And because the fixed point m is in the first quadrant, according to the symmetry axis is greater than zero, b=3.
(2)① y =-x 2+3x from (1)
Vertex M(3/2, 9/4)
So the length of AB is an integer value of 1 or 2.
When AB= 1, BC is not an integer (truncation);
When AB=2, B (1, 0) c (2,0) is BC= 1.
So the circumference of the rectangular ABCD is 6.
(2) Let the abscissa of point A be X by the analytical formula E (3 3,0), then:
Circumference = 2 (3-2x)+2 (-x 2+3x)
=-2x^2+2x+6
=-2(x- 1/2)^2+ 13/2
So when x= 1/2, the maximum perimeter is 13/2, and at this time A( 1/2, 5/4).
③ Area = (3-2x) (-x 2+3x)
=x(x-3)(2x-3)