖,∵SKIPIF1<0,SKIPIF1<0∴其中點(diǎn)E的坐標(biāo)為SKIPIF1<0設(shè)直線OB的解析式為y=kx把SKIPIF1<0代入得SKIPIF1<0解得k=SKIPIF1<0∴直線OB的解析式為y=SKIPIF1<0x,∵直線CD垂直平分OB,∴可設(shè)直線CD的解析式為y=-SKIPIF1<0x+m,把ESKIPIF1<0代入得SKIPIF1<0解得m=SKIPIF1<0∴直線CD的解析式為y=-SKIPIF1<0x+SKIPIF1<0;(3)聯(lián)立SKIPIF1<0得到SKIPIF1<0解得x1=-SKIPIF1<0,x2=1,設(shè)P的橫坐標(biāo)為t,則P(t,SKIPIF1<0),∵過(guò)點(diǎn)P作SKIPIF1<0軸,交直線CD于Q,∴Q(t,-SKIPIF1<0t+SKIPIF1<0)∴PQ=(-SKIPIF1<0t+SKIPIF1<0)-(SKIPIF1<0)=-SKIPIF1<0故當(dāng)t=-SKIPIF1<0時(shí)PQ有最大值SKIPIF1<0此時(shí)P的坐標(biāo)為(-SKIPIF1<0,SKIPIF1<0).【點(diǎn)評(píng)】此題主要考查二次函數(shù)綜合,解題的關(guān)鍵是熟知待定系數(shù)法、二次函數(shù)的圖像與性質(zhì).19.如圖,二次函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0,以SKIPIF1<0為邊在SKIPIF1<0軸上方作正方形SKIPIF1<0,點(diǎn)SKIPIF1<0是SKIPIF1<0軸上一動(dòng)點(diǎn),連接SKIPIF1<0,過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0的垂線與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0.(1)求拋物線的函數(shù)關(guān)系式;(2)當(dāng)點(diǎn)SKIPIF1<0在線段SKIPIF1<0(點(diǎn)SKIPIF1<0不與SKIPIF1<0、SKIPIF1<0重合)上運(yùn)動(dòng)至何處時(shí),線段SKIPIF1<0的長(zhǎng)有最大值?并求出這個(gè)最大值.【答案】(1)SKIPIF1<0;(2)當(dāng)SKIPIF1<0運(yùn)動(dòng)到SKIPIF1<0處時(shí),SKIPIF1<0有最大值SKIPIF1<0【分析】(1)將點(diǎn)A、B的坐標(biāo)代入二次函數(shù)表達(dá)式,即可求解;

(2)設(shè)OP=a,則PB=3-a,由△POE∽△CBP得出比例線段,可表示OE的長(zhǎng),利用二次函數(shù)的性質(zhì)可求出線段OE的最大值;【解

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