Question 979279
There is a typo, or at least a failure to understand and/or apply the time-honored conventions on order of operations that are the basis of algebra.

F (x) = - 6x-6/5x+4 ={{{- 6x-6/"5 x"+4}}} graphs like this {{{graph(300,300,-1,4,-9,1,- 6x-6/5/x+4)
}}}
That function is negative for x >1, so that does not make sense.
If you just forgot that you needed parentheses, and meant
F (x) = - (6x-6)/(5x+4) ={{{- (6x-6)/(5x+4)}}} {{{graph(300,300,-1,9,-1,9,- (6x-6)/(5x+4))
}}} or F (x) = (-6x-6)/(5x+4) ={{{(-6x-6)/(5x+4)}}} {{{graph(300,300,-2,3,-9,1,(-6x-6)/(5x+4))
}}} ,
those functions are also negative for x >1, so they does not make sense either.

If you meant F (x) = (6x-6)/(5x+4) ={{{(6x-6)/(5x+4)}}} {{{drawing(500,200,-1,19,-0.5,1.5,locate(2,1.2,green(y=1.2)),locate(4,0.8,red(y=F(x))),graph(500,200,-1,19,-0.5,1.5,(6x-6)/(5x+4),1.2))}}} , that could make some sense for some drug, at least for the first few hours, or if it is being continuously infused.
{{{F(x)=(6x-6)/(5x+4)=(6x+4.8-10.8)/(5x+4)=(6x+4.8)/(5x+4)-10.8/(5x+4)=(6(x+0.8))/(5(x+0.8))-10.8/(5x+4)=6/5-10.8/(5x+4)=1.2-10.8/(5x+4)}}} ,
so {{{lim(x->infinity,F(x)=6/5)}}} , so {{{y=1.2}}} is the horizontal asymptote of {{{F(x)}}} ,
and that is the only asymptote that matters for {{{x>1}}} .
Looking at {{{F(x)=(6x-6)/(5x+4)}}} as a function without restrictions to its domain,
we may want to graph the function over more of its domain,
and there is, of course, a vertical asymptote for
{{{5x+4=0}}}<-->{{{5x=-4}}}<-->{{{x=-4/5}}} :
{{{drawing(500,500,-10,10,-9,11,green(line(-0.8,-9,-0.8,11)),locate(-3.2,-3,green(x=-0.8)),locate(2,2,green(y=1.2)),locate(4,0.8,red(y=F(x))),locate(-5,3,red(y=F(x))),graph(500,500,-10,10,-9,11,(6x-6)/(5x+4),1.2))}}}