Substitute x=t and y=0 into y=x2+bx to get t2+bt=0.

∫t > 0，

∴b=﹣t；

(2)① unchanged.

As shown in fig. 6, when x= 1 and y = 1-t, then M( 1, 1-t),

∫tan∠AMP = 1，

∴∠amp=45；

②S=S quadrilateral amnp-s △ PAM = s △ dpn+s trapezoid ndam-s △ PAM =

(t﹣4)(4t﹣ 16)+

[(4t﹣ 16)+(t﹣ 1)]×3﹣

(t﹣ 1)(t﹣ 1)=

t2﹣

t+6。

solve

t2﹣

t+6=

De: t 1=

，t2=

∫4 < t < 5，

∴t 1=

Give it up,

∴t=

.

(3)

< t