第2章 二次函數(shù)專題訓(xùn)練3 二次函數(shù)的最值及自變量的取值范圍(含答案)

專題訓(xùn)練三二次函數(shù)的最值及自變量的取值范圍由自變量的取值范圍求函數(shù)值的取值范圍1.二次函數(shù)y=-x2+bx+c的部分圖象如圖所示,若y>3,則x的取值范圍是 ()A.-4<x<1B.-2<x<0C.x<-4或x>1D.x<-2或x>02.二次函數(shù)y=-x2+bx+c,若y≥2時(shí),x的取值范圍為n-3≤x≤n+1(n為常數(shù)),則當(dāng)n-4≤x≤n時(shí),y的取值范圍為()A.-3≤y≤5 B.-3≤y≤6C.0≤y≤5 D.0≤y<63.已知二次函數(shù)y=-x2+2x+3,當(dāng)-1≤x≤2時(shí),y的取值范圍為.

4.已知二次函數(shù)y=-x2+bx+c,函數(shù)值y與自變量x之間的部分對(duì)應(yīng)值如表:x…-4-101…y…-21-2-7…(1)寫(xiě)出二次函數(shù)圖象的對(duì)稱軸;(2)求二次函數(shù)的表達(dá)式;(3)當(dāng)-4<x<-1時(shí),寫(xiě)出函數(shù)值y的取值范圍.由自變量取值范圍下的函數(shù)最值求字母系數(shù)5.(2024西安臨潼區(qū)二模)已知拋物線y=-(x-n)2-1(n為常數(shù)),當(dāng)1≤x≤4時(shí),其對(duì)應(yīng)的函數(shù)值最大為-10,則n的值為 ()A.4 B.-2或7C.1或7 D.-2或46.如圖,拋物線y=12x2-x-32的頂點(diǎn)為D點(diǎn),與y軸交于C點(diǎn),點(diǎn)M(m,0)是x軸上的一個(gè)動(dòng)點(diǎn),當(dāng)MC+MD的值最小時(shí),m的值是 (A.13 B.12 C.27.(2024蘇州期末)如圖,Rt△ABC中,∠ABC=90°,AB=6cm,BC=8cm,點(diǎn)P從點(diǎn)A出發(fā)沿邊AB向點(diǎn)B以1cm/s的速度移動(dòng),點(diǎn)Q從B出發(fā)沿邊BC向點(diǎn)C以2cm/s的速度移動(dòng),P,Q兩點(diǎn)同時(shí)出發(fā),當(dāng)一點(diǎn)到達(dá)終點(diǎn)時(shí)另一點(diǎn)也停止運(yùn)動(dòng),設(shè)運(yùn)動(dòng)時(shí)間為t(s).(1)若P,Q兩點(diǎn)的距離為42cm時(shí),求t的值;(2)當(dāng)t為何值時(shí),△BPQ的面積最大?并求出最大面積.8.(2024廊坊大城縣期中)已知拋物線y=x2+ax+a+1經(jīng)過(guò)點(diǎn)A(-2,3).(1)求a的值;(2)已知點(diǎn)P(m,yP),Q(m-4,yQ)均在該拋物線上.①若m=0,請(qǐng)比較yP與yQ的大小關(guān)系;②當(dāng)-3≤x≤m時(shí),函數(shù)y的最大值是6,最小值是2,求m的取值范圍.9.(2024葫蘆島綏中縣月考)如圖,已知拋物線y=ax2+bx+3(a≠0)經(jīng)過(guò)點(diǎn)A(1,0)和點(diǎn)B(3,0),與y軸相交于點(diǎn)C.(1)求此拋物線的表達(dá)式;(2)若點(diǎn)P是直線BC下方的拋物線上一動(dòng)點(diǎn)(不與點(diǎn)B、C重合),過(guò)點(diǎn)P作y軸的平行線交直線BC于點(diǎn)D,設(shè)點(diǎn)P的橫坐標(biāo)為m.①用含有m的代數(shù)式表示線段PD的長(zhǎng);②連接PB,PC,求△PBC的面積最大時(shí)點(diǎn)P的坐標(biāo).

【詳解答案】1.B解析:由題圖可得:拋物線對(duì)稱軸為直線x=-1,當(dāng)x=0時(shí),y=3,根據(jù)拋物線的對(duì)稱性可得:當(dāng)x=-2時(shí),y=3,∴若y>3,則x的取值范圍是-2<x<0,故選B.2.B解析:由題意知,當(dāng)y≥2時(shí),x的取值范圍為n-3≤x≤n+1,且拋物線開(kāi)口向下,∴對(duì)稱軸是直線x=n-3+n+12=∴b=2(n-1).∴拋物線為y=-x2+2(n-1)x+c.又當(dāng)x=n+1時(shí),y=-(n+1)2+2(n-1)·(n+1)+c=2,∴c=-n2+2n+5.∴二次函數(shù)為y=-x2+2(n-1)x-n2+2n+5.∵拋物線開(kāi)口向下,∴拋物線上的點(diǎn)離對(duì)稱軸越近函數(shù)值越大.∵n-1-(n-4)=3>n-(n-1)=1,n-4<n-1<n,又n-4≤x≤n,∴當(dāng)x=n-1時(shí),y取最大值為y=-(n-1)2+2(n-1)2-n2+2n+5=6;當(dāng)x=n-4時(shí),y取最小值為y=-(n-4)2+2(n-4)(n-1)-n2+2n+5=-3.∴當(dāng)n-4≤x≤n時(shí),-3≤y≤6.故選B.3.0≤y≤4解析:∵二次函數(shù)y=-x2+2x+3=-(x-1)2+4,∴該函數(shù)圖象開(kāi)口向下,對(duì)稱軸為直線x=1.∵-1≤x≤2,∴當(dāng)x=-1時(shí),y取得最小值0;當(dāng)x=1時(shí),y取得最大值4;∴當(dāng)-1≤x≤2時(shí),y的取值范圍為0≤y≤4.4.解:(1)∵x=-4和x=0時(shí)的函數(shù)值相等,都是-2,∴此函數(shù)圖象的對(duì)稱軸為直線x=-4+02=-(2)將(-1,1),(0,-2)代入y=-x2+bx+c,得-解得b∴二次函數(shù)的表達(dá)式為y=-x2-4x-2.(3)∵y=-x2-4x-2=-(x+2)2+2,∴當(dāng)x=-2時(shí),y取得最大值2.由表可知當(dāng)x=-4時(shí)y=-2,當(dāng)x=-1時(shí)y=1,∴當(dāng)-4<x<-1時(shí),-2<y≤2.5.B解析:①當(dāng)n≥4時(shí),當(dāng)x=4,y=-10時(shí),代入拋物線y=-(x-n)2-1(n為常數(shù)),得-10=-(4-n)2-1,整理,得n2-8n+7=0,解得n=7或1(舍去),②當(dāng)n≤1時(shí),當(dāng)x=1,y=-10時(shí),代入拋物線y=-(x-n)2-1(n為常數(shù)),得-10=-(1-n)2-1,整理,得n2-2n-8=0,解得n=-2或4(舍去).故n的值為7或-2.故選B.6.D解析:當(dāng)x=0時(shí),y=12x2-x-32=-32,則點(diǎn)C∴C點(diǎn)關(guān)于x軸的對(duì)稱點(diǎn)C'的坐標(biāo)為0,∵y=12x2-x-32=12(∴點(diǎn)D的坐標(biāo)為(1,-2).連接C'D交x軸于M,如圖,∵M(jìn)C+MD=MC'+MD=C'D,∴此時(shí)MC+MD的值最小.設(shè)直線C'D的表達(dá)式為y=kx+32把D(1,-2)代入,得-2=k+32,解得k=-7∴直線C'D的表達(dá)式為y=-72x+3當(dāng)y=0時(shí),-72x+32=0,解得x=∴此時(shí)M點(diǎn)的坐標(biāo)為37,0,即m故選D.7.解:(1)由題意知,BP=(6-t)cm,BQ=2tcm.在Rt△BPQ中,PQ2=PB2+BQ2=(6-t)2+(2t)2.又∵P,Q兩點(diǎn)的距離為42cm,∴(6-t)2+(2t)2=(42)2,解得t1=2,t2=25又∵0≤t≤4,∴上述兩解都符合題意,故t的值為2或25(2)由(1)知,BP=(6-t)cm,BQ=2tcm,∴S△BPQ=12BP·BQ=12·2t(6-t)=t(6-t)=-t2+6t=-(t2-6t+9)+9=-(t-3)2又∵0≤t≤4,且-1<0,∴當(dāng)t=3時(shí),S△BPQ有最大值為9cm2.8.解:(1)將點(diǎn)A(-2,3)代入y=x2+ax+a+1中,得3=4-2a+a+1,解得a=2.(2)①∵a=2,∴拋物線為y=x2+2x+3,當(dāng)m=0時(shí),點(diǎn)P(m,yP),Q(m-4,yQ)為P(0,yP),Q(-4,yQ),∴yP=0+0+3=3,yQ=16-8+3=11,∴yP與yQ的大小關(guān)系為yP<yQ;②y=x2+2x+3=(x+1)2+2.當(dāng)x2+2x+3=6時(shí),x1=-3,x2=1.如圖,根據(jù)圖象和題意可得m的取值范圍是-1≤m≤1.9.解:(1)∵拋物線y=ax2+bx+3(a≠0)經(jīng)過(guò)點(diǎn)A(1,0)和點(diǎn)B(3,0),∴將A,B點(diǎn)坐標(biāo)代入,得a解得a∴拋物線表達(dá)式為y=x2-4x+3.(2)①由y=x2-4x+3可知,拋物線對(duì)稱軸為直線x=2,點(diǎn)C(0,3),設(shè)直線BC的表達(dá)式為y=kx+c.將點(diǎn)B(3,0),C(0,3)代入直線BC表達(dá)式y(tǒng)=kx+c,則3k+∴直線BC的表達(dá)式為yBC=-x+3.設(shè)P(m,m2-4m+3),如圖,過(guò)點(diǎn)