Then the point (2,1) is obtained by linear function: 2a+b =1.

( 1)

∵Q(x+ 1, y+3) is on the image of f(x).

∴a(x+ 1)+b=y+3 ……①

∫P(x, y) on f(x) image

∴y=ax+b substitute into formula ①: ax+a+b=ax+b+3.

Solution: a=3

∫2a+b = 1 ∴b=-5

The analytical formula of the function f(x) is: f(x)=3x-5.

(2)

∵ right x∈, f(x)>=0 holds.

∴①a>； 0, f(x) monotonically increases on r.

∴f(x) Minimum value =f(0)=b≥0

②a & lt； 0, f(x) monotonically decreases on r.

∴f(x) Minimum value =f(4)=4a+b≥0

∫2a+b = 1 ∴a=( 1-b)/2: 2-b≥0； b≤2。

∴ To sum up: 0≤b≤2

∫F(x) and the coordinate of the intersection of the y axis are (0, b).

∴ f (x), the value range of the ordinate of the intersection point of the y axis is 0,2.