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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