Solution:

∴ A (- 1, 0), B (3 3,0) 2 points.

∵ = ,

The symmetry axis of parabola is the straight line x= 1,

Substitute x= 1 to get y=2.

∴ C (1, 2) .3 point

(2) ① In Rt△ACE, tan∠CAE=,

∴∠CAE=60？ ,

According to the symmetry of parabola, l is the middle vertical line of line segment ab,

∴AC=BC，

∴△ABC is an equilateral triangle with four points.

∴AB= BC =AC = 4，∠ABC=∠ACB= 60？ ,

AM = AP，BN=BP，

∴BN = cm,

∴△ABN≌△BCM，

∴ An = bm.5 o'clock

② The minimum area of quadrilateral AMNB is 0.6.

Let AP=m and the area of quadrilateral AMNB be s,

It can be seen from ① that AB= BC= 4, BN = CM=BP, S△ABC= ×42=,

∴CM=BN= BP=4-m，CN=m，

M is MF⊥BC, and the vertical foot is F.

So MF=MC? sin60？ = ,

∴S△CMN= =？ =, 7 points

∴S=S△ABC-S△CMN