【淘寶店鋪：向陽(yáng)百分百】【淘寶店鋪：向陽(yáng)百分百】備戰(zhàn)2024中考數(shù)學(xué)一輪復(fù)習(xí)備戰(zhàn)2024中考數(shù)學(xué)一輪復(fù)習(xí)第6講二次函數(shù)的圖像與性質(zhì)№第6講二次函數(shù)的圖像與性質(zhì)№考向解讀?考點(diǎn)精析?真題精講?題型突破?專題精練第三章函數(shù)第6講二次函數(shù)的圖像與性質(zhì)→?考點(diǎn)精析←→?真題精講←考向一二次函數(shù)的最值考向二二次函數(shù)平移考向三二次函數(shù)圖像對(duì)稱考向四二次函數(shù)綜合性質(zhì)考向五二次函數(shù)參數(shù)問(wèn)題考向六二次函數(shù)交點(diǎn)問(wèn)題第6講二次函數(shù)的圖像與性質(zhì)二次函數(shù)是非常重要的函數(shù),年年都會(huì)考查,總分值為18~20分，預(yù)計(jì)2024年各地中考還會(huì)考,它經(jīng)常以一個(gè)壓軸題獨(dú)立出現(xiàn)，有的地區(qū)也會(huì)考察二次函數(shù)的應(yīng)用題，小題的考察主要是二次函數(shù)的圖象和性質(zhì)及或與幾何圖形結(jié)合來(lái)考查.→?考點(diǎn)精析←一、二次函數(shù)的概念一般地，形如y=ax2+bx+c（a，b，c是常數(shù)，a≠0）的函數(shù)，叫做二次函數(shù)．二、二次函數(shù)解析式的三種形式（1）一般式：y=ax2+bx+c（a，b，c為常數(shù)，a≠0）．（2）頂點(diǎn)式：y=a（x–h）2+k（a，h，k為常數(shù)，a≠0），頂點(diǎn)坐標(biāo)是（h，k）．（3）交點(diǎn)式：y=a（x–x1）（x–x2），其中x1，x2是二次函數(shù)與x軸的交點(diǎn)的橫坐標(biāo)，a≠0．三、二次函數(shù)的圖象及性質(zhì)1．二次函數(shù)的圖象與性質(zhì)解析式二次函數(shù)y=ax2+bx+c（a，b，c是常數(shù)，a≠0）對(duì)稱軸x=–SKIPIF1<0頂點(diǎn)（–SKIPIF1<0，SKIPIF1<0）a的符號(hào)a>0a<0圖象開口方向開口向上開口向下最值當(dāng)x=–SKIPIF1<0時(shí)，y最小值=SKIPIF1<0當(dāng)x=–SKIPIF1<0時(shí)，y最大值=SKIPIF1<0最點(diǎn)拋物線有最低點(diǎn)拋物線有最高點(diǎn)增減性當(dāng)x<–SKIPIF1<0時(shí)，y隨x的增大而減小；當(dāng)x>–SKIPIF1<0時(shí)，y隨x的增大而增大當(dāng)x<–SKIPIF1<0時(shí)，y隨x的增大而增大；當(dāng)x>–SKIPIF1<0時(shí)，y隨x的增大而減小2.二次函數(shù)圖象的特征與a，b，c的關(guān)系字母的符號(hào)圖象的特征aa>0開口向上a<0開口向下bb=0對(duì)稱軸為y軸ab>0（a與b同號(hào)）對(duì)稱軸在y軸左側(cè)ab<0（a與b異號(hào)）對(duì)稱軸在y軸右側(cè)cc=0經(jīng)過(guò)原點(diǎn)c>0與y軸正半軸相交c<0與y軸負(fù)半軸相交b2–4acb2–4ac=0與x軸有唯一交點(diǎn)（頂點(diǎn)）b2–4ac>0與x軸有兩個(gè)交點(diǎn)b2–4ac<0與x軸沒(méi)有交點(diǎn)四、拋物線的平移1．將拋物線解析式化成頂點(diǎn)式y(tǒng)=a（x–h）2+k，頂點(diǎn)坐標(biāo)為（h，k）．2．保持y=ax2的形狀不變，將其頂點(diǎn)平移到（h，k）處，具體平移方法如下：3．注意二次函數(shù)平移遵循“上加下減，左加右減”的原則，據(jù)此，可以直接由解析式中常數(shù)的加或減求出變化后的解析式；二次函數(shù)圖象的平移可看作頂點(diǎn)間的平移，可根據(jù)頂點(diǎn)之間的平移求出變化后的解析式．五、二次函數(shù)與一元二次方程的關(guān)系1．二次函數(shù)y=ax2+bx+c（a≠0），當(dāng)y=0時(shí)，就變成了一元二次方程ax2+bx+c=0（a≠0）．2．a(chǎn)x2+bx+c=0（a≠0）的解是拋物線y=ax2+bx+c（a≠0）的圖象與x軸交點(diǎn)的橫坐標(biāo)．3．（1）b2–4ac>0?方程有兩個(gè)不相等的實(shí)數(shù)根，拋物線與x軸有兩個(gè)交點(diǎn)；（2）b2–4ac=0?方程有兩個(gè)相等的實(shí)數(shù)根，拋物線與x軸有且只有一個(gè)交點(diǎn)；（3）b2–4ac<0?方程沒(méi)有實(shí)數(shù)根，拋物線與x軸沒(méi)有交點(diǎn)．六、二次函數(shù)的綜合1、函數(shù)存在性問(wèn)題解決二次函數(shù)存在點(diǎn)問(wèn)題，一般先假設(shè)該點(diǎn)存在，根據(jù)該點(diǎn)所在的直線或拋物線的表達(dá)式，設(shè)出該點(diǎn)的坐標(biāo)；然后用該點(diǎn)的坐標(biāo)表示出與該點(diǎn)有關(guān)的線段長(zhǎng)或其他點(diǎn)的坐標(biāo)等；最后結(jié)合題干中其他條件列出等式，求出該點(diǎn)的坐標(biāo)，然后判別該點(diǎn)坐標(biāo)是否符合題意，若符合題意，則該點(diǎn)存在，否則該點(diǎn)不存在．2、函數(shù)動(dòng)點(diǎn)問(wèn)題（1）函數(shù)壓軸題主要分為兩大類：一是動(dòng)點(diǎn)函數(shù)圖象問(wèn)題；二是與動(dòng)點(diǎn)、存在點(diǎn)、相似等有關(guān)的二次函數(shù)綜合題．（2）解答動(dòng)點(diǎn)函數(shù)圖象問(wèn)題，要把問(wèn)題拆分，分清動(dòng)點(diǎn)在不同位置運(yùn)動(dòng)或不同時(shí)間段運(yùn)動(dòng)時(shí)對(duì)應(yīng)的函數(shù)表達(dá)式，進(jìn)而確定函數(shù)圖象；解答二次函數(shù)綜合題，要把大題拆分，做到大題小做，逐步分析求解，最后匯總成最終答案．（3）解決二次函數(shù)動(dòng)點(diǎn)問(wèn)題，首先要明確動(dòng)點(diǎn)在哪條直線或拋物線上運(yùn)動(dòng)，運(yùn)動(dòng)速度是多少，結(jié)合直線或拋物線的表達(dá)式設(shè)出動(dòng)點(diǎn)的坐標(biāo)或表示出與動(dòng)點(diǎn)有關(guān)的線段長(zhǎng)度，最后結(jié)合題干中與動(dòng)點(diǎn)有關(guān)的條件進(jìn)行計(jì)算．→?真題精講←考向一二次函數(shù)最值1．（2023·甘肅蘭州·統(tǒng)考中考真題）已知二次函數(shù)SKIPIF1<0，下列說(shuō)法正確的是（

）A．對(duì)稱軸為SKIPIF1<0 B．頂點(diǎn)坐標(biāo)為SKIPIF1<0 C．函數(shù)的最大值是－3 D．函數(shù)的最小值是－3【答案】C【分析】根據(jù)二次函數(shù)的圖象及性質(zhì)進(jìn)行判斷即可．【詳解】二次函數(shù)SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0∵SKIPIF1<0∴二次函數(shù)圖象開口向下，函數(shù)有最大值，為SKIPIF1<0∴A、B、D選項(xiàng)錯(cuò)誤，C選項(xiàng)正確故選：C.【點(diǎn)睛】本題考查二次函數(shù)的圖象及性質(zhì)，熟練掌握二次函數(shù)圖象和性質(zhì)是解題的關(guān)鍵．2．（2023·遼寧大連·統(tǒng)考中考真題）已知拋物線SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，函數(shù)的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．0 D．2【答案】D【分析】把拋物線SKIPIF1<0化為頂點(diǎn)式，得到對(duì)稱軸為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)的最小值為SKIPIF1<0，再分別求出SKIPIF1<0和SKIPIF1<0時(shí)的函數(shù)值，即可得到答案．【詳解】解：∵SKIPIF1<0，∴對(duì)稱軸為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)的最小值為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，函數(shù)的最大值為2，故選：D.【點(diǎn)睛】此題考查了二次函數(shù)的最值，熟練掌握二次函數(shù)的性質(zhì)是解題的關(guān)鍵．3．（2023·浙江杭州·統(tǒng)考中考真題）設(shè)二次函數(shù)SKIPIF1<0是實(shí)數(shù)SKIPIF1<0，則（

）A．當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的最小值為SKIPIF1<0 B．當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的最小值為SKIPIF1<0C．當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的最小值為SKIPIF1<0 D．當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的最小值為SKIPIF1<0【答案】A【分析】令SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0，從而求得拋物線對(duì)稱軸為直線SKIPIF1<0，再分別求出當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)函數(shù)y的最小值即可求解．【詳解】解：令SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0，∴拋物線對(duì)稱軸為直線SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，拋物線對(duì)稱軸為直線SKIPIF1<0，把SKIPIF1<0代入SKIPIF1<0，得SKIPIF1<0，∵SKIPIF1<0∴當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，y有最小值，最小值為SKIPIF1<0．故A正確，B錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，拋物線對(duì)稱軸為直線SKIPIF1<0，把SKIPIF1<0代入SKIPIF1<0，得SKIPIF1<0，∵SKIPIF1<0∴當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，y有最小值，最小值為SKIPIF1<0，故C、D錯(cuò)誤，故選：A．【點(diǎn)睛】本題考查拋物線的最值，拋物線對(duì)稱軸．利用拋物線的對(duì)稱性求出拋物線對(duì)稱軸是解題的關(guān)鍵．考向二二次函數(shù)平移4．（2023·廣西·統(tǒng)考中考真題）將拋物線SKIPIF1<0向右平移3個(gè)單位，再向上平移4個(gè)單位，得到的拋物線是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)“左加右減，上加下減”的法則進(jìn)行解答即可．【詳解】解：將拋物線SKIPIF1<0向右平移3個(gè)單位，再向上平移4個(gè)單位，得到的拋物線的函數(shù)表達(dá)式為：SKIPIF1<0．故選：A．【點(diǎn)睛】本題考查了二次函數(shù)圖象的平移，熟知二次函數(shù)圖象平移的法則是解答此題的關(guān)鍵．考向三二次函數(shù)圖像對(duì)稱5.（2023·湖南·統(tǒng)考中考真題）如圖所示，直線l為二次函數(shù)SKIPIF1<0的圖像的對(duì)稱軸，則下列說(shuō)法正確的是（

）

A．b恒大于0 B．a(chǎn)，b同號(hào) C．a(chǎn)，b異號(hào) D．以上說(shuō)法都不對(duì)【答案】C【分析】先寫出拋物線的對(duì)稱軸方程，再列不等式，再分SKIPIF1<0，SKIPIF1<0兩種情況討論即可．【詳解】解：∵直線l為二次函數(shù)SKIPIF1<0的圖像的對(duì)稱軸，∴對(duì)稱軸為直線SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，∴a，b異號(hào)，故選：C．【點(diǎn)睛】本題考查的是二次函數(shù)的性質(zhì)，熟練的利用對(duì)稱軸在y軸的右側(cè)列不等式是解本題的關(guān)鍵．考向四二次函數(shù)綜合性質(zhì)6．（2023·四川南充·統(tǒng)考中考真題）拋物線SKIPIF1<0與x軸的一個(gè)交點(diǎn)為SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是(

)A．SKIPIF1<0 B．SKIPIF1<0SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0SKIPIF1<0【答案】B【分析】根據(jù)拋物線有交點(diǎn)，則SKIPIF1<0有實(shí)數(shù)根，得出SKIPIF1<0或SKIPIF1<0，分類討論，分別求得當(dāng)SKIPIF1<0和SKIPIF1<0時(shí)SKIPIF1<0的范圍，即可求解．【詳解】解：∵拋物線SKIPIF1<0與x軸有交點(diǎn)，∴SKIPIF1<0有實(shí)數(shù)根，∴SKIPIF1<0即SKIPIF1<0解得：SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，如圖所示，

依題意，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得：SKIPIF1<0∴SKIPIF1<0

綜上所述，SKIPIF1<0SKIPIF1<0或SKIPIF1<0，故選：B．【點(diǎn)睛】本題考查了二次函數(shù)的性質(zhì)，熟練掌握二次函數(shù)的性質(zhì)是解題的關(guān)鍵．7．（2023·四川廣安·統(tǒng)考中考真題）如圖所示，二次函數(shù)SKIPIF1<0為常數(shù)，SKIPIF1<0的圖象與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0．有下列結(jié)論：①SKIPIF1<0；②若點(diǎn)SKIPIF1<0和SKIPIF1<0均在拋物線上，則SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0．其中正確的有（）

A．1個(gè) B．2個(gè) C．3個(gè) D．4個(gè)【答案】C【分析】根據(jù)二次函數(shù)圖像的性質(zhì)、二次函數(shù)圖像與系數(shù)的關(guān)系以及與SKIPIF1<0軸交點(diǎn)問(wèn)題逐項(xiàng)分析判斷即可.【詳解】解：由圖可知，二次函數(shù)開口方向向下，與SKIPIF1<0軸正半軸交于一點(diǎn)，SKIPIF1<0，SKIPIF1<0.SKIPIF1<0，SKIPIF1<0.SKIPIF1<0.故①正確.SKIPIF1<0SKIPIF1<0是關(guān)于二次函數(shù)對(duì)稱軸對(duì)稱，SKIPIF1<0.SKIPIF1<0在對(duì)稱軸的左邊，SKIPIF1<0在對(duì)稱軸的右邊，如圖所示，

SKIPIF1<0.故②正確.SKIPIF1<0圖象與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.SKIPIF1<0.SKIPIF1<0.故③正確.SKIPIF1<0SKIPIF1<0，SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0.SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故④不正確.綜上所述，正確的有①②③.故選：C.【點(diǎn)睛】本題考查了二次函數(shù)圖像與系數(shù)之間的關(guān)系，解題的關(guān)鍵在于通過(guò)圖像判斷對(duì)稱軸，開口方向以及與SKIPIF1<0軸交點(diǎn).8．（2023·浙江寧波·統(tǒng)考中考真題）已知二次函數(shù)SKIPIF1<0，下列說(shuō)法正確的是(

)A．點(diǎn)SKIPIF1<0在該函數(shù)的圖象上B．當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，SKIPIF1<0C．該函數(shù)的圖象與x軸一定有交點(diǎn)D．當(dāng)SKIPIF1<0時(shí)，該函數(shù)圖象的對(duì)稱軸一定在直線SKIPIF1<0的左側(cè)【答案】C【分析】根據(jù)二次函數(shù)的圖象和性質(zhì)，逐一進(jìn)行判斷即可．【詳解】解：∵SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)：SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，即：點(diǎn)SKIPIF1<0不在該函數(shù)的圖象上，故A選項(xiàng)錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴拋物線的開口向上，對(duì)稱軸為SKIPIF1<0，∴拋物線上的點(diǎn)離對(duì)稱軸越遠(yuǎn)，函數(shù)值越大，∵SKIPIF1<0，SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最大值為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最小值為SKIPIF1<0，∴SKIPIF1<0，故B選項(xiàng)錯(cuò)誤；∵SKIPIF1<0，∴該函數(shù)的圖象與x軸一定有交點(diǎn)，故選項(xiàng)C正確；當(dāng)SKIPIF1<0時(shí)，拋物線的對(duì)稱軸為：SKIPIF1<0，∴該函數(shù)圖象的對(duì)稱軸一定在直線SKIPIF1<0的右側(cè)，故選項(xiàng)D錯(cuò)誤；故選：C．【點(diǎn)睛】本題考查二次函數(shù)的圖象和性質(zhì)．熟練掌握二次函數(shù)的性質(zhì)，是解題的關(guān)鍵．考向五二次函數(shù)參數(shù)問(wèn)題9．（2023·內(nèi)蒙古通遼·統(tǒng)考中考真題）如圖，拋物線SKIPIF1<0與x軸交于點(diǎn)SKIPIF1<0，其中SKIPIF1<0，下列四個(gè)結(jié)論：①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0；④不等式SKIPIF1<0的解集為SKIPIF1<0．其中正確結(jié)論的個(gè)數(shù)是（

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A．1 B．2 C．3 D．4【答案】C【分析】根據(jù)函數(shù)圖象可得出a，b，c的符號(hào)即可判斷①，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0即可判斷②；根據(jù)對(duì)稱軸為SKIPIF1<0，SKIPIF1<0可判斷③；SKIPIF1<0，SKIPIF1<0數(shù)形結(jié)合即可判斷④．【詳解】解：∵拋物線開口向上，對(duì)稱軸在y軸右邊，與y軸交于正半軸，∴SKIPIF1<0，∴SKIPIF1<0，故①正確．∵當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0，故②錯(cuò)誤．∵拋物線SKIPIF1<0與x軸交于兩點(diǎn)SKIPIF1<0，其中SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，故③正確；設(shè)SKIPIF1<0，SKIPIF1<0，如圖：

由圖得，SKIPIF1<0時(shí)，SKIPIF1<0，故④正確．綜上，正確的有①③④，共3個(gè)，故選：C．【點(diǎn)睛】本題考查了二次函數(shù)的圖象及性質(zhì)，根據(jù)二次函數(shù)的圖象及性質(zhì)巧妙借助數(shù)學(xué)結(jié)合思想解決問(wèn)題是解題的關(guān)鍵．10．（2023·四川瀘州·統(tǒng)考中考真題）已知二次函數(shù)SKIPIF1<0（其中SKIPIF1<0是自變量），當(dāng)SKIPIF1<0時(shí)對(duì)應(yīng)的函數(shù)值SKIPIF1<0均為正數(shù)，則SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】D【分析】首先根據(jù)題意求出對(duì)稱軸SKIPIF1<0，然后分兩種情況：SKIPIF1<0和SKIPIF1<0，分別根據(jù)二次函數(shù)的性質(zhì)求解即可．【詳解】∵二次函數(shù)SKIPIF1<0，∴對(duì)稱軸SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，∵當(dāng)SKIPIF1<0時(shí)對(duì)應(yīng)的函數(shù)值SKIPIF1<0均為正數(shù)，∴此時(shí)拋物線與x軸沒(méi)有交點(diǎn)，∴SKIPIF1<0，∴解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，∵當(dāng)SKIPIF1<0時(shí)對(duì)應(yīng)的函數(shù)值SKIPIF1<0均為正數(shù)，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴解得SKIPIF1<0，∴SKIPIF1<0，∴綜上所述，當(dāng)SKIPIF1<0時(shí)對(duì)應(yīng)的函數(shù)值SKIPIF1<0均為正數(shù)，則SKIPIF1<0的取值范圍為SKIPIF1<0或SKIPIF1<0．故選：D．【點(diǎn)睛】此題考查了二次函數(shù)的圖象和性質(zhì)，解題的關(guān)鍵是分兩種情況討論．11．（2023·山東煙臺(tái)·統(tǒng)考中考真題）如圖，拋物線SKIPIF1<0的頂點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，與SKIPIF1<0軸的一個(gè)交點(diǎn)位于0合和1之間，則以下結(jié)論：①SKIPIF1<0；②SKIPIF1<0；③若圖象經(jīng)過(guò)點(diǎn)SKIPIF1<0，則SKIPIF1<0；④若關(guān)于SKIPIF1<0的一元二次方程SKIPIF1<0無(wú)實(shí)數(shù)根，則SKIPIF1<0．其中正確結(jié)論的個(gè)數(shù)是（

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A．1 B．2 C．3 D．4【答案】C【分析】根據(jù)圖象，分別得出a、b、c的符號(hào)，即可判斷①；根據(jù)對(duì)稱軸得出SKIPIF1<0，再根據(jù)圖象得出當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即可判斷②；分別計(jì)算兩點(diǎn)到對(duì)稱軸的距離，再根據(jù)該拋物線開口向下，在拋物線上的點(diǎn)離對(duì)稱軸越遠(yuǎn)，函數(shù)值越小，即可判斷③；將方程SKIPIF1<0移項(xiàng)可得SKIPIF1<0，根據(jù)該方程無(wú)實(shí)數(shù)根，得出拋物線SKIPIF1<0與直線SKIPIF1<0沒(méi)有交點(diǎn)，即可判斷④．【詳解】解：①∵該拋物線開口向下，∴SKIPIF1<0，∵該拋物線的對(duì)稱軸在y軸左側(cè)，∴SKIPIF1<0，∵該拋物線于y軸交于正半軸，∴SKIPIF1<0，∴SKIPIF1<0，故①正確，符合題意；②∵SKIPIF1<0，∴該拋物線的對(duì)稱軸為直線SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，把SKIPIF1<0得：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由圖可知：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0，故②不正確，不符合題意；③∵該拋物線的對(duì)稱軸為直線SKIPIF1<0，∴SKIPIF1<0到對(duì)稱軸的距離為SKIPIF1<0，SKIPIF1<0到對(duì)稱軸的距離為SKIPIF1<0，∵該拋物線開口向下，∴在拋物線上的點(diǎn)離對(duì)稱軸越遠(yuǎn)，函數(shù)值越小，∵SKIPIF1<0，∴SKIPIF1<0，故③正確，符合題意；④將方程SKIPIF1<0移項(xiàng)可得SKIPIF1<0，∵SKIPIF1<0無(wú)實(shí)數(shù)根，∴拋物線SKIPIF1<0與直線SKIPIF1<0沒(méi)有交點(diǎn)，∵SKIPIF1<0，∴SKIPIF1<0．故④正確綜上：正確的有：①③④，共三個(gè)．故選：C．【點(diǎn)睛】本題主要考查了二次函數(shù)的圖象和性質(zhì)，解題的關(guān)鍵是掌握根據(jù)二次函數(shù)圖象判斷各系數(shù)的方法，熟練掌握二次函數(shù)的圖象和性質(zhì)．考向六二次函數(shù)交點(diǎn)問(wèn)題11．（2023·四川成都·統(tǒng)考中考真題）如圖，二次函數(shù)SKIPIF1<0的圖象與x軸交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，下列說(shuō)法正確的是（

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A．拋物線的對(duì)稱軸為直線SKIPIF1<0 B．拋物線的頂點(diǎn)坐標(biāo)為SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0兩點(diǎn)之間的距離為SKIPIF1<0 D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的值隨SKIPIF1<0值的增大而增大【答案】C【分析】待定系數(shù)法求得二次函數(shù)解析式，進(jìn)而逐項(xiàng)分析判斷即可求解．【詳解】解：∵二次函數(shù)SKIPIF1<0的圖象與x軸交于SKIPIF1<0，SKIPIF1<0兩點(diǎn)，∴SKIPIF1<0∴SKIPIF1<0∴二次函數(shù)解析式為SKIPIF1<0SKIPIF1<0，對(duì)稱軸為直線SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0，故A，B選項(xiàng)不正確，不符合題意；∵SKIPIF1<0，拋物線開口向上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的值隨SKIPIF1<0值的增大而減小，故D選項(xiàng)不正確，不符合題意；當(dāng)SK