如图.二次函数y=ax2+bx+c(a>0)与坐标轴交于点A.B.C且OA=1.OB=OC=3 . (1)求此二次函数的解析式. (2)写出顶点坐标和对称轴方程. (3)点M.N在y=ax2+bx+c的图像上.且MN∥x轴.求以MN为直径且与x轴相切的圆的半径. (1)依题意分别代入 1分 解方程组得所求解析式为······································································ 4分 (2)··············································································· 5分 顶点坐标.对称轴················································································· 7分 (3)设圆半径为.当在轴下方时.点坐标为····························· 8分 把点代入得································································· 9分 同理可得另一种情形 圆的半径为或 10分 查看更多

 

题目列表(包括答案和解析)

如图,二次函数yax2bxc(a>0)与坐标轴交于点A、B、C且OA=1,OB=OC=3 .

(1)求此二次函数的解析式.

(2)写出顶点坐标和对称轴方程.

(3)点M、N在yax2bxc的图像上(点N在点M的右边),且MN∥ x轴,求以MN为直径且与x轴相切的圆的半径.

查看答案和解析>>

如图是二次函数y1=ax2+bx+c和一次函数y2=mx+n的图象,观察图象写出y2 ≥ y1时,x的取值范围 (    )

A.x≥0         B.0≤x≤1       C.-2≤x≤1           D.x≤-2或x≥1

 

查看答案和解析>>

如图,二次函数y=ax2+bx+c的图象与y轴正半轴相交,其顶点坐标为,下列结论:

①ac<0;②a+b=0;③4ac-b2=4a;④a+b+c<0.其中正确的个数是(    )

A. 1    B. 2     C. 3     D. 4

 

查看答案和解析>>

如图是二次函数y1=ax2+bx+c和一次函数y2=mx+n的图象,观察图象写出y2 ≥ y1时,x的取值范围 (   )
A.x≥0 B.0≤x≤1C.-2≤x≤1D.x≤-2或x≥1

查看答案和解析>>

如图是二次函数y=ax2+bx+c(a≠0)的图象的一部分,给出下列命题:①a+b+c=0;②b>2a;③ax2+bx+c=0的两根分别为-3和1;④a-2b+c>0.其中正确的命题是______.(只要求填写正确命题的序号)

查看答案和解析>>


同步练习册答案