第七講：函數(shù)圖像、函數(shù)與方程【考點梳理】1、函數(shù)的圖象(1)平移變換：SKIPIF1<0SKIPIF1<0(2)伸縮變換：SKIPIF1<0SKIPIF1<0(3)對稱變換：SKIPIF1<0SKIPIF1<0SKIPIF1<0(4)翻折變換：SKIPIF1<0SKIPIF1<02、函數(shù)與方程(1)判斷二次函數(shù)SKIPIF1<0在SKIPIF1<0上的零點個數(shù)，一般由對應(yīng)的二次方程SKIPIF1<0的判別式SKIPIF1<0來完成；對于一些不便用判別式判斷零點個數(shù)的二次函數(shù)，則要結(jié)合二次函數(shù)的圖象進行判斷．(2)對于一般函數(shù)零點個數(shù)的判斷，不僅要用到零點存在性定理，還必須結(jié)合函數(shù)的圖象和性質(zhì)才能確定，如三次函數(shù)的零點個數(shù)問題．(3)若函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象是連續(xù)不斷的一條曲線，且是單調(diào)函數(shù)，又SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有唯一零點．【典型題型講解】考點一：函數(shù)的圖像【典例例題】例1．（多選題）在同一直角坐標系中，函數(shù)SKIPIF1<0的圖象可能是（

）A． B．C． D．【方法技巧與總結(jié)】1.熟練掌握高中八個基本初等函數(shù)的圖像的畫法2.函數(shù)的圖像變換：平移，對稱、翻折變換【變式訓練】1.已知圖①中的圖象是函數(shù)SKIPIF1<0的圖象，則圖②中的圖象對應(yīng)的函數(shù)可能是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02.已知函數(shù)SKIPIF1<0無最小值，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03.若函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）在R上為減函數(shù)，則函數(shù)SKIPIF1<0的圖象可以是（）A． B．C． D．4.函數(shù)SKIPIF1<0的圖象如圖所示，則函數(shù)SKIPIF1<0的圖象為（）A． B．C． D．考點二：求函數(shù)的零點或零點所在區(qū)間判斷【典例例題】例1.已知函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0是SKIPIF1<0的一個零點，則SKIPIF1<0一定是下列函數(shù)的零點的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0例2.函數(shù)SKIPIF1<0的零點所在的區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【方法技巧與總結(jié)】求函數(shù)SKIPIF1<0零點的方法：（1）代數(shù)法，即求方程SKIPIF1<0的實根，適合于宜因式分解的多項式；（2）幾何法，即利用函數(shù)SKIPIF1<0的圖像和性質(zhì)找出零點，適合于宜作圖的基本初等函數(shù).【變式訓練】1.已知函數(shù)SKIPIF1<0，則SKIPIF1<0的所有零點之和為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的零點分別是a，b，c，則a，b，c的大小順序是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03.函數(shù)SKIPIF1<0的所有零點之和為__________．4．若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則x、y、z由小到大的順序是___________.5.函數(shù)SKIPIF1<0的零點所在的區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0考點三：函數(shù)零點個數(shù)的判斷【典例例題】例1.函數(shù)SKIPIF1<0的零點個數(shù)為___________.例2.定義在R上的偶函數(shù)SKIPIF1<0滿足SKIPIF1<0，且當SKIPIF1<0時，SKIPIF1<0，若關(guān)于x的方程SKIPIF1<0恰有5個解，則m的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【方法技巧與總結(jié)】1.利用函數(shù)圖像判斷方程解的個數(shù).由題設(shè)條件作出所研究對象的圖像，利用圖像的直觀性得到方程解的個數(shù).2.利用函數(shù)圖像求解不等式的解集及參數(shù)的取值范圍.先作出所研究對象的圖像，求出它們的交點，根據(jù)題意結(jié)合圖像寫出答案3.利用函數(shù)圖像求函數(shù)的最值，先做出所涉及到的函數(shù)圖像，根據(jù)題目對函數(shù)的要求，從圖像上尋找取得最值的位置，計算出結(jié)果，這體現(xiàn)出了數(shù)形結(jié)合的思想?！咀兪接柧殹?.已知函數(shù)SKIPIF1<0是偶函數(shù)，且SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，則方程SKIPIF1<0在區(qū)間SKIPIF1<0上的解的個數(shù)是________2.已知函數(shù)f（x）＝SKIPIF1<0和函數(shù)g（x）＝SKIPIF1<0，則函數(shù)h（x）＝f（x）－g（x）的零點個數(shù)是________．3.已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0有6個零點，則m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04.已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有4個零點，則實數(shù)k的取值范圍為_______________.5.已知函數(shù)SKIPIF1<0，SKIPIF1<0若關(guān)于x的方程SKIPIF1<0有四個不同的解，則實數(shù)m的取值集合為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06.已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0有且僅有三個不同的實數(shù)解，則實數(shù)SKIPIF1<0的取值范圍是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【鞏固練習】一、單選題1．函數(shù)SKIPIF1<0的圖象大致為（

）A． B．C． D．2．聲音是由物體振動產(chǎn)生的．我們平時聽到的聲音幾乎都是復合音．復合音的產(chǎn)生是由于發(fā)音體不僅全段在振動，它的各部分如二分之一、三分之一、四分之一等也同時在振動．不同的振動的混合作用決定了聲音的音色，人們以此分辨不同的聲音．己知刻畫某聲音的函數(shù)為SKIPIF1<0，則其部分圖象大致為（

）A． B．C． D．3．若函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）在R上既是奇函數(shù)，又是減函數(shù)，則SKIPIF1<0的大致圖象是（

）A． B．C． D．4．已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0與SKIPIF1<0的圖象恰有5個不同公共點，則實數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0二、多選題5．設(shè)函數(shù)SKIPIF1<0，則下列命題中正確的是（

）A．若方程SKIPIF1<0有四個不同的實根SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍是SKIPIF1<0B．若方程SKIPIF1<0有四個不同的實根SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍是SKIPIF1<0C．若方程SKIPIF1<0有四個不同的實根，則SKIPIF1<0的取值范圍是SKIPIF1<0D．方程SKIPIF1<0的不同實根的個數(shù)只能是1，2，3，66．已知SKIPIF1<0為常數(shù)，函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0恰有四個零點，則實數(shù)SKIPIF1<0的值可以是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．已知函數(shù)SKIPIF1<0若關(guān)于x的方程SKIPIF1<0有5個不同的實根，則實數(shù)a的取值可以為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．已知SKIPIF1<0