Answer
We start with the graph of $f_{1}(x)=\log_{6}(x)$ (dashed blue).
Since g$(x)=\log_{6}(x-4)=f(x-$3$)$
its graph (red) is obtained by
shifting $f_{1}(x)$ to the right by 3 units.
Domain :$\qquad($3$,+\infty)$
Range :$\qquad(-\infty,+\infty)$
Vertical asymptote: $\qquad x=$3
Work Step by Step
We start with the graph of $f_{1}(x)=\log_{6}(x)$ (dashed blue).
Since g$(x)=\log_{6}(x-3)=f(x-$3$)$
its graph (red) is obtained by
shifting $f_{1}(x)$ to the right by 3 units.
Domain :$\qquad($3$,+\infty)$
Range :$\qquad(-\infty,+\infty)$
Vertical asymptote: $\qquad x=$3
