Question

如图1,抛物线y=ax2+bx+3与x轴相交于点A（-3,0）,B（-1,0）,与y轴相交于点C,⊙O1为△ABC的外接

Answer 1

(1)

A(-3, 0): 9a - 3b + 3 = 0

B(-1, 0): a-b+ 3 = 0

a = 1, b = 4

y = x² +4x + 3 = (x + 2)² - 1

对称轴x = -2

C(0, 3)

O₁(-2, c), O₁B² = O₁C²

(-2+ 1)² + (c - 0)² = (-2 - 0)² + (c - 3)²

c = 2

O₁(-2, 2)

O₁C斜率p= (3 - 2)/(0 + 2) = 1/2

O₁B斜率q = (2 - 0)/(-2 + 1) = -2

pq = -1

△BO₁C为直角三角形

(2) O₁(-2, 2)

(3)

EF与 AB平行, 且EF = AB = (-1)- (-3) = 2

设F(f, f² +4f + 3), 则E(-2, f² +4f + 3)

|f + 2| = 2

f = 0, F(0, 3)

f = -4, F(-4, 3)