Question 728570
{{{y=2x-3+7/(x^2-1)=(2x^3-3x^2-2x-4)/(x^2-1)}}} {{{graph(300,300,-5,5,-15,10,2x-3,2x-3-7/(x^2-1))}}}
If we add to {{{y=2x-3}}} a term {{{f(x)}}} such that {{{lim( x->infinity, f(x) ) =0}}} ,
we get a function that has {{{y=2x-3}}} for an asymptote.
Since we want {{{y=4}}} for {{{x=0}}} (y-intercept=4),
we need
{{{2*0-3+f(0)=4}}} --> {{{-3+f(0)=4}}} --> {{{f(0)=4+3=7}}}

{{{g(x)=1/((x+1)(x-1))=1/(x^2-1)}}} is a function with vertical asymptotes {x=-1}}} and {{{x=1}}}  and
{{{lim( x->infinity,1/(x^2-1)) =0}}} .
{{{g(x)}}} is promising, but {{{g(0)=-1}}} ,
so we use {{{f(x)=-7g(x)}}} which makes {{{f(0)=-7*(-1)=7}}}.