函數(shù)-二次最值

專題研究,制作：金一中金建軍,二次函數(shù)在區(qū)間上的最值問(wèn)題,-7,問(wèn)題1求函數(shù)y=3x2-12x+5當(dāng)自變量x在下列范圍內(nèi)取值時(shí)的最值，并求此函數(shù)取最值時(shí)的x值（1）xR；（2）0x3（3）-1x1,解：,配方得y=3(x-2)27，如圖：,(1)xR時(shí),當(dāng)x=2時(shí),ymin=7,3,問(wèn)題1求函數(shù)y=3x2-12x+5當(dāng)自變量x在下列范圍內(nèi)取值時(shí)的最值，并求此函數(shù)取最值時(shí)的x值（1）xR；（2）0x3（3）-1x1,解：,配方得y=3(x-2)27，如圖：,(1)xR時(shí),當(dāng)x=2時(shí),ymin=7,(2)0x3時(shí)，函數(shù)在0,2上單調(diào)遞減,在2,3上單調(diào)遞增當(dāng)x=0時(shí),ymax=5當(dāng)x=2時(shí)，ymin=7,-7,-4,20,問(wèn)題1求函數(shù)y=3x2-12x+5當(dāng)自變量x在下列范圍內(nèi)取值時(shí)的最值，并求此函數(shù)取最值時(shí)的x值（1）xR；（2）0x3（3）-1x1,解：,配方得y=3(x-2)27，如圖：,(1)xR時(shí),當(dāng)x=2時(shí),ymin=7,(2)0x3時(shí)，函數(shù)在0,2上單調(diào)遞減,在2,3上單調(diào)遞增當(dāng)x=0時(shí),ymax=5當(dāng)x=2時(shí)，ymin=7,(3)-1x1時(shí)，函數(shù)在-1,1上單調(diào)遞減當(dāng)x=-1時(shí)，ymax=20當(dāng)x=1時(shí)，ymin=-4,解:函數(shù)圖象的對(duì)稱軸為直線x=1,拋物線開(kāi)口向上,問(wèn)題2求函數(shù)y=x2-2x+3在區(qū)間0，a上的最值，并求此時(shí)x的值。,2,當(dāng)x=0時(shí)，ymax=3當(dāng)x=a時(shí)，ymin=a2-2a+3,1.當(dāng)a1時(shí)，函數(shù)在0，a上單調(diào)遞減，,當(dāng)x=0時(shí)，ymax=3當(dāng)x=a時(shí)，ymin=a2-2a+3,函數(shù)在0,1上單調(diào)遞減,在1,a上單調(diào)遞增,當(dāng)x=1時(shí),ymin=2當(dāng)x=0時(shí)，ymax=3,解:函數(shù)圖象的對(duì)稱軸為直線x=1,拋物線開(kāi)口向上,問(wèn)題2求函數(shù)y=x2-2x+3在區(qū)間0，a上的最值，并求此時(shí)x的值。,2.當(dāng)1a2時(shí),1.當(dāng)a1時(shí)，函數(shù)在0，a上單調(diào)遞減，,函數(shù)在0,1上單調(diào)遞減,在1，a上單調(diào)遞增,當(dāng)x=1時(shí),ymin=2,當(dāng)x=a時(shí),ymax=a2-2a+3,問(wèn)題2求函數(shù)y=x2-2x+3在區(qū)間0，a上的最值，并求此時(shí)x的值。,3.當(dāng)a2時(shí),函數(shù)在0,1上單調(diào)遞減,在1,a上單調(diào)遞增,當(dāng)x=1時(shí),ymin=2當(dāng)x=0時(shí)，ymax=3,解:函數(shù)圖象的對(duì)稱軸為直線x=1,拋物線開(kāi)口向上,1.當(dāng)a1時(shí)，函數(shù)在0，a上單調(diào)遞減，,當(dāng)x=0時(shí)，ymax=3當(dāng)x=a時(shí)，ymin=a2-2a+3,2.當(dāng)1-,對(duì)稱軸在x=-的右邊.,(1)當(dāng)-1a時(shí),即a0時(shí),由二次函數(shù)圖象,可知:ymax=f()=,(2)當(dāng)a時(shí),即-1a0時(shí),綜上所述:當(dāng)-1a0時(shí),ymax=0當(dāng)a0時(shí),ymax=,問(wèn)題3求函數(shù)y=-x(x-a)在x-1,a上的最大值,(2)當(dāng)a時(shí),即-1a0時(shí),由二次函數(shù)的圖象可知:ymax=f(a)=0,思考關(guān)于x的方程x2-(k-2)x+k2+3k+5=0有兩個(gè)實(shí)根,.求2+2的最值。,解:由題意,可得=(k-2)2-4(k2+3k+5)0即:3k2+16k+160,-4k-,由方程根與系數(shù)的關(guān)系知+=k-2,=k2+3k+5，2