How do you graph #y = 1/(x+2) - 1#?

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Answer 1 · 2015-06-06T18:12:11

This is a hyperbolic function with asymptotes #y = -1# and #x = -2#

When #x = -2+epsilon# for small #epsilon > 0#, #y# is large and positive.

When #x = -2-epsilon# for small #epsilon > 0#, #y# is large and negative.

As #x->-oo# we have #y->-1_-#

As #x->oo# we have #y->-1_+#

The line of slope #1# through the intersection #(-2, -1)# of the asymptotes cuts the hyperbola at its two vertices #(-1, 0)# and #(-3, -2)#

graph{1/(x+2)-1 [-12.08, 7.92, -5.68, 4.32]}