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状元成才路,第,3,课时 二次函数,y,=,a,(,x,+,h,)+,k,的图象和性质,沪科版九年级数学上册,状元成才路,状元成才路,新课导入,导入课题,问题:说说,抛物线,y,=,ax,2,的,平移规律,.,y,=,ax,2,y,=,ax,2,+,k,y=a,(,x+h,),2,y=a,(,x+h,),2,+k,h,0,,向,平移,个单位,h,0,,向,平移,个单位,左,|,h,|,右,|,h,|,k,0,,向,平移,个单位,k,0,,向,平移,个单位,上,|,k,|,下,|,k,|,状元成才路,状元成才路,(,1,)会用描点法画二次函数,y=a,(,x+h,),2,+k,的图象,.,(,2,)能说出抛物线,y=a,(,x+h,),2,+k,与抛物线,y,=,ax,2,的相互关系,.,(,3,)能说出抛物线,y=a,(,x+h,),2,+k,的开口方向、对称轴、顶点,.,学习目标,状元成才路,状元成才路,推进新课,知识点,1,二次函数,y=a,(,x+h,),2,+k,的图象的画法,问题,3,方法一,方法二,状元成才路,状元成才路,2,4,y,-2,6,方法一,O,-2,2,x,4,-4,6,问题,3,x,-1,0,2,4,5,5.5,3,1,3,5.5,描点作图法,状元成才路,状元成才路,方法二,问题,3,1,上,1,状元成才路,状元成才路,先向,平移,个单位,再向,平移,个单位,向,平移,个单位,上,1,向,平移,个单位,右,2,?,右,2,上,1,方法二,平移法,状元成才路,状元成才路,问题,3,2,4,y,-2,6,O,-2,2,x,4,-4,6,状元成才路,状元成才路,问题,3,2,4,y,-2,6,O,-2,2,x,4,-4,6,先向,平移,个单位,再向,平移,个单位,右,2,上,1,状元成才路,状元成才路,-4,-2,y,-6,O,-2,2,x,4,-4,画一画,填出下表,:,状元成才路,状元成才路,-4,-2,y,-6,O,-2,2,x,4,-4,向左平移一个单位,向下平移一个单位,向左平移一个单位,,再向下平移一个单位,还有其他平移方法吗?,状元成才路,状元成才路,做一做,抛物线 的开口方向是,,顶点坐标是,(,,,),,对称轴是,.,当,x,时,函数,y,随,x,的增大而增大;,当,x,时,函数,y,随,x,的增大而减小;,当,x,=,时,函数取得最,值,,y,最,=,.,向上,1,-1,x,=1,1,1,1,小,小,-1,状元成才路,状元成才路,y,O,x,开口方向,对称轴,顶点坐标,上,下,x=,-h,x=,-h,(,-,h,k,),y=a,(,x+h,),2,+k,y=a,(,x+h,),2,+k,a,0,a0,a,0,图象,h,0,开口方向,对称轴,顶点坐标,函数的增减性,最值,当,x,-,h,时,,y,随,x,增大而减小,.,当,x,-,h,时,,y,随,x,增大而增大,.,向上,向下,直线,x=-h,直线,x=-h,(,-,h,k,),x=-h,时,,y,最小值,=k,x=-h,时,,y,最大值,=k,(,-,h,k,),状元成才路,状元成才路,y=a,(,x+h,),2,+k,y=ax,2,平移关系,?,二次函数,y=a,(,x+h,),2,+k,的几种图象:,这些图象与抛物线,y=ax,2,有什么关系?,状元成才路,状元成才路,结论,:,h,0,,将抛物线,y=ax,2,向左平移,,h,0,,将抛物线,y=ax,2,向上平移;,k,0),或向,右,(,h,0),或向,下,(,k,0),或向,右,(,h,0),或向,下,(,k,0),或向,右,(,h,0),或向,下,(,k,0),平移,|,k,|,个单位,y,O,x,y,=,ax,2,y=a,(,x+h,),2,+k,h,k,状元成才路,状元成才路,课后作业,1.,从课后习题中选取;,2.,完成练习册本课时的习题。,状元成才路,状元成才路,
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