Solution: y=x? -2x-3

2. (1) Let y=0, that is:

x？ -2x-3=0

∴(x-3)(x+ 1)=0

x=3，x=- 1，

Because b is on the positive semi-axis

So-1 is discarded.

∴B coordinate (3,0)

∴ Let BC be y = KX+B.

Substitute in the b and c coordinates.

D: y=x-3

(2)OB=3，OC=3

∠BOC=90

∴△BOC is an isosceles right triangle.

∴∠ABC=∠OBC=45

(3) The existence of point P can be found by vector method. It's simple. Try it yourself.