冪函數(shù)與二次函數(shù)課件

2-1-知識(shí)梳理雙基自測(cè)3415答案答案關(guān)閉(1)×

(2)×

(3)×

(4)√

(5)√

2-1-知識(shí)梳理雙基自測(cè)3415答案答案關(guān)閉(1)×-2-知識(shí)梳理雙基自測(cè)23415A.(-1,1)B.(-∞,-1)∪(1,+∞)C.(-∞,-2)∪(2,+∞)D.(-2,2)答案解析解析關(guān)閉答案解析關(guān)閉-2-知識(shí)梳理雙基自測(cè)23415A.(-1,1)答案解析-3-知識(shí)梳理雙基自測(cè)23415答案解析解析關(guān)閉答案解析關(guān)閉A.b<a<c B.a<b<cC.b<c<a D.c<a<b-3-知識(shí)梳理雙基自測(cè)23415答案解析解析關(guān)閉答案-4-知識(shí)梳理雙基自測(cè)234154.若冪函數(shù)

的圖象不經(jīng)過原點(diǎn),則實(shí)數(shù)m的值為

.

答案解析解析關(guān)閉答案解析關(guān)閉-4-知識(shí)梳理雙基自測(cè)234154.若冪函數(shù)-5-知識(shí)梳理雙基自測(cè)234155.(教材習(xí)題改編P82T10)已知冪函數(shù)y=f(x)的圖象過點(diǎn)

則此函數(shù)的解析式為

;在區(qū)間

上單調(diào)遞減.

答案解析解析關(guān)閉答案解析關(guān)閉-5-知識(shí)梳理雙基自測(cè)234155.(教材習(xí)題改編P82T1-6-考點(diǎn)1考點(diǎn)2考點(diǎn)3例1(1)若冪函數(shù)y=f(x)的圖象過點(diǎn)(4,2),則冪函數(shù)y=f(x)的圖象是(

)(2)已知冪函數(shù)(n∈Z)在(0,+∞)上是減函數(shù),則n的值為(

)A.-3 B.1 C.2 D.1或2思考冪函數(shù)與指數(shù)函數(shù)有怎樣的區(qū)別?冪函數(shù)有哪些重要的性質(zhì)?答案解析解析關(guān)閉答案解析關(guān)閉-6-考點(diǎn)1考點(diǎn)2考點(diǎn)3例1(1)若冪函數(shù)y=f(x)的圖象-7-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得1.冪函數(shù)中底數(shù)是自變量,指數(shù)是常數(shù),而指數(shù)函數(shù)中底數(shù)是常數(shù),指數(shù)是自變量.2.冪函數(shù)的主要性質(zhì):(1)冪函數(shù)在(0,+∞)上都有定義,冪函數(shù)的圖象都過定點(diǎn)(1,1).(2)當(dāng)α>0時(shí),冪函數(shù)的圖象都過點(diǎn)(1,1)和(0,0),且在(0,+∞)上單調(diào)遞增.(3)當(dāng)α<0時(shí),冪函數(shù)的圖象都過點(diǎn)(1,1),且在(0,+∞)上單調(diào)遞減.(4)冪函數(shù)圖象在第一象限的特點(diǎn):當(dāng)α>1時(shí),曲線下凸;當(dāng)0<α<1時(shí),曲線上凸;當(dāng)α<0時(shí),曲線下凸.-7-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得1.冪函數(shù)中底數(shù)是自變量,指-8-考點(diǎn)1考點(diǎn)2考點(diǎn)3答案解析解析關(guān)閉答案解析關(guān)閉-8-考點(diǎn)1考點(diǎn)2考點(diǎn)3答案解析解析關(guān)閉答案解析關(guān)閉-9-考點(diǎn)1考點(diǎn)2考點(diǎn)3例2已知二次函數(shù)f(x)滿足f(2)=-1,f(-1)=-1,且f(x)的最大值是8,試確定此二次函數(shù)的解析式.思考求二次函數(shù)解析式時(shí)如何選取恰當(dāng)?shù)谋磉_(dá)形式?-9-考點(diǎn)1考點(diǎn)2考點(diǎn)3例2已知二次函數(shù)f(x)滿足f(2)-10-考點(diǎn)1考點(diǎn)2考點(diǎn)3-10-考點(diǎn)1考點(diǎn)2考點(diǎn)3-11-考點(diǎn)1考點(diǎn)2考點(diǎn)3-11-考點(diǎn)1考點(diǎn)2考點(diǎn)3-12-考點(diǎn)1考點(diǎn)2考點(diǎn)3解法三

(利用交點(diǎn)式)由已知f(x)+1=0的兩根為x1=2,x2=-1,故可設(shè)f(x)+1=a(x-2)(x+1),即f(x)=ax2-ax-2a-1.又函數(shù)有最大值ymax=8,因此所求函數(shù)的解析式為f(x)=-4x2+4x+7.-12-考點(diǎn)1考點(diǎn)2考點(diǎn)3解法三(利用交點(diǎn)式)-13-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得根據(jù)已知條件確定二次函數(shù)的解析式,一般用待定系數(shù)法,選擇規(guī)律如下:(1)已知三個(gè)點(diǎn)坐標(biāo),宜選用一般式.(2)已知頂點(diǎn)坐標(biāo)、對(duì)稱軸、最大(小)值等,宜選用頂點(diǎn)式.(3)已知圖象與x軸的兩個(gè)交點(diǎn)坐標(biāo),宜選用交點(diǎn)式.-13-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得根據(jù)已知條件確定二次函數(shù)的-14-考點(diǎn)1考點(diǎn)2考點(diǎn)3對(duì)點(diǎn)訓(xùn)練2已知二次函數(shù)f(x)有兩個(gè)零點(diǎn)0