Greetings from Brasil...
A first degree function is given by the expression:
f(x) = ax + b ⇔ y = ax + b
where
a = Δy/Δx ⇒ a = (y₂ - y₁)/(x₂ - x₁)
choosing 2 values arbitrarily from the table
x₁ = 8 and y₁ = 1
x₂ = 12 and y₂ = 3
So
a = Δy/Δx = (y₂ - y₁)/(x₂ - x₁)
a = (3 - 1)/(12 - 8)
a = 2/4
a = 1/2
for b:
f(x) = ax + b ⇒ f(x) = x/2 + b ⇔ y = x/2 + b
Remember that when x = 8, y = 1, so
y = x/2 + b
1 = 8/2 + b
b = 1 - 4
b = - 3
So,
y = ax + b (a = 1/2 and b = - 3)
y = (x/2) - 3
or
f(x) = (x/2) - 3