Question 778620
F(x) = (x+2)/(x+3) where x is not equal to 0. Parentheses are not in the original problem, I included them to show what is over what.
<pre>
I'm glad you put those parentheses there.  Most students would
have written  " x+2/x+3 " which really means {{{x+2/x+3}}}, an
altogether different expression.

But you didn't state what you were to find.  I assume it is 
the domain and range.

The denominator x+3 must not equal 0, so

x+3 &#8800; 0
  x &#8800; -3,  so we have a vertical asymptote at x = -3

The degree of the numerator and denominator are both 1, so
the horizontal asymptote is

y = {{{(LEADING_COEFFICIENT_OF_THE_NUMERATOR)/(LEADING_COEFFICIENT_OF_THE_DENOMINATOR)}}}

y = {{{1/1}}}

y = 1 , so we have a horizontal asymptote at y = 1

You also state "x is not equal to 0", so we must leave out the
point (0,{{{2/3}}}) and put an open circle there. The two asymptotes
are in green:

{{{drawing(800,4000/7,-10,4,-5,5, graph(800,4000/7,-10,4,-5,5,((x+2)/(x+3))*sqrt(x-.15)/sqrt(x-.15)),  graph(800,4000/7,-10,4,-5,5,((x+2)/(x+3))*sqrt(-x-.15)/sqrt(-x-.15)),locate(0,2/3,"(0,2/3)"),

green(line(-20,1,20,1),line(-3,-20,-3,20)),red(circle(0,2/3,.15)) )}}}

So the domain is (-&#8734;,-3)U(-3,0)U(0,&#8734;) and

the range is (-&#8734;,{{{2/3}}})U({{{2/3}}},1)U(1,&#8734;)

Edwin</pre>