How do you find the asymptote and graph #y=1/(x+3)#?

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Answer 1 · 2018-06-06T04:14:30

#y# has a vertical asymptote at #x=-3#
#y# has a horizontal asymptote at #y=0#
Graph below.

#y=1/(x+3)#

#y# is defined #forall x !=-3#

Let's examine #y# as #x# approaches #-3# from above and from below.

#lim_(x->-3^+) 1/(x+3) ->+oo#

#lim_(x->-3^-) 1/(x+3) ->-oo#

Therefore, #y# is discontinuous at #x=-3# and has a vertical asymptote at that point.

Now consider:

#lim_(x->-oo) y = 0 and lim_(x->+oo) y =0#

Hence the #x-#axis forms a horizontal asymptote.

Therefore, #y# has a horizontal asymptote at #y=0#

#y# is a rectangular hyperbola with the graph below.

graph{1/(x+3) [-7.58, 4.904, -3.4, 2.843]}