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.