题目内容

如图,某公路隧道横截面为抛物线,其最大高度为6米,底部宽度OM为12米. 现以O点为原点,OM所在直线为x轴建立直角坐标系.

1.直接写出点M及抛物线顶点P的坐标;

2.求这条抛物线的解析式;

3.若要搭建一个矩形“支撑架”AD- DC- CB,

使C、D点在抛物线上,A、B点在地面OM上,

 

【答案】

 

1.M(12,0),P(6,6).

2.设抛物线解析式为:.  ································································· 3分

∵抛物线经过点(0,0),

,即                  4分

∴抛物线解析式为:

  .        

3.设A(m,0),则

B(12-m,0),. ···································· 7分

∴“支撑架”总长AD+DC+CB =

=.  ··································································· 10分

 ∵ 此二次函数的图象开口向下.

∴ 当m = 3米时,AD+DC+CB有最大值为15米.

 【解析】略

 

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