Its image passe through points A(-2, 5) and B(-5, p),
∴5=k-2,
∴k=- 10,
∴ The analytical formula of inverse proportional function is y =- 10x.
∴p=- 10-5,
The coordinates of point B are (-5,2).
Let the expression of straight line AB be y=mx+n,
Then 5=-2m+n2=-5m+n,
∴m= 1n=7,
∴ The expression of straight line AB is y = x+7;
(2) Let the expression of CD be y=x+c from AB∨CD in□ABCD.
∴C(0,c),D(-c,0),
CD = AB,
∴CD2=AB2
∴c2+c2=(-5+2)2+(2-5)2,
∴c=-3,
∴ The coordinates of point C and point D are (0, -3) and (3, 0) respectively;
(3) Let the analytic expression of quadratic function be y=ax2+bx-3, and the image of quadratic function passes through points A and D. 。
∴5=4a-2b-30=9a+3b-3,
∴a= 1b=-2,
∴ The analytic formula of quadratic function is y = x2-2x-3.
Make EF⊥y axis and BG⊥y axis, and the vertical feet are f and g respectively.
OC = OD,BG=CG,
∴∠BCG=∠OCD=∠ODC=45。
∴∠BCD=90,
∠∠DCE =∠BDO,
∴∠ECF=∠BDC,
∴tan∠ECF=tan∠BDC=BCCD