Question 1194411
assuming you have:

a) 

{{{f(x) = (4x+3)/(x+2)}}}

domain:

{{{x<-2 }}}or {{{x>-2}}}

Interval Notation:

({{{-infinity}}},{{{-2}}}) U ({{{-2}}},{{{infinity}}})

range: 
range of given function is equat to domain of inverse
so, find inverse of {{{(4x+3)/(x+2)}}} which is {{{(3-2x)/(x-4) }}} =>domain is {{{x<4 }}}or {{{x>4}}}
then we say the range of (4x+3)/(x+2) is:

{{{f(x)<4}}} or{{{ f (x )>4}}}

asympthotes:

Vertical: {{{x=-2}}}
Horizontal: {{{y=4}}}


b) 

{{{f(x) =(x+2)/(x^2-1)}}}

domain:

{{{x<-1 }}}or {{{-1<x<1}}} or {{{x>1}}}

Interval Notation:

({{{-infinity}}},{{{-1}}}) U ({{{-1}}},{{{1}}}) U ({{{1}}},{{{infinity}}})  

range:
{{{f(x) =y}}}
{{{y=(x+2)/(x^2-1)}}}
{{{y(x^2-1)=x+2}}}
{{{yx^2-y-x-2=0}}}
{{{yx^2-x-(y+2)=0}}}

range is where discriminant is >=0
discriminant is {{{b^2-4ac}}}
in your case {{{a=y}}}, {{{b=-1}}},{{{ c=-(y+2)}}}
{{{(-1)^2-4y(-(y+2))>=0}}}
{{{1+4y^2+8y>=0}}}
{{{4y^2+8y+1>=0}}}
{{{y>=1/2 (sqrt(3) - 2)}}}=> {{{y>=sqrt(3) - 1}}}
{{{y<=1/2 (-2 - sqrt(3)) }}}=> {{{y<=-1 - sqrt(3)/2}}}

 {{{f(x)<= -sqrt(3)/2-1}}} or  {{{f(x)>=sqrt(3)/2-1}}}

asymptotes:

Vertical: {{{x=-1}}},{{{x=1}}}
Horizontal: {{{y=0}}}