高考數(shù)學(xué)總復(fù)習(xí)——二次函數(shù)、二次方程與二次不等式-預(yù)覽頁(yè)
【正文】 ?????????aaaaabb結(jié)合因過原點(diǎn)個(gè)圖形故應(yīng)是第從后兩個(gè)圖形看不是前兩個(gè)圖形?[解析 ] .10,01,3.0,02,0 2?????????????aaaaabb結(jié)合因過原點(diǎn)個(gè)圖形故應(yīng)是第從后兩個(gè)圖形看不是前兩個(gè)圖形?[解析 ] [答案 ] B ) ( )(,2)2(),0()4( ,0 ,20 ,)()2020( 2解的個(gè)數(shù)為的的方程則關(guān)于若設(shè)年湖南高考題xxfxfffxxcbxxxf??????????????[例 2] 4 D. 3 C. 2 B. 1 A.,)0( 2)0(24)(,2,4:2 2)2()2(2)2(1 )4()4()0()4( 222??????????????????????????????xxxxxfcbcbfccbff則解之得由[解析 ] .)( ,2,0。)()(39。 )1( 2 axaxxf ????[解析 ] 。,.,0]43)21[(4)1(40)1(230)(39。,21221是極小值點(diǎn)是極大值點(diǎn)因此時(shí)當(dāng)時(shí)當(dāng)xxxfxxxfxxx?????.,0)(39。1)()( )2( )( )1( .2)(,)()( )( 20 05 2的取值范圍求實(shí)數(shù)上是增函數(shù)在若解不等式的解析式；求函數(shù)且的圖象關(guān)于原點(diǎn)對(duì)稱和已知函數(shù)年浙江高考題???????????xfxgxhxxfxgxgxxxfxgxf[例 2] ,2,2,)(),(.,0202),(),()( 1)( 2200000000xxyxxyxfyyxQyyxxyyxxyxPyxQxfy???????????????????????????即的圖象上在函數(shù)點(diǎn)即則關(guān)于原點(diǎn)的對(duì)稱點(diǎn)一點(diǎn)的圖象上任設(shè)函數(shù)?[解析 ] .2)( 2 xxxg ???故.2)( 2 xxxg ???故].21,1[,211,012,1。 )2( )(39。),21)(4()(,21:0)1(39。 )3( 2??????????????????????的取值范圍為即由條件得的拋物線為開口向上且過點(diǎn)的圖象aaaaaffaxx
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