How do you find any asymptotes and intercepts for this function?

Question

#1##-##log_3#(#x#+1)

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Answer 1 · 2018-01-23T13:32:02

Function has intercepts #(0,1) and (2,0)# and has a vertical asymptote at #x=-1#.

#f(x) = 1-log_3 (x+1)#

#f(x)# is defined for #(x+1)>0# and #lim_(x->-1) f(x) = +oo#

Hence, #f(x)# has a vertical asymptote at #x=-1#.

At the #y-#intercept #x=0 ->f(0) = 1-log_3(0+1)#

#f(0) = 1 - log_3 1 #

#=1 -0 =1#

Hence, the #y-# intercept is #(0,1)#

At the #x-#intercept #f(x )=0 -> 1-log_3 (x+1) =0#

Then, #log_3(x+1) = 1#

#(x+1) = 3^1 -> x=3-1#

#x=2#

Hence, the #x-# intercept is #(2,0)#

We can see these results from the graph of #f(x)# below.

graph{1-ln(x+1)/ln3 [-2.236, 8.863, -1.493, 4.056]}