1、3.4 一元二次函數(shù)的圖象和性質(zhì)復(fù)習(xí)目標(biāo)1 1 . .掌握一元二次函數(shù)圖象的畫(huà)法及圖象的特征2 2 . .掌握一元二次函數(shù)的性質(zhì),能利用性質(zhì)解決實(shí)際問(wèn)題3 3 . .會(huì)求二次函數(shù)在指定區(qū)間上的最大(小)值4 4 . .掌握一元二次函數(shù)、一元二次方程的關(guān)系.知識(shí)回憶1 .函數(shù)yax2bxc(a0)叫做一元二次函數(shù).2. .一元二次函數(shù)的圖象是一條拋物線.23 .任何一個(gè)一次函數(shù)yaxbxc(a0)都可把它的解析式配萬(wàn)為頂點(diǎn)式：(b2ya(x丁丁) )2a性質(zhì)如下:(2)(2)最大(小)值4acb當(dāng)a0,函數(shù)圖象開(kāi)口向下,y有最大值,ymax,無(wú)最小值.4a(3)當(dāng)a0,函數(shù)在區(qū)間(,包)上是減函

2、數(shù),在(,)上是增函數(shù).2a2abb當(dāng)a0,函數(shù)在區(qū)間上(,)是減函數(shù),在(上-)上是增函數(shù).2a2a【說(shuō)明】【說(shuō)明】1 1.我們研究二次函數(shù)的性質(zhì)常用的方法有兩種：配方法和公式法.2 2.無(wú)論是利用公式法還是配方法我們都可以直接得出二次函數(shù)的頂點(diǎn)坐標(biāo)與對(duì)稱軸;但我們討論函數(shù)的最值以及它的單調(diào)區(qū)間時(shí)一定要考慮它的開(kāi)口方向.例題精解一、一元二次函數(shù)的圖象的畫(huà)法【例【例 1 1】求作函數(shù)y1x24x6的圖象21c1c【斛】【斛】yx4x6(x8x12)2224acb4a,(1)(1)圖象的頂點(diǎn)坐標(biāo)為b4acb2a4a2-),對(duì)稱軸是直線b.2a0,函數(shù)圖象開(kāi)口向上,y有最小值,ymin4acb,無(wú)

3、取大值.4a(x24)2-4(x24)2-222以x4為中間值,取x的一些值,列表如下:x-7-6-5-4-3-2-1y52032-232052【例【例 2 2】求作函數(shù)yx24x3的圖象.(x2)27(x2)273 3描點(diǎn)成圖；也可利用圖象的對(duì)稱性,出右左局部就可.元二次函數(shù)性質(zhì)【例【例 3 3】求函數(shù)yx26x9的最小值及圖象的對(duì)稱軸和頂點(diǎn)坐標(biāo),并求它的單調(diào)區(qū)間.【解】【解】yx26x2x26x97x327由配方結(jié)果可知：頂點(diǎn)坐標(biāo)為3,7,對(duì)稱軸為x3;10當(dāng)x3時(shí),ymin725x3x1圖象的頂點(diǎn)坐標(biāo)、對(duì)稱軸、最值及它的單調(diào)區(qū)間.-2一2b334acb4(5)1329-2a2(5)104

4、a4(5)20【點(diǎn)評(píng)】要研究二次函數(shù)頂點(diǎn)、對(duì)稱軸、最值、單調(diào)區(qū)間等性質(zhì)時(shí),方法有兩個(gè):1 1配方法；如例 3 3【解】【解】yx24x3(x24x3)x2的右邊局部,列表x-2-1012y76543【點(diǎn)評(píng)】畫(huà)二次函數(shù)圖象步驟：1 1配方；2 2列表；先畫(huà)出函數(shù)的左右邊局部圖象,再利用對(duì)稱性描函數(shù)在區(qū)間3上是減函數(shù),在區(qū)間3,上是增函數(shù).【例【例 4 4】求函數(shù)y329.函數(shù)圖象的頂點(diǎn)坐標(biāo)為士,29,對(duì)稱軸為x10203,50,當(dāng)x一時(shí),函數(shù)取得最大值103.函數(shù)在區(qū)間,一上是增函數(shù),在區(qū)間3,102920ymaz2920上是減函數(shù).2 2公式法：適用于不容易配方題目二次項(xiàng)系數(shù)為負(fù)數(shù)或分?jǐn)?shù)如例

5、4,4,可防止出錯(cuò).先畫(huà)出圖角在對(duì)稱軸任何一個(gè)函數(shù)都可配方成如下形式:b2ya(x)2a三、二次函數(shù)性質(zhì)的應(yīng)用【例【例 5 5】如果fxx2bxc對(duì)于任意實(shí)數(shù)t都有f3t)f(3t),那么(【解】【解】解法y(n2)x2nx1是偶函數(shù),f(x)的圖像關(guān)于x3對(duì)稱又a10拋物線開(kāi)口向上.f(3)是f(x)的最小值.f(3)f(4)f(1)2(2)(2)如果f(x)xbxc對(duì)于任意實(shí)數(shù)t都有f(2t)f(2t),那么f(1)f(1).(用或“填空) )【解】【解】f(2t)f(2t)對(duì)于一切的tR均成立.f(x)的圖像關(guān)于x2對(duì)稱又a10拋物線開(kāi)口向下.1(2)1(2),f(1)f(1)【點(diǎn)評(píng)】1

6、 1.當(dāng)a0時(shí),對(duì)稱軸通過(guò)它的最低點(diǎn)(此時(shí)函數(shù)有最小值) ),如果這時(shí)有一個(gè)點(diǎn)離圖象對(duì)稱軸越遠(yuǎn),那么對(duì)應(yīng)的函數(shù)彳苴就越大.如例 5(1)5(1)中當(dāng)x1所對(duì)應(yīng)的點(diǎn)比當(dāng)x4所對(duì)應(yīng)的點(diǎn)離對(duì)稱軸遠(yuǎn),所以x1時(shí)對(duì)應(yīng)的函數(shù)值也比擬大.2 2. .1 1.當(dāng)a0時(shí),對(duì)稱軸通過(guò)它的最高點(diǎn)(此時(shí)函數(shù)有最大值) ),如果這時(shí)有一個(gè)點(diǎn)離圖象對(duì)稱軸越遠(yuǎn),那么對(duì)應(yīng)的函數(shù)值就越小.如例 5(2)5(2)中當(dāng)x1所對(duì)應(yīng)的點(diǎn)比當(dāng)x1所對(duì)應(yīng)的點(diǎn)離對(duì)稱軸遠(yuǎn),所以x1對(duì)應(yīng)的函數(shù)值也比擬小.【例 6 6】求函數(shù)yx22x5在給定區(qū)間1,5上的最值.【解】【解】(1)(1)原函數(shù)化為yx22x5x126a10當(dāng)x1時(shí),ymin6又1

7、151.當(dāng)x5時(shí),ymax(51)26101o10(2)原函數(shù)可化為：y(x-)2一,圖象的對(duì)稱軸是直線x39注意到當(dāng)1x2時(shí),函數(shù)為減函數(shù)22413yminf(2)2-214-1333(n2)x2nx1是偶函數(shù),試比擬f(2),f(6),f(北)的大小.(A)f(3)f(1)f(4)(B)f(1)f(3)f(4)(C)f(3)f(4)f(1)(D)f(4)f(3)f(1)f(3t)f(3t)對(duì)于一切的tR均成立【例 7 7】函數(shù)yA.4,)B.(4,)C.(,4D.(,4)2.2.如果二次函數(shù) y y5x2mx4在區(qū)間(1)上是減函數(shù),在區(qū)間1,)上是增函數(shù),那么mn0, ,y2x21可知函

8、數(shù)的對(duì)稱軸為直線x0又a20,|5020卜丘0f(、.2)f(2)f(、,5)解法二：.y(m1)x22mx3是偶函數(shù),n0, ,y2x21可知y2x21在(0,)上單調(diào)遞減2又y(n2)xnx1是偶函數(shù),f(.5)f(,5)而.52,2f(.2)f(2)f(,5)f(-.2)f(2)f(,5)三、一元二次函數(shù)、一元二次方程的關(guān)系.【例 8 8】求當(dāng)k為何值時(shí),函數(shù)y2x24xk的圖象與x軸(1)(1)只有一個(gè)公共點(diǎn)；(2)(2)有兩個(gè)公共點(diǎn)；(3)(3)沒(méi)有公共點(diǎn). .【解】令2x24xk0,那么2x2xk0的判別式b24ac168k(1)(1)當(dāng)0,即168k0,k2時(shí),方程有兩個(gè)相等的實(shí)

9、根,這時(shí)圖象與x軸只有一個(gè)公共點(diǎn)；(2)(2)當(dāng)0,即168k0,k2時(shí),方程有兩個(gè)不相等的實(shí)根,這時(shí)圖象與x軸有兩個(gè)公共點(diǎn)；(3)(3)當(dāng)0,即168k0,k2時(shí),方程有兩個(gè)不相等的實(shí)根,這時(shí)圖象與x軸無(wú)公共點(diǎn)；同步練習(xí)一.選擇題21.一次函數(shù)yx2x5的值域是()3.如果二次函數(shù)yx2mxm3有兩個(gè)不相等的實(shí)數(shù)根,那么m的聚值范圍是中,對(duì)稱軸是直線x1的是的是A.收B.2C.4i二.填空A.2C.10D.-10A.(,2)(6,)B.(2,6)C.2,6)0D.2,6一、“,10-4.函數(shù)yxX3的最小值是(21A.-3.B.3.25.函數(shù)y2x24x2具有性質(zhì)(31.2A,開(kāi)口方向向上,

10、對(duì)稱軸為B,開(kāi)口方向向上,對(duì)稱軸為C.開(kāi)口方向向下,對(duì)稱軸為D,開(kāi)口方向向下,對(duì)稱軸為6.以下命題正確的選項(xiàng)是x1,頂點(diǎn)坐標(biāo)為-1,0 x1,頂點(diǎn)坐標(biāo)為1,0 x1,頂點(diǎn)坐標(biāo)為-1,0 x1,頂點(diǎn)坐標(biāo)為1,0A.函數(shù)y2x26x3的最小值是322152x6x3的最小值是4C.函數(shù)yx24x3的最小值為72D.函數(shù)yx4x3的最大值為77.函數(shù)1y222x4x3；y2x4x3；(3)y223x6x3；(4)y3x6x3A.(1)與(2)B.(2)與(3)C.(1)與(3)D.(2)與(4)8.對(duì)于二次函數(shù)y2x28x,以下結(jié)論正確的選項(xiàng)是A.當(dāng)x2時(shí),y有最大值8C.當(dāng)x2時(shí),y有最小值8B.當(dāng)x2時(shí),y有最大值8D.當(dāng)x2時(shí),y有最小值829.如果函數(shù)yaxbxc(a0,對(duì)于任意實(shí)數(shù)t都有f2tf2t,那么以下選項(xiàng)中正確A.f(2)f(1)f(4)B-f(1)f(2)f(4)C.f(2)f(4)f(1)D.f(4)f(2)f(1)10.假設(shè)二次函數(shù)y22ax4x1有最小值,那么實(shí)數(shù)a=D.21.假設(shè)函數(shù)fx2x2x1,那么fx的對(duì)稱軸是直線3 .函數(shù)y2x23x9的圖象與y軸的交點(diǎn)坐標(biāo)是,與x軸的交點(diǎn)坐標(biāo)是、4.y9x26x6,那么y有最值為5 .y4x228x1,那么y有最值為三.解做題1 .二次函數(shù)yx24x3,(1)指出函數(shù)圖象的開(kāi)口方向；(2)當(dāng)x為何值時(shí)y0;(