摘要：例1一座拱桥的轮廓是抛物线型,拱高6m,跨度20m,相邻两支柱间的距离均为5m. (1)将抛物线放在所给的直角坐标系中,求抛物线的解析式; (2)求支柱的长度, (3)拱桥下地平面是双向行车道(正中间是一条宽2m的隔离带),其中的一条行车道能否并排行驶宽2m.高3m的三辆汽车?请说明你的理由. 解:(1)根据题目条件,的坐标分别是．······························ 1分 设抛物线的解析式为y=ax2+bx+c,············································································· 2分 将的坐标代入y=ax2+bx+c.得 100a-10b+c=0 100a+10b+c=0 c=6··············· 3分 解得a=-,b=0,c=6．··············································· 4分 所以抛物线的表达式是.····················· 5分 (2)可设,于是 ························································································· 6分 从而支柱EF的长度是米.··································································· 7分 (3)设是隔离带的宽,是三辆车的宽度和, 则点坐标是.······························································································· 8分 过点作垂直交抛物线于,则.················ 9分 根据抛物线的特点,可知一条行车道能并排行驶这样的三辆汽车.····························· 10分 例2.如图4,小明的父亲在相距2米的两棵树间拴了一根绳子,给他做了一个简易的秋千,拴绳子的地方距地面高都是2.5米,绳子自然下垂呈抛物线状,身高1米的小明距较近的那棵树0.5米时,头部刚好接触到绳子,则绳子的最低点距地面的距离为 米. B A 分析:如果把左边的树子看成纵轴,地平线看成横轴,则 A , C C 可设函数解析式为y=ax2+bx+c,把A.B.C三点分别代入这个 解析式可得一个方程组 c=2.5 4a+2b+c=2.50.25a+0.5b+c=1 解之得:a=2, b=-4, c=2.5 所以y= 2x2-4x+2.5 当x=1时, y=2-4+2.5=0.5

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