题目内容
(2004•青海)一男生在校运会的比赛中推铅球,铅球的行进高度y(m)与水平距离x(m)之间的关系用如图所示的二次函数图象表示.(铅球从A点被推出,实线部分表示铅球所经过的路线)(1)由已知图象上的三点,求y与x之间的函数关系式;
(2)求出铅球被推出的距离;
(3)若铅球到达的最大高度的位置为点B,落地点为C,求四边形OABC的面积.
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【答案】分析:(1)由已知图象上的三点坐标,设一般式y=ax2+bx+c,列方程组,求解析式;
(2)求OC长,令y=0,求x的值;
(3)求面积要抓住A、B、C三点坐标,把四边形分割成一个直角梯形和一个直角三角形,求面积和.
解答:
解:(1)设y与x之函数关系式为y=ax2+bx+c
由图象得,图象经过(-2,0),(0,
),(2,
)三点,则:
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解得:a=-
,b=
,c=
∴y与x之间的函数关系式为y=-
x2+
x+
;
(2)令y=0,则-
x2+
x+
=0
解得:x1=10,x2=-2(不合题意,舍去)
∴铅球被推出的距离是10米;
(3)过B作BD⊥OC于D
∵y=-
(x2-8x-20))=-
(x-4)2+3
∴B点坐标(4,3)
由(2)得C点坐标是(10,0)
∴S四边形OABC=S梯形OABD+S△BDC=
×(
+3)×4+
×6×3=18
.
答:四边形OABC的面积为18
.
点评:题考查点的坐标的求法及二次函数的实际应用.此题为数学建模题,借助二次函数解决实际问题.
(2)求OC长,令y=0,求x的值;
(3)求面积要抓住A、B、C三点坐标,把四边形分割成一个直角梯形和一个直角三角形,求面积和.
解答:
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由图象得,图象经过(-2,0),(0,
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解得:a=-

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∴y与x之间的函数关系式为y=-
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(2)令y=0,则-


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解得:x1=10,x2=-2(不合题意,舍去)
∴铅球被推出的距离是10米;
(3)过B作BD⊥OC于D
∵y=-


∴B点坐标(4,3)
由(2)得C点坐标是(10,0)
∴S四边形OABC=S梯形OABD+S△BDC=

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答:四边形OABC的面积为18
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点评:题考查点的坐标的求法及二次函数的实际应用.此题为数学建模题,借助二次函数解决实际问题.
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