[tex]f:R \rightarrow R,~f(x)=\frac{1}{3}x+2\\~
G_f = \{(x,f(x)):x \in R\}\\~\\~
M(a,2a+1) \in G_f \iff f(a) = 2a+1\\~
\frac{1}{3}a+2 = 2a+1\\~
\frac{1}{3}a-2a+1 = 0\\~
\frac{-5a}{3}+1=0\\~
\frac{-5}{3}a=-1\\~
\frac{5}{3}a=1\\~
a = \frac{3}{5}[/tex]
Am obținut punctul [tex]M(a,2a+1)=M(\frac{3}{5},\frac{6}{5}+1)=M(\frac{3}{5},\frac{11}{5})[/tex]
