#y=2/(x+1)-5#
#y# is defined for all real x except where #x=-1# because #2/(x+1)# is undefined at #x=-1#
N.B. This can be written as: #y# is defined #forall x in RR: x!=-1#
Let's consider what happens to #y# as #x# approaches #-1# from below and from above.
#lim_(x->-1^-) 2/(x+1)-5=-oo#
and
#lim_(x->-1^+) 2/(x+1)-5=+oo#
Hence, #y# has a vertical asymptote at #x=-1#
Now let's see what happens as #x-> +-oo#
#lim_(x->+oo) 2/(x+1)-5 =0-5 =-5#
and
#lim_(x->-oo) 2/(x+1)-5 =0-5 =-5#
Hence, #y# has a horizontal asymptote at #y=-5#
#y# is a rectangular hyperbola with "parent" graph #2/x#, shifted 1 unit negative on the #x-#axis and 5 units negative on the #y-#axis.
To find the intercepts:
#y(0) = 2/1-5 -> (0,-3)# is the #y-#intercept.
#2/(x+1)-5 =0 -> 2-5(x+1)=0#
#-5x=3 -> (-0.6,0)# is the #x-#intercept.
The graph of #y# is shown below.
graph{2/(x+1)-5 [-20.27, 20.29, -10.13, 10.14]}
