How do you graph #f(x)=x/(x+2)#?

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Answer 1 · 2015-07-01T22:53:18

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

As #x -> -oo# then #f(x) -> 1_+#

As #x -> -2_-# then #f(x) -> oo#

As #x -> -2_+# then #f(x) -> -oo#

#f(0) = 0#

As #x -> oo# then #f(x) -> 1_-#

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

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

with exclusion #x != -2#

So it is clear that for large #x#, #f(x)# is asymptotic to #1# and that #f(x)# has a simple pole at #x = -2#

graph{x/(x+2) [-10, 10, -5, 5]}