SOLUTION: I'm trying to figure out the horizontal aymptote of the following equation, or if there is one: h(x) = [1/(x+2)] - 1 I think it is 0 because 0/1 = n<m, y = 0 Am I correc

Algebra ->  Algebra  -> Graphs -> SOLUTION: I'm trying to figure out the horizontal aymptote of the following equation, or if there is one: h(x) = [1/(x+2)] - 1 I think it is 0 because 0/1 = n<m, y = 0 Am I correc      Log On

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Question 22289: I'm trying to figure out the horizontal aymptote of the following equation,
or if there is one:
h(x) = [1/(x+2)] - 1
I think it is 0 because 0/1 = n Am I correct?
Thanks for your help!
Sandy
Answer by stanbon(57412) About Me  (Show Source):
You can put this solution on YOUR website!
The vertical asymptote occurs when the denominator is zero.
In your problem that happens when x+2=0 or when x=-2
Cheers,
Stan H.