Question 687244
I believe you mean g(x) = 1/(x-4) + 2,
which I can write as
{{{g(x)=1/(x-4)+2}}}.
 
The function {{{f(x)=1/x}}}  {{{graph(200,200,-8,8,-8,8,1/x)}}} is sort of an ancestor of your function.
It has the x- and y-axes for asymptotes.
The y-axis, {{{x=0}}} is a vertical asymptote
because the function does not exist for {{{x=0}}} (it is not continuous at {{{x=0}}})
but as you approach {{{x=0}}}, from either side,
{{{abs(f(x))=abs(1/x)}}} grows without bounds.
The horizontal x-axis, {{{y=0}}}, is an asymptote because as {{{abs(x)}}} grows larger (at both ends of the x-axis),
{{{f(x)=1/x}}} approaches zero.
 
The function {{{h(x)=1/(x-4)}}} {{{graph(200,200,-6,10,-8,8,1/(x-4))}}}
is the same, but moved 4 units to the right, a closer ancestor.
Its asymptotes are {{{x=4}} and {{{y=0}}}.
{{{x=4}}} (vertical asymptote, where the function is not continuous), and
{{{y=0}}} (again).
 
Adding {{{2}}} to that function gives you {{{g(x)=1/(x-4)+2}}}.
The graph is {{{graph(200,200,-6,10,-4,12,1/(x-4)+2,2)}}}
and the asymptotes are {{{y=2}}} and
{{{x=4}}}.
 
Just five points seems too little for this function, but the symmetry might help.
For {{{x=4.1}}}, {{{g(x)=1/(4.1-4)+2=1/0.1+2=10+2=12}}}.
(Symmetrically, for {{{x=3.9}}}, {{{g(x)=1/(3.9-4)+2=1/(-0.1)+2=-10+2=-8}}}.
For {{{x=4.5}}}, {{{g(x)=1/(4.5-4)+2=1/0.5+2=2+2=4}}}.
(Symmetrically, for {{{x=3.5}}}, {{{g(x)=1/(3.5-4)+2=1/(-0.5)+2=-2+2=0}}}.
For {{{x=5}}}, {{{g(x)=1/(5-4)+2=1/1+2=1+2=3}}}.
(Symmetrically, for {{{x=3}}}, {{{g(x)=1/(3-4)+2=1/(-1)+2=-1+2=1}}}.
For{{{x=9}}}, {{{g(x)=1/(9-4)+2=1/5+2=0.2+2=2.2}}}.
(Symmetrically, for {{{x=-1}}}, {{{g(x)=1/(-1-4)+2=1/(-5)+2=-0.2+2=1.8}}}.
For{{{x=14}}}, {{{g(x)=1/(14-4)+2=1/10+2=0.1+2=2.1}}}.
(Symmetrically, for {{{x=-6}}}, {{{g(x)=1/(-6-4)+2=1/(-10)+2=-0.1+2=1.9}}}.
We can plot those points as blue dots and connect them.{{{drawing(300,300,-10,14,-10,14,
grid(1),
green(line(-10,2,14,2)),green(line(4,-10,4,16)),
blue(circle(4.1,12,0.2)),blue(circle(3.9,-8,0.2)),
blue(circle(4.5,4,0.2)),blue(circle(3.5,0,0.2)),
blue(circle(5,3,0.2)),blue(circle(3,1,0.2)),
blue(circle(9,2.2,0.2)),blue(circle(-1,1.8,0.2)),
blue(circle(14,2.1,0.2)),blue(circle(-6,1.9,0.2))
)}}}
We can draw the asymptotes in green too.