（山西專版）2020年中考數(shù)學(xué)復(fù)習(xí) 第三單元 函數(shù)及其圖象 課時(shí)訓(xùn)練10 一次函數(shù)的圖象與性質(zhì)

課時(shí)訓(xùn)練(十)一次函數(shù)的圖象與性質(zhì)(限時(shí):40分鐘)|夯實(shí)基礎(chǔ)|1.直線y=2x-4與y軸的交點(diǎn)的坐標(biāo)是()A.(4,0)B.(0,4)C.(-4,0)D.(0,-4)2.2018·常德若一次函數(shù)y=(k-2)x+1的函數(shù)值y隨x的增大而增大,則k的取值范圍是()A.k<2B.k>2C.k>0D.k<03.2019·廣安一次函數(shù)y=2x-3的圖象經(jīng)過的象限是()A.一、二、三B.二、三、四C.一、三、四D.一、二、四4.2019·山西模擬如圖K10-1,直線y=kx+b(k0)經(jīng)過點(diǎn)A(-2,4),則不等式kx+b<4的解集為()圖K10-1A.x>-2B.x<-2C.x>4D.x<45.2018·太原模擬若正比例函數(shù)y=3x的圖象經(jīng)過A(-2,y1),B(-1,y2)兩點(diǎn),則y1與y2的大小關(guān)系為()A.y1<y2B.y1>y2C.y1y2D.y1y26.如圖K10-2,ABC的頂點(diǎn)坐標(biāo)分別為A(-1,0),B(-4,0),C(-1,4),將ABC沿x軸向左平移,當(dāng)點(diǎn)C落在直線y=-2x-6上時(shí),線段BC掃過的面積為()圖K10-2A.16B.82C.8D.47.2018·武漢武昌區(qū)期末已知一次函數(shù)y=(m-4)x+2m+1的圖象不經(jīng)過第三象限,則m的取值范圍是()A.m<4B.-12m<4C.-12m4D.m-128.2019·山西省適應(yīng)性訓(xùn)練若直線l1經(jīng)過點(diǎn)(0,4),l2經(jīng)過點(diǎn)(3,2),且l1與l2關(guān)于x軸對(duì)稱,則l1與l2的交點(diǎn)坐標(biāo)為()A.(-2,0)B.(2,0)C.(-6,0)D.(6,0)9.如圖K10-3所示,已知點(diǎn)A的坐標(biāo)為(6,0),直線y=x+b(b>0)與y軸交于點(diǎn)B,連接AB,=75°,則b的值為()圖K10-3A.23B.33C.3D.6310.若一次函數(shù)y=2x+b(b為常數(shù))的圖象經(jīng)過點(diǎn)(1,5),則b=. 11.2018·太原模擬已知點(diǎn)A,B的坐標(biāo)分別為(-2,3),(1,-2).將線段AB平移得到線段A'B',其中點(diǎn)A與點(diǎn)A'對(duì)應(yīng),點(diǎn)B與點(diǎn)B'對(duì)應(yīng),若點(diǎn)A'的坐標(biāo)為(2,-3),則點(diǎn)B'的坐標(biāo)為. 12.2019·大同模擬如圖K10-4,已知函數(shù)y=2x+b與函數(shù)y=kx-6的圖象交于點(diǎn)P,則不等式kx-6<2x+b的解集是. 圖K10-413.2018·白銀如圖K10-5,一次函數(shù)y=-x-2與y=2x+m的圖象交于點(diǎn)P(n,-4),則關(guān)于x的不等式組2x+m<-x-2,-x-2<0的解集為. 圖K10-514.2018·東營在平面直角坐標(biāo)系內(nèi)有兩點(diǎn)A,B,其坐標(biāo)為A(-1,-1),B(2,7),點(diǎn)M為x軸上的一個(gè)動(dòng)點(diǎn),若要使MB-MA的值最大,則點(diǎn)M的坐標(biāo)為. 15.2017·杭州在平面直角坐標(biāo)系中,一次函數(shù)y=kx+b(k,b都是常數(shù),且k0)的圖象經(jīng)過點(diǎn)(1,0)和(0,2).(1)當(dāng)-2<x3時(shí),求y的取值范圍;(2)已知點(diǎn)P(m,n)在該函數(shù)的圖象上,且m-n=4,求點(diǎn)P的坐標(biāo).16.如圖K10-6,直線y=-2x+3與x軸相交于點(diǎn)A,與y軸相交于點(diǎn)B.(1)求A,B兩點(diǎn)的坐標(biāo);(2)過點(diǎn)B作直線BP與x軸相交于點(diǎn)P,且使OP=2OA,求ABP的面積.圖K10-6|拓展提升|17.2019·郴州若一個(gè)函數(shù)當(dāng)自變量在不同范圍內(nèi)取值時(shí),函數(shù)表達(dá)式不同,我們稱這樣的函數(shù)為分段函數(shù).下面我們參照學(xué)習(xí)函數(shù)的過程與方法,探究分段函數(shù)y=-2x(x-1),|x-1|(x>-1)的圖象與性質(zhì).列表:x-3-52-2-32-1-120121322523y23451432321120121322描點(diǎn):在平面直角坐標(biāo)系中,以自變量x的取值為橫坐標(biāo),以相應(yīng)的函數(shù)值y為縱坐標(biāo),描出相應(yīng)的點(diǎn),如圖K10-7所示.(1)如圖K10-7,在平面直角坐標(biāo)系中,觀察描出的這些點(diǎn)的分布,作出函數(shù)圖象.(2)研究函數(shù)并結(jié)合圖象與表格,回答下列問題:點(diǎn)A(-5,y1),B-72,y2,Cx1,52,D(x2,6)在函數(shù)圖象上,則y1y2,x1x2;(填“>”“=”或“<”) 當(dāng)函數(shù)值y=2時(shí),求自變量x的值;在直線x=-1的右側(cè)的函數(shù)圖象上有兩個(gè)不同的點(diǎn)P(x3,y3),Q(x4,y4),且y3=y4,求x3+x4的值;若直線y=a與函數(shù)圖象有三個(gè)不同的交點(diǎn),求a的取值范圍.圖K10-7【參考答案】1.D2.B3.C4.B5.A6.A7.B解析根據(jù)題意,得m-4<0,2m+10.解得-12m<4.故選B.8.B解析直線l1經(jīng)過點(diǎn)(0,4),l2經(jīng)過點(diǎn)(3,2),且l1與l2關(guān)于x軸對(duì)稱,兩直線相交于x軸上,直線l1經(jīng)過點(diǎn)(3,-2),l2經(jīng)過點(diǎn)(0,-4),把(0,4)和(3,-2)分別代入直線l1的解析式y(tǒng)=kx+b,則b=4,3k+b=-2,解得k=-2,b=4,故直線l1的解析式為y=-2x+4,可得l1與l2的交點(diǎn)坐標(biāo)為l1與l2與x軸的交點(diǎn),令y=0,得-2x+4=0,解得x=2,即l1與l2的交點(diǎn)坐標(biāo)為(2,0).9.A解析如圖,設(shè)直線y=x+b與x軸交于點(diǎn)C.將y=0代入y=x+b,得x=-b;將x=0代入y=x+b,得y=b.OB=OC=b,OBC=OCB.BOC=90°,OBC=OCB=45°.=75°,=OCB+BAC,BAC=30°,即BAO=30°.點(diǎn)A(6,0),BOA=90°,點(diǎn)B(0,b),OA=6,OB=b,tan30°=b6,解得b=23.10.3解析把(1,5)代入y=2x+b,得5=2×1+b,解得b=3.11.(5,-8)12.x>2解析函數(shù)y=2x+b與函數(shù)y=kx-6的圖象交于點(diǎn)P(2,-8),不等式kx-6<2x+b的解集是x>2.13.-2<x<2解析y=-x-2的圖象過點(diǎn)P(n,-4),-n-2=-4,解得n=2.點(diǎn)P的坐標(biāo)是(2,-4).觀察圖象可知,2x+m<-x-2的解集為x<2.解不等式-x-2<0,得x>-2.不等式組2x+m<-x-2,-x-2<0的解集是-2<x<2.14.-32,0解析如圖,作點(diǎn)A關(guān)于x軸的對(duì)稱點(diǎn)A',連接A'B,直線A'B交x軸于點(diǎn)M,則點(diǎn)M為符合條件的點(diǎn).易得A'的坐標(biāo)為(-1,1).設(shè)直線A'B的解析式為y=kx+b,將A'(-1,1),B(2,7)分別代入解析式中,得-k+b=1,2k+b=7,解得k=2,b=3,直線A'B的解析式為y=2x+3.當(dāng)y=0時(shí),2x+3=0,解得x=-32,點(diǎn)M的坐標(biāo)是-32,0.15.解:(1)由題意知,y=kx+2.圖象過點(diǎn)(1,0),0=k+2.解得k=-2.y=-2x+2.當(dāng)x=-2時(shí),y=6;當(dāng)x=3時(shí),y=-4.k=-2<0,函數(shù)值y隨x的增大而減小.y的取值范圍為-4y<6.(2)根據(jù)題意,得n=-2m+2,m-n=4.解得m=2,n=-2.點(diǎn)P的坐標(biāo)為(2,-2).16.解:(1)令y=0,得x=32;令x=0,得y=3.A32,0,B(0,3).(2)OP=2OA,A32,0,P(3,0)或(-3,0).AP=32或92.SABP=12AP·OB=12×32×3=94或SABP=12AP·OB=12×92×3=274.17.解:(1)根據(jù)列表、描點(diǎn),可以作出函數(shù)圖象,如圖.(2)<,<解析由圖象可知,當(dāng)x-1時(shí),函數(shù)值y隨x值的增大而增大.因?yàn)辄c(diǎn)A,B在函數(shù)圖象上,且-5<-72<-1,所以y1<y2.因?yàn)?2>2,6>2,點(diǎn)C,D在函數(shù)圖象上,所以C,D在函數(shù)y=x-1(x>1)圖象上,且函數(shù)值y隨x值的增大而增大,因?yàn)?2<6,所以x1<x2.故填:<,<.當(dāng)y=2時(shí),若x-1,則有-2x=2,解得x=-1;若x>-1,則有|x-1|=2,即x-1=±2,解得x=3或x=-1(舍去).綜上所述,當(dāng)y=2時(shí),自變量x的值為-1或3.若點(diǎn)P(x3,y3),Q(x4,y4)是直線x=-1的右側(cè)的函數(shù)圖象上的兩個(gè)不同的點(diǎn),且y3=y4,則|x3-1|=|x4-1|,所以x3-1=-(x4-1),所以x3+x4=2.若直線y=a與函數(shù)圖象有三個(gè)不同的交點(diǎn),通過觀察函數(shù)圖象可知0<a<2.9