Question 117539
To move a function up, just subtract the amount OUTSIDE of the function.  If I want to move h(x) up 2 units to make f(x), then f(x) = h(x) + 3.  To move a function to the right subtract the amount INSIDE the function.  If I want to move h(x) to the right 3 units, then f(x) = h(x - 3).


So the f(x) that we are looking for, given {{{y=1/x}}}, is {{{y-2=1/(x-3)}}}.


Now we need this to be in the form of the quotient of two polynomials, so let's solve for y and simplify:


{{{y-2=1/(x-3)}}}
{{{y=(1/(x-3))+2}}}
{{{y=(1+2(x-3))/(x-3)}}}
{{{y=(2x-5)/(x-3)}}}

Vertical Asymptotes.
Look for values of x not in the domain of the function.  In this case the domain of the function is all reals except x = 3, therefore there is one vertical asymptote at x = 3.


Horizontal Asymptote.
Since the degree of the numerator polynomial and the denominator polynomial are equal, there is a horizontal asymptote at {{{y=a/b}}} where a is the lead coefficient on the numerator and b is the lead coefficient on the denominator.  In this case, {{{y=2/1=2}}}

There is no slant asymptote.  Slant asymptotes only occur where the numerator polynomial is of larger degree than the denominator polynomial.


{{{graph(600,600,-6,6,-6,6,(2x-5)/(x-3))}}}