Asymptotes
Find the vertical asymptote of this rational function by setting its denominator to #0# and solving for #x#.
Let #x-1=0#
#x=1#
Which means that there's a vertical asymptote passing through the point #(1,0)#.
*FYI you can make sure that #x=1# does give a vertical asymptote rather than a removable point of discontinuity by evaluating the numerator expression at #x=1#. You can confirm the vertical asymptote if the result is a non-zero value. However if you do end up with a zero, you'll need to simplify the function expression, remove the factor in question, for example #(x-1)#, and repeat those steps. *
You may find the horizontal asymptote (a.k.a "end behavior") by evaluating #lim_{x to infty}4/(x-1)# and #lim_{x to -infty}4/(x-1)#.
If you haven't learned limits yet, you'll still able to find the asymptote by plugging in large values of #x# (e.g., by evaluating the function at #x=11#, #x=101#, and #x=1001#.) You'll likely find that as the value of #x# increase towards positive infinity, the value of #y# getting closer and closer to- but never reaches #0#. So is the case as #x# approaches negative infinity.
By definition , we see that the function has a horizontal asymptote at #y=0#
Graph
You might have found the expression of #y=1/x#, the #x#-reciprocal function similar to that of #y=4/(x-1)#. It is possible to graph the latter based on knowledge of the shape of the first one.
Consider what combination of transformations (like stretching and shifting) will convert the first function we are likely familiar with, to the function in question.
We start by converting
#y=1/x# to #y=1/(x-1)#
by shifting the graph of the first function to the right by #1# unit. Algebraically, that transformation resembles replacing #x# in the original function with the expression #x-1#.
Finally we'll vertically stretch the function #y=1/(x-1)# by a factor of #4# to obtain the function we're looking for, #y=4/(x-1)#. (For rational functions with horizontal asymptotes the stretch would effectively shifts the function outwards.)
