**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.)