【答案】
分析:先對(duì)字母a的取值進(jìn)行分類(lèi)討論::①當(dāng)a=0時(shí);②當(dāng)a>0時(shí);③當(dāng)a<0時(shí).再針對(duì)二次函數(shù)圖象,找對(duì)稱(chēng)軸,利用開(kāi)口向上(或向下)的二次函數(shù)在對(duì)稱(chēng)軸右邊遞增(減),左邊遞減(增)即可研究其單調(diào)性.
解答:解:①當(dāng)a=0時(shí),y=-2x+3,是一次函數(shù),在[-3,3]上單調(diào)遞減;
②當(dāng)a>0時(shí),函數(shù)y=ax
2-2(3a+1)x+3的圖象是對(duì)稱(chēng)軸為x=3+
>3,開(kāi)口向上的拋物線,
所以在[-3,3]上是減函數(shù);
③當(dāng)a<0時(shí),函數(shù)y=ax
2-2(3a+1)x+3的圖象是對(duì)稱(chēng)軸為x=3+
<3,開(kāi)口向下的拋物線,
(i)當(dāng)-
≤a<0時(shí),函數(shù)y=ax
2-2(3a+1)x+3的圖象是對(duì)稱(chēng)軸為x=3+
≤-3,開(kāi)口向下的拋物線,
所以在[-3,3]上是減函數(shù);
(ii)當(dāng)a<-
時(shí),函數(shù)y=ax
2-2(3a+1)x+3的圖象是對(duì)稱(chēng)軸為x=3+
∈[-3,3],開(kāi)口向下的拋物線,
所以在[-3,3+
]上是增函數(shù);在(3+
,3]上是減函數(shù);
綜上,a≥
時(shí),在[-3,3]上是減函數(shù);當(dāng)a<-
時(shí),在[-3,3+
)上是增函數(shù);在(3+
,3]上是減函數(shù).
點(diǎn)評(píng):本題主要考查了二次函數(shù)的單調(diào)性,二次函數(shù)的單調(diào)區(qū)間有對(duì)稱(chēng)軸和開(kāi)口方向二者決定,開(kāi)口向上的二次函數(shù)在對(duì)稱(chēng)軸右邊遞增,左邊遞減;開(kāi)口向下的二次函數(shù)在對(duì)稱(chēng)軸左邊遞增,右邊遞減.