Answer :
[tex]f(x)=\frac{5x-1}{x-3}=\frac{5x-15+14}{x-3}=\frac{5x-15}{x-3}+\frac{14}{x-3}=\frac{14}{x-3}+\frac{5(x-3)}{x-3}=\frac{14}{x-3}+5\\\\y=\frac{14}{x}\\\\\Downarrow T_{\vec{a}=[3;\ 5]-translate\ 3\ units\ right\ and\ 5\ units\ up}\\\\f(x)=\frac{14}{x-3}+5[/tex]
[tex]y=\frac{14}{x}\\\\\underline{x|-28|-14|-7|-2|\ -1\ |\ 1\ |\ 2\ |\ 7\ |14|28|}\\y|-\frac{1}{2}\ |\ -1\ |-2|-7|-14|14\ |\ 7\ |\ 2\ |\ 1|\ \frac{1}{2}\ |[/tex]
[tex]y=\frac{14}{x}\\\\\underline{x|-28|-14|-7|-2|\ -1\ |\ 1\ |\ 2\ |\ 7\ |14|28|}\\y|-\frac{1}{2}\ |\ -1\ |-2|-7|-14|14\ |\ 7\ |\ 2\ |\ 1|\ \frac{1}{2}\ |[/tex]
