a) A € Gf <=> f(1) = 5
f(1) = a•1 + 3 = a+3
a+3 = 5 => a = 5-3 => a = 2
b) pt. a = 2 fcț. devine f(x) = 2x+3
f(x) = 0 => 2x+3 = 0 => 2x = -3 => x = -3/2
c) pt. a = -2 fcț. devine f(x) = -2x+3
f(x) > 0 => -2x+3 >0
-2x > (-3) | • (-1)
2x < 3
x < 3/2
x € (-infinit, 3/2)