Răspuns :
a) A(1;0)€Gf => f(1)=0
m-3+2m=0
3m-3=0
3m=3
m=1
b) m=1 => f(x)=(1-3)x+2×1
f(x)=-2x+2
Gf intersectat cu Oy: f(0)=-2×0+2=2 => B(0;2)
Gf intersectat cu Ox: f(x)=0 <=> -2x+2=0 <=> -2x=-2 <=> x=1 => A(1;0)
m-3+2m=0
3m-3=0
3m=3
m=1
b) m=1 => f(x)=(1-3)x+2×1
f(x)=-2x+2
Gf intersectat cu Oy: f(0)=-2×0+2=2 => B(0;2)
Gf intersectat cu Ox: f(x)=0 <=> -2x+2=0 <=> -2x=-2 <=> x=1 => A(1;0)