题干

已知二次函数y=ax2+bx+c中自变量x和函数值y的部分对应值如下表:

x

﹣1

0

1

2

3

y

10

5

2

1

2

(1)求该二次函数的函数关系式;

(2)在所给的直角坐标系中画出此函数的图象;

(3)写出y≤5时自变量x的取值范围(可以结合图象说明).

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答案(点此获取答案解析)

解:(1)由图表可知抛物线y=ax2+bx+c过点(1,2),(3,2),求出对称轴为x=2;∴顶点坐标为:(2,1),∴设y=a(x﹣2)2+1,将(1,2)代入可得:a+1=2,解得:a=1,∴二次函数的解析式为:y=(x﹣2)2+1=x2﹣4x+5.(2)由表格中的值可以判断:图象的对称点为:(1,2),(3,2),顶点坐标为:(2,1),画出函数的图象如图:

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