廣西柳州市2019年中考數(shù)學(xué)專(zhuān)題訓(xùn)練05函數(shù)與幾何圖形的綜合.docx

專(zhuān)題訓(xùn)練(五)函數(shù)與幾何圖形的綜合1.2017濟(jì)寧已知函數(shù)y=mx2-(2m-5)x+m-2的圖象與x軸有兩個(gè)公共點(diǎn).(1)求m的取值范圍,并寫(xiě)出當(dāng)m取范圍內(nèi)最大整數(shù)時(shí)函數(shù)的解析式;(2)題(1)中求得的函數(shù)記為C1.當(dāng)nx-1時(shí),y的取值范圍是1y-3n,求n的值;函數(shù)C2:y=m(x-h)2+k的圖象由函數(shù)C1的圖象平移得到,其頂點(diǎn)P落在以原點(diǎn)為圓心,半徑為5的圓內(nèi)或圓上.設(shè)函數(shù)C1的圖象頂點(diǎn)為M,求點(diǎn)P與點(diǎn)M距離最大時(shí)函數(shù)C2的解析式.2.2017攀枝花改編如圖ZT5-1,拋物線(xiàn)y=x2+bx+c與x軸交于A,B兩點(diǎn),B點(diǎn)坐標(biāo)為(3,0),與y軸交于點(diǎn)C(0,3).圖ZT5-1 (1)求拋物線(xiàn)的解析式;(2)點(diǎn)P在x軸下方的拋物線(xiàn)上,過(guò)點(diǎn)P的直線(xiàn)y=x+m與直線(xiàn)BC交于點(diǎn)E,與y軸交于點(diǎn)F,求PE+EF的最大值.(3)點(diǎn)D為拋物線(xiàn)對(duì)稱(chēng)軸上一點(diǎn).當(dāng)BCD是以BC為直角邊的直角三角形時(shí),求點(diǎn)D的坐標(biāo).3.2017無(wú)錫如圖ZT5-2,以原點(diǎn)O為圓心,3為半徑的圓與x軸分別交于A,B兩點(diǎn)(點(diǎn)B在點(diǎn)A的右邊),P是半徑OB上一點(diǎn),過(guò)點(diǎn)P且垂直于AB的直線(xiàn)與O分別交于C,D兩點(diǎn)(點(diǎn)C在點(diǎn)D的上方),直線(xiàn)AC,DB交于點(diǎn)E.若ACCE=12.圖ZT5-2 (1)求點(diǎn)P的坐標(biāo);(2)求過(guò)點(diǎn)A和點(diǎn)E,且頂點(diǎn)在直線(xiàn)CD上的拋物線(xiàn)的函數(shù)表達(dá)式.4.2018柳北區(qū)三模如圖ZT5-3,拋物線(xiàn)y=a(x-2)2-1過(guò)點(diǎn)C(4,3),交x軸于A,B兩點(diǎn)(點(diǎn)A在點(diǎn)B的左側(cè)).圖ZT5-3 (1)求拋物線(xiàn)的解析式,并寫(xiě)出頂點(diǎn)M的坐標(biāo);(2)連接OC,CM,求tanOCM的值;(3)若點(diǎn)P在拋物線(xiàn)的對(duì)稱(chēng)軸上,連接BP,CP,BM,當(dāng)CPB=PMB時(shí),求點(diǎn)P的坐標(biāo).5.2018柳北區(qū)4月模擬如圖ZT5-4,在平面直角坐標(biāo)系xOy中,直線(xiàn)l:y=34x+m與x軸,y軸分別交于點(diǎn)A和點(diǎn)B(0,-1),拋物線(xiàn)y=12x2+bx+c經(jīng)過(guò)點(diǎn)B,且與直線(xiàn)l的另一個(gè)交點(diǎn)為C(4,n).圖ZT5-4 (1)求n的值和拋物線(xiàn)的解析式.(2)點(diǎn)D在拋物線(xiàn)上,且點(diǎn)D的橫坐標(biāo)為t(0t0.解得:m2512,且m0.當(dāng)m=2時(shí),函數(shù)解析式為y=2x2+x.(2)函數(shù)y=2x2+x圖象開(kāi)口向上,對(duì)稱(chēng)軸為直線(xiàn)x=-14,當(dāng)x-14時(shí),y隨x的增大而減小.當(dāng)nx-1時(shí),y的取值范圍是1y-3n,2n2+n=-3n.n=-2或n=0(舍去).n=-2.y=2x2+x=2x+142-18,函數(shù)C1的圖象頂點(diǎn)M的坐標(biāo)為-14,-18.由圖形可知當(dāng)P為射線(xiàn)MO與圓的交點(diǎn)時(shí),距離最大.點(diǎn)P在直線(xiàn)OM上,由O(0,0),M-14,-18可求得直線(xiàn)的解析式為y=12x.設(shè)P(a,b),則有a=2b.根據(jù)勾股定理可得PO2=(2b)2+b2=(5)2,解得b=1(負(fù)值已舍).a=2.PM最大時(shí)函數(shù)C2的解析式為y=2(x-2)2+1.2.解:(1)由題意得32+3b+c=0,c=3,解得b=-4,c=3.拋物線(xiàn)的解析式為y=x2-4x+3.(2)方法1(代數(shù)法):如圖,過(guò)點(diǎn)P作PGCF交CB于點(diǎn)G,由題意知BCO=CFE=45,F(0,m),C(0,3),CFE和GPE均為等腰直角三角形,EF=22CF=22(3-m),PE=22PG.又易知直線(xiàn)BC的解析式為y=-x+3.設(shè)xP=t(1t3),則PE=22PG=22(-t+3-t-m)=22(-m-2t+3).又t2-4t+3=t+m,m=t2-5t+3.PE+EF=22(3-m)+22(-m-2t+3)=22(-2t-2m+6)=-2(t+m-3)=-2(t2-4t)=-2(t-2)2+42,當(dāng)t=2時(shí),PE+EF取最大值42.方法2:(幾何法)如圖,由題易知直線(xiàn)BC的解析式為y=-x+3,OC=OB=3,OCB=45.同理可知OFE=45,CEF為等腰直角三角形.以BC為對(duì)稱(chēng)軸將FCE對(duì)稱(chēng)得到FCE,作PHCF于點(diǎn)H則PE+EF=PF=2PH.又PH=yC-yP=3-yP.當(dāng)yP最小時(shí),PE+EF取最大值.拋物線(xiàn)的頂點(diǎn)坐標(biāo)為(2,-1),當(dāng)yP=-1時(shí),(PE+EF)max=2(3+1)=42.(3)由(1)知對(duì)稱(chēng)軸為直線(xiàn)x=2,設(shè)D(2,n),如圖.當(dāng)BCD是以BC為直角邊的直角三角形,且D在BC上方D1位置時(shí),由勾股定理得CD12+BC2=BD12,即(2-0)2+(n-3)2+(32)2=(3-2)2+(0-n)2,解得n=5;當(dāng)BCD是以BC為直角邊的直角三角形,且D在BC下方D2位置時(shí),由勾股定理得BD22+BC2=CD22,即(2-3)2+(n-0)2+(32)2=(2-0)2+(n-3)2,解得n=-1.當(dāng)BCD是以BC為直角邊的直角三角形時(shí),D點(diǎn)坐標(biāo)為(2,5)或(2,-1).3.解:(1)過(guò)點(diǎn)E作EFx軸于點(diǎn)F,CDAB,CDEF,PC=PD.ACPAEF,BPDBFE.ACCE=12,ACAE=13.APAF=CPEF=13,DPEF=PBBF=13.AF=3AP,BF=3PB.AF-BF=AB.3AP-3PB=AB.又O的半徑為3,設(shè)P(m,0),3(3+m)-3(3-m)=6,m=1.P(1,0).(2)P(1,0),OP=1,A(-3,0).OA=3,AP=4,BP=2.AF=12.連接BC.AB是直徑,ACB=90.CDAB,ACPCBP,APCP=CPBP.CP2=APBP=42=8.CP=22(負(fù)值已舍).EF=3CP=62.E(9,62).拋物線(xiàn)的頂點(diǎn)在直線(xiàn)CD上,CD是拋物線(xiàn)的對(duì)稱(chēng)軸,拋物線(xiàn)過(guò)點(diǎn)(5,0).設(shè)拋物線(xiàn)的函數(shù)表達(dá)式為y=ax2+bx+c.根據(jù)題意得0=9a-3b+c,0=25a+5b+c,62=81a+9b+c,解得a=28,b=-24,c=-1528,拋物線(xiàn)的函數(shù)表達(dá)式為y=28x2-24x-1528.4.解:(1)由拋物線(xiàn)y=a(x-2)2-1過(guò)點(diǎn)C(4,3),得3=a(4-2)2-1,解得a=1,拋物線(xiàn)的解析式為y=(x-2)2-1,頂點(diǎn)M的坐標(biāo)為(2,-1).(2)如圖,連接OM,OC2=32+42=25,OM2=22+12=5,CM2=22+42=20,CM2+OM2=OC2,OMC=90.OM=5,CM=25,tanOCM=OMCM=525=12.(3)如圖,過(guò)C作CN垂直于對(duì)稱(chēng)軸,垂足N在對(duì)稱(chēng)軸上,取一點(diǎn)E,使EN=CN=2,連接CE,EM=6.當(dāng)y=0時(shí),(x-2)2-1=0,解得x1=1,x2=3,A(1,0),B(3,0).CN=EN,CEP=PMB=CPB=45,EPB=EPC+CPB=PMB+PBM,EPC=PBM,CEPPMB,EPMB=CEPM,易知MB=2,CE=22,6-PM2=22PM,解得PM=35,P點(diǎn)坐標(biāo)為(2,2+5)或(2,2-5).5.解:(1)直線(xiàn)l:y=34x+m經(jīng)過(guò)點(diǎn)B(0,-1),m=-1,直線(xiàn)l的解析式為y=34x-1.直線(xiàn)l:y=34x-1經(jīng)過(guò)點(diǎn)C(4,n),n=344-1=2.拋物線(xiàn)y=12x2+bx+c經(jīng)過(guò)點(diǎn)C(4,2)和點(diǎn)B(0,-1),1242+4b+c=2,c=-1,解得b=-54,c=-1,拋物線(xiàn)的解析式為y=12x2-54x-1.(2)令y=0,則34x-1=0,解得x=43,點(diǎn)A的坐標(biāo)為43,0,OA=43.在RtOAB中,OB=1,AB=OA2+OB2=(43)2+12=53.DEy軸,ABO=DEF,在矩形DFEG中,EF=DEcosDEF=DEOBAB=35DE,DF=DEsinDEF=DEOAAB=45DE,p=2(DF+EF)=245+35DE=145DE,點(diǎn)D的橫坐標(biāo)為t(0t4),Dt,12t2-54t-1,Et,34t-1,DE=34t-1-12t2-54t-1=-12t2+2t,p=145-12t2+2t=-75t2+285t,p=-75(t-2)2+285,且-750,當(dāng)t=2時(shí),p有最大值285.(3)AOB繞點(diǎn)M沿逆時(shí)針?lè)较蛐D(zhuǎn)90,A1O1y軸,B1O1x軸.設(shè)點(diǎn)A1的橫坐標(biāo)為x,如圖,點(diǎn)O1,B1在拋物線(xiàn)上時(shí),點(diǎn)O1的橫坐標(biāo)為x,點(diǎn)B1的橫坐標(biāo)為x+1,12x2-54x-1=12(x+1)2-54(x+1)-1,解得x=34.如圖,點(diǎn)A1,B1在拋物線(xiàn)上時(shí),點(diǎn)B1的橫坐標(biāo)為x+1,點(diǎn)A1的縱坐標(biāo)比點(diǎn)B1的縱坐標(biāo)大43,12x2-54x-1=12(x+1)2-54(x+1)-1+43,解得x=-712.綜上所述,點(diǎn)A1的橫坐標(biāo)為34或-712.6.解:(1)令y=0,得33x2-233x-3=0,解得x1=-1,x2=3,點(diǎn)A(-1,0),B(3,0).點(diǎn)E(4,n)在拋物線(xiàn)上,n=3342-2334-3=533,即點(diǎn)E4,533,設(shè)直線(xiàn)AE的解析式為y=kx+b,則-k+b=0,4k+b=533,解得k=33,b=33,直線(xiàn)AE的解析式為y=33x+33.(2)令y=33x2-233x-3中x=0,得y=-3,C(0,-3).由(1)得點(diǎn)E4,533,直線(xiàn)CE的解析式為y=233x-3.過(guò)點(diǎn)P作PHy軸,交CE于點(diǎn)H,如圖,設(shè)點(diǎn)Pt,33t2-233t-3,則Ht,233t-3,PH=233t-3-33t2-233t-3=-33t2+433t,SPCE=SPHC+SPHE=12PHxE-xC=12-33t2+433t4=-233t2+833t=-233(t2-4t)=-233(t-2)2+833.-2330,當(dāng)t=2時(shí),SPCE最大,此時(shí)點(diǎn)P(2,-3).C(0,-3),PCx軸.B(3,0),K為BC的中點(diǎn),K32,-32.如圖,作點(diǎn)K關(guān)于CP,CD的對(duì)稱(chēng)點(diǎn)K1,K2,連接K1K2,分別交C