Răspuns:
Explicație pas cu pas:
a) f:R->R, f(x )= 2x-1
∩ Ox <=> f(x) = 0 <=> 2x-1 = 0 <=> 2x = 1 => x = 1/2 => A(1/2 ; 0)
∩ Oy <=> x = 0 => f(0) = 2·0-1 = -1 => B (0 ; -1)
b) f:R->R, f(x) = -3x+2
∩ Ox <=> f(x) = 0 <=> -3x+2 = 0 => 3x = 2 => x = 2/3 => C(2/3 ; 0)
∩ Oy <=> x = 0 => f(0) = -3·0+2 = 2 => D(0 ; 2)