How should this relationship be classified?

Domain = -2, range = -5

Domain =>0, range = 0

domain = 2, range = 5

let y =>f(x)

It can be seen that Function can be linear having common difference 2

F(x) = 2x + 1 for x>0

is satisfied for above values

let's check

F(2) = 2(2)+1 = 4+1 = 5 As given

For x<0 , F(x) = 2x -1

f(-2)= 2(-2) -1= -4-1= -5 As given

For X= 0, F(x)= 2x

So , F(0) = 0 As given

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong

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step-by-step explanation: