知2次函数y=ax平方+bx+c与y交与a、b,a在b左边与y交与c三角形aoc与boc面积和为6顶点坐标（2,-a）求解析式

y=ax^2+bx+c
顶点坐标（2,-a）
2 = -b/(2a)
-a = (4ac-b^2)/4a
∴b = - 4a,c=3a
∴y = ax^2 - 4ax + 3a = a(x^2-4x+3) = a(x-2)^2 -a
与x交与A、B,A在B左边：
a(x-2)^2-a=0
a(x-2)^2 = a
(x-2)^2 = 1
x = 2 ± 1
xA = 2-1 = 1
xB = 2+1 = 3
与y交与C
yc = 0-0+3a = 3a
三角形AOC与BOC面积和为6
1/2 * |xB-xA| * |yC| = 1/2 * (3-1) * |3a| = 6
a = ±2
解析式：
y = ax^2 - 4ax + 3a = -2x^2+8x-6
或
y = ax^2 - 4ax + 3a = 2x^2-8x+6