# 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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 Click here to see ALL problems on Graphs 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! SandyAnswer by stanbon(57412)   (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.