1. Point A (6 6,0) is brought into 36a+6b=0. The ordinate of vertex B is -3, and -b 2/4a is -3. The solutions are b=0, a=0 or b=-2. a= 1/3。 Analytical formula Y =

2. Vertex B(3, -3) m is the intersection of a straight line and the Y axis. The equation of -3=3k+m k=-(3+m) /3 is obtained by pulling the vertices into a straight line.

Y = kx+m. point D (x 1, 0) brings in 0 = kx1+m0 =-(3+m)/3 * x1+MX1= 3m/(3+m) =

≤3 1/3≤ 1/(3/m+ 1)≤2/3 1≤x 1≤2。

The height of OD side of △BOD is point B, the absolute value of ordinate 3 OD is point D, and the absolute value of abscissa x 1.

△ BOD =1/2 * x 1 * 3 = 3/2x1.. On the linear function of x1(monotonically increasing), 1≤x 1≤2.

X 1= 1 minimum area S=3/2. X 1=2 Maximum area S=3.