[tex]\displaystyle\\
I)\\
I.1)\\
f(-2)=2\cdot(-2)-3=-4-3=-7~\Longrightarrow~(-2,~-7)~\text{ este corect.}\\\\
f(0)=2\cdot0-3=0-3=-3~\Longrightarrow~(0,~-3)~\text{ este corect.}\\\\
f(2)=2\cdot2-3=4-3=1~\Longrightarrow~(2,~1)~\text{ este corect.}\\\\
\Longrightarrow~G_f=\{(-2,~-7);~(0,~-3);~(2,~1)\}~\text{ este corect.}\\\\\\
I.2)\\
\text{Verificam daca punctul }~N(-3,~1)\in G_f\\\\
f(x)=-\frac{1}{3}x+1\\
f(-3)=-\frac{1}{3}\cdot(-3)+1=-\frac{-3}{3}+1=\frac{3}{3}+1=1+1=2\\
2\neq1\\
Concluzie:~~N(-3,~1)\notin G_f[/tex]
