The image of the ∫ linear function intersects the X axis at points A (-3,0) and B (0,3 root number 3).

∫C(3，0)

∴OA=OC

Y axis ⊥AC

∴AB=BC

In Rt△AOB,

∴∠BAC=60

△ ABC is an equilateral triangle.

(2)① Answer: ∠ AEP = 120

② Solution:

Connect DC,

The y axis bisects AC vertically, and △ABC is an equilateral triangle.

∴DA=DC,∠BDA=∠BDC=,∠DBP=30

∴∠BDH=60

∫DH vertical division CP

∴ SAR =DP

∴ DA=DC=DP

In△△ CDP, ∠CDH=∠PDH.

≈BDH =≈BDC+≈CDH =+= 60

∴∠ADB=∠ADC+∠PDC= 120

(3) Make PG⊥x axis at G point.

In Rt△PGC, PC=t, CG=t/2 PG= root 3t/2.

In Rt△BDH, DH= root number 3/3 (t/2 +6).

∴DO=B-BO= root number 3/3 t+ root number 3

y=S 1-S2

=S△DAC-S△PAC

Replace it.

∴y=- root number 3/2 t+root number 3 (t > 0)