Question

stie cineva rezolvarea?

Answer 1

$A(1;2) \in \ G_f \ ===\ \textgreater \ f(1)=2 \\\\ f(x)=x+a \ \ \ --\ \textgreater \ f(1)=1+a \ \ \ dar \ f(1)=2 \\\\\\ ===\ \textgreater \ 1+a=2 \longrightarrow \boxed{a= 1}$