二次函数周长最小问题

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周长最小问题基本解题方法:1.如图,已知抛物线y=ax2-4x+c经过点A(0,-6)和B(3,-9).(1)求抛物线的解析式;(2)写出抛物线的对称轴方程及顶点坐标;(3)点P(m,m)与点Q均在抛物线上(其中m>0),且这两点关于抛物线的对称轴对称,求m的值及点Q的坐标;(4)在满足(3)的情况下,在抛物线的对称轴上寻找一点M,使得△QMA的周长最小.OABxy-6-93解:(1)依题意有a×02-4×0+c=-6a×32-4×3+c=-9即c=-69a-12+c=-9·····································································2分∴a=1c=-6·················································································4分∴抛物线的解析式为:y=x2-4x-6···············································5分(2)把y=x2-4x-6配方,得y=(x-2)2-10∴对称轴方程为x=2··································································7分顶点坐标(2,-10)··································································10分(3)由点P(m,m)在抛物线上得m=m2-4m-6······································································12分即m2-5m-6=0∴m1=6或m2=-1(舍去)························································13分∴P(6,6)∵点P、Q均在抛物线上,且关于对称轴x=2对称∴Q(-2,6)···················································································15分(4)连接AP、AQ,直线AP与对称轴x=2相交于点M由于P、Q两点关于对称轴对称,由轴对称性质可知,此时的交点M能够使得△QMA的周长最小·············································································17分设直线AP的解析式为y=kx+b则b=-66k+b=6∴k=2b=-6∴直线AP的解析式为:y=2x-618分设点M(2,n)则有n=2×2-6=-219分此时点M(2,-2)能够使得△QMA的周长最小20分OABxy-6-93PQM2.如图,在平面直角坐标系中,直线y=-3x-3与x轴交于点A,与y轴交于点C,抛物线y=ax2-332x+c(a≠0)经过点A、C,与x轴交于另一点B.(1)求抛物线的解析式及顶点D的坐标;(2)若P是抛物线上一点,且△ABP为直角三角形,求点P的坐标;(3)在直线AC上是否存在点Q,使得△QBD的周长最小,若存在,求出Q点的坐标;若不存在,请说明理由.(1)∵直线y=-3x-3与x轴交于点A,与y轴交于C∴A(-1,0),C(0,-3)∵点A,C都在抛物线上∴a+332+c=0c=-3解得a=33c=-3∴抛物线的解析式为y=33x2-332x-3=33(x-1)2-334∴顶点D的坐标为(1,-334)(2)令33x2-332x-3=0,解得x1=-1,x2=3∴B(3,0)∴AB2=(1+3)2=16,AC2=12+(3)2=4,BC2=32+(3)2=12∴AC2+BC2=AB2,∴△ABC是直角三角形∴P1(0,-3)由抛物线的对称性可知P2的纵坐标为-3,代入抛物线的解析式求得:P2(2,-3)(3)存在.延长BC到点B′,使B′C=BC,连接B′D交直线AC于点Q,则Q点就是所求的点过点B′作B′H⊥x轴于H在Rt△BOC中,∵BC=12=32,∴BC=2OC∴∠OBC=30°∴B′H=21BB′=BC=32,BH=3B′H=6,∴OH=3∴B′(-3,-32)设直线B′D的解析式为y=kx+b,则:yBOACDxyBOACDxB′QH-32=-3k+b-334=k+b解得k=63b=-233联立y=-3x-3y=63x-233解得x=73y=-7310∴Q(73,-7310)故在直线AC上存在点Q,使得△QBD的周长最小,Q点的坐标为(73,-7310)3.在平面直角坐标系中,矩形OACB的顶点O在坐标原点,顶点A、B分别在x轴、y轴的正半轴上,OA=3,OB=4,D为边OB的中点.(Ⅰ)若E为边OA上的一个动点,当△CDE的周长最小时,求点E的坐标;(Ⅱ)若E、F为边OA上的两个动点,且EF=2,当四边形CDEF的周长最小时,求点E、F的坐标.解:(Ⅰ)如图1,作点D关于x轴的对称点D′,连接CD′与x轴交于点E,连接DE若在边OA上任取点E′(与点E不重合),连接CE′、DE′、D′E′由DE′+CE′=D′E′+CE′>CD′=D′E+CE=DE+CE可知△CDE的周长最小∵在矩形OACB中,OA=3,OB=4,D为边OB的中点∴BC=3,D′O=DO=2,D′B=6∵OE∥BC,∴Rt△D′OE∽Rt△D′BC,∴BCOE=BDOD∴OE=BDOD·BC=62×3=1∴点E的坐标为(1,0)························································6分(Ⅱ)如图2,作点D关于x轴的对称点D′,在CB边上截取CG=2,连接D′G与x轴交于点E,在EA上截取EF=2,则四边形GEFC为平行四边形,得GE=CF又DC、EF的长为定值,∴此时得到的点E、F使四边形CDEF的周长最小∵OE∥BC,∴Rt△D′OE∽Rt△D′BG,∴BGOE=BDOD∴OE=BDOD·BG=BDOD·(BC-CG)=62×1=31∴OF=OE+EF=31+2=37∴点E的坐标为(31,0),点F的坐标为(37,0)···················10分OABxyCDOABxyCDED′(备用图)OABxyCDED′图1E′OABxyCDED′图2FG3.如图,抛物线y=ax2+bx+4与x轴的两个交点分别为A(-4,0)、B(2,0),与y轴交于点C,顶点为D.E(1,2)为线段BC的中点,BC的垂直平分线与x轴、y轴分别交于F、G.(1)求抛物线的函数解析式,并写出顶点D的坐标;(2)在直线EF上求一点H,使△CDH的周长最小,并求出最小周长;(3)若点K在x轴上方的抛物线上运动,当K运动到什么位置时,△EFK的面积最大?并求出最大面积.解:(1)由题意,得16a-4b+4=04a+2b+4=0解得a=-21,b=-1∴抛物线的函数解析式为y=-21x2-x+4,顶点D的坐标为(-1,29)··············································································4分(2)设抛物线的对称轴与x轴交于点M.因为EF垂直平分BC,即C关于直线EG的对称点为B,连结BD交EF于一点,则这一点为所求点H,使DH+CH最小,即最小为:DH+CH=DH+HB=BD=22DMBM+=1323而CD=224291)(+=25∴△CDH的周长最小值为CD+DR+CH=21335·······6分设直线BD的解析式为y=k1x+b1,则2k1+b1=0-k1+b1=29解得k1=-23,b1=3∴直线BD的解析式为y=-23x+3由于BC=52,CE=21BC=5,Rt△CEG∽Rt△COBCBAOEFxyDGCBAOEFxyDGH得CE:CO=CG:CB,∴CG=25,GO=23,∴G(0,23)同理可求得直EF的解析式为y=21x+23联立y=-23x+3y=21x+23解得x=43y=815故使△CDH的周长最小的点H坐标为(43,815)(3)设K(t,-21t2-t+4),xF<t<xE.过K作x轴的垂线交EF于N则KN=yK-yN=-21t2-t+4-(21t+23)=-21t2-23t+25∴S△EFK=S△KFN+S△KNE=21KN(t+3)+21KN(1-t)=2KN=-t2-3t+5=-(t+23)2+429··········································10分∴当t=-23时,△EFK的面积最大,最大面积为429,此时K(-23,835)·············································································14分CBAOEFxyDGHKN4.如图,在平面直角坐标系中放置一矩形ABCO,其顶点为A(0,1)、B(-33,1)、C(-33,0)、O(0,0).将此矩形沿着过E(-3,1)、F(-334,0)的直线EF向右下方翻折,B、C的对应点分别为B′、C′.(1)求折痕所在直线EF的解析式;(2)一抛物线经过B、E、B′三点,求此二次函数解析式;(3)能否在直线EF上求一点P,使得△PBC周长最小?如能,求出点P的坐标;若不能,说明理由.解:(1)由于折痕所在直线EF过E(-3,1)、F(-334,0)∴tan∠EFO=3,直线EF的倾斜角为60°∴直线EF的解析式为:y-=tan60°[x-(-3)]化简得:y=3x+4.·············································································3分(2)设矩形沿直线EF向右下方翻折后,B、C的对应点为B′(x1,y1),C′(x2,y2)过B′作B′A′⊥AE交AE所在直线于A′点∵B′E=BE=32,∠B′EF=∠BEF=60°∴∠B′EA′=60°,∴A′E=3,B′A′=3∴A与A′重合,B′在y轴上,∴x1=0,y1=-2,即B′(0,-2)【此时需说明B′(x1,y1)在y轴上】·························································6分设二次函数的解析式为:y=ax2+bx+c∵抛物线经过B(-33,1)、E(-3,1)、B′(0,-2)∴27a-33b+c=13a-3b+c=1c=-2解得233431---===cbaxyOBCEFA∴该二次函数解析式为:y=-31x2-334x-2··············································9分(3)能,可以在直线EF上找到P点,连接B′C交EF于P点,再连接BP由于B′P=BP,此时点P与C、B′在一条直线上,故BP+PC=B′P+PC的和最小由于为BC定长所以满足△PBC周长最小.·················································10分设直线B′C的解析式为:y=kx+b则bkb+==--3302解得2932--==bk∴直线B′C的解析式为:y=-932x-2··········································12分又∵点P为直线B′C与直线EF的交点∴432932+==--xxyy解得111031118--==yx∴点P的坐标为(-31118,-11

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