#f(x) = (3x+3)/(2x+4) = (3(x+1))/(2(x+2)) = 3/2*(x+1)/(x+2)#

#=3/2*(x+2-1)/(x+2)#

#=3/2((x+2)/(x+2)-1/(x+2))#

#=3/2(1-1/(x+2))#

#=3/2-3/(2(x+2))#

with exclusion #x != -2#

So as #x->-oo# the term #3/(2(x+2)) ->0_-# so #f(x)->(3/2)_+#

As #x->(-2)_-# the term #3/(2(x+2))->-oo# so #f(x)->oo#

As #x->(-2)_+# the term #3/(2(x+2))->oo# so #f(x)->-oo#

As #x->oo# the term #3/(2(x+2)) -> 0_+# so #f(x)->(3/2)_-#

The intersection with the #x# axis occurs at #(-1, 0)# since the numerator of the original expression is #0# for #x = -1#

The intersection with the #y# axis can be found by substituting #x=0# into the original equation to derive #(0, 3/4)#

graph{ (3x+3)/(2x+4) [-12.29, 7.71, -3.44, 6.56]}