(2014?绵阳)如图,抛物线y=ax2+bx+c(a≠0)的图象过点M(-2,3),顶点坐标为N(-1,433),且与x轴交

2024-12-23 03:07:23
推荐回答(1个)
回答1:

(1)由抛物线顶点坐标为N(-1,

4
3
3
),可设其解析式为y=a(x+1)2+
4
3
3

将M(-2,
3
)代入,得
3
=a(-2+1)2+
4
3
3

解得a=-
3
3

故所求抛物线的解析式为y=-
3
3
x2-
2
3
3
x+
3


(2)∵y=-
3
3
x2-
2
3
3
x+
3

∴x=0时,y=
3

∴C(0,
3
).
y=0时,-
3
3
x2-
2
3
3
x+
3
=0,
解得x=1或x=-3,
∴A(1,0),B(-3,0),
∴BC=
OB2+OC2
=2
3

设P(-1,m),
当CP=CB时,有CP=