Question 866602
The expected description is whatever you were taught in your math class, and we do not know what that was.
Also, I am not sure if you really mean
{{{green(y= 4/x + 1)}}} as you wrote,
or if you really meant
{{{blue(y=4/(x+1))}}} which should be written as y=4/(x+1).
Either function is derived from {{{red(y=1/x)}}} .
translated and dilated vertically:
{{{graph(300,300,-10,10,-10,10,1/x,4/x + 1,4/(x+1))}}}
 
If you are studying functions and their graphs:
you would be expected to say that there is a vertical asymptote, and a horizontal asymptote.
The vertical asymptote is {{{x=0}}} for {{{green(y= 4/x + 1)}}} and {{{red(y=1/x)}}} ,
and it is {{{x+1=0}}}<--->{{{x=-1}}} for {{{blue(y=4/(x+1))}}} .
The horizontal asymptote is {{{y=0}}} for
{{{blue(y=4/(x+1))}}} and {{{red(y=1/x)}}} ,
and it is {{{y=1}}} for {{{green(y= 4/x + 1)}}} .
You may be expected to say more about the function, but there are many possibilities, and I may not think if every one.
 
If you are studying conic sections:
you may be expected to sat that
{{{y=1/x}}}<--->{{{xy=1}}} graphs as a hyperbola,
with the transverse axis along the line {{{y=x}}}.
The graphs of the other functions,
{{{blue(y=4/(x+1))}}} and {{{green(y= 4/x + 1)}}} ,
are also hyperbolas.