Question 904333
{{{f(x)=1/(x+3)}}} does not exist when {{{x+3=0}}}<--->{{{x=-3}}} .
For any other value of {{{x}}} , {{{f(x)=1/(x+3)}}} can be calculated,
so the domain is all the real numbers except {{{-3}}} .
{{{x=-3}}} is the equation of the vertical asymptote.
For {{{x<-3}}} , {{{x+3<0}}} and {{{f(x)=1/(x+3)<0}}} .
For {{{x>-3}}} , {{{x+3>0}}} and {{{f(x)=1/(x+3)>0}}} .
As {{{x}}} approaches {{{-3}}} (from either side),
{{{abs(x+3)}}} approaches zero, and {{{abs(f(x))=abs(1/(x+3))=1/abs(x+3)}}} grows without bounds.
As {{{abs(x)}}} increases, {{{abs(f(x))=abs(1/(x+3))=1/abs(x+3)}}} approaches zero,
so {{{y=0}}} is the horizontal asymptote.
{{{f(x)=1/(x+3)}}} can take any value, except zero,
so its range is all the real numbers except zero.
{{{drawing(300,300,-8,3,-20,20,
graph(300,300,-8,3,-20,20,1/(x+3)),
green(line(-3,-20,-3,20))
)}}}