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点A是反比例函数图象上一点,它到原点的距离为10,到x轴的距离为8,则此函数表达式可能为______
点A是反比例函数图象上一点,它到原点的距离为10,到x轴的距离为8,则此函数表达式可能为______
2025-03-16 10:17:52
推荐回答(1个)
回答1:
设反比例函数的解析式为:y=
k
x
,
设A点为(a,b),
∵点A是反比例函数图象上一点,它到原点的距离为10,
∴a
2
+b
2
=100①,
∵点A到x轴的距离为8,
∴|b|=8,把b值代入①得,
∴|a|=6,
∴A(6,8)或(-6,-8)或(-6,8)或(6,-8),
把A点代入函数解析式y=
k
x
,
得k=±48,
∴函数表达式为:y=
48
x
或y=-
48
x
,
.故答案为y=
48
x
或y=-
48
x
.
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