#f(x)=log_(1/3) (x+5)#

The vertical asymptote is found by setting #x+5# equal to zero.

#x+5=0 => x=-5#

The base of the log is #1/3#. A base that is less than one indicates that the graph is a decreasing log.

To find the #x# intercept, set #f(x)=0#

#0=log_(1/3) (x+5)#

Rewrite as an exponential and solve.

#(1/3) ^0 = x +5#

#1=x+5#

#x=-4# when #f(x)=0 =># the #x# intercept is #(-4,0)#

The #y# intercept is found by setting #x=0#

#y=log_(1/3) (0+5)#

#y=log_(1/3) 5#

Use the change of base formula # log_a b = logb/loga#

#y=log5/(log (1/3))=-1.46 =>#the #y# intercept is #(0, -1.46)#

See the graph below.

Alternatively, the function can be rewritten as an exponential and an #xy# table can be constructed by choosing values of #y# and finding corresponding #x# values.

#y=log_(1/3) (x+5)#

#x+5 = (1/3)^y#

#x=(1/3)^y -5#

Then choose values of #y# such as #-2,-1,0,1,2# and find the corresponding values of #x#. Plot the resulting #xy# coordinates.