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find asymptotes of f(x) = (x+1)/(x^2+x)
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@danish071996 @ganeshie8
factor denominator and see if anything cancels out
x=0
so is there a hole at x=-1
and what are the horizontal and vertical asymptotes
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you're right, x = -1 is just a hole cuz the factor (x+1) cancels out
the only realy vertical asymptote is x=0
do you know the process to find horizontal asymptotes ?
why isn't there a vertical asymptote at x=-1
the numerator ate the factor x+1
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\[\large f(x) = \dfrac{x+1}{x^2+x} = \dfrac{x+1}{x(x+1)} = \dfrac{1}{x}~~, x\ne -1\]
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