Question 1033903
I will explain just some of it.


Notice degree 1 for both numerator and denominator, equal, so that you can have horizontal asymptote.  {{{a/1=-4}}} and similarly  {{{r/2=-4}}}.



The given x-intercept indicates  {{{ax+b=0}}} and {{{rx+s=0}}}.


You must have denominator equal to zero if x is 3, but x<>3 is the requirement for having given vertical asymptote of x=3.