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Mathematics 24 Online
OpenStudy (anonymous):

find asymptotes of f(x) = (x+1)/(x^2+x)

OpenStudy (anonymous):

@danish071996 @ganeshie8

ganeshie8 (ganeshie8):

factor denominator and see if anything cancels out

OpenStudy (anonymous):

x=0

OpenStudy (anonymous):

so is there a hole at x=-1

OpenStudy (anonymous):

and what are the horizontal and vertical asymptotes

ganeshie8 (ganeshie8):

you're right, x = -1 is just a hole cuz the factor (x+1) cancels out

ganeshie8 (ganeshie8):

the only realy vertical asymptote is x=0

ganeshie8 (ganeshie8):

do you know the process to find horizontal asymptotes ?

OpenStudy (anonymous):

why isn't there a vertical asymptote at x=-1

ganeshie8 (ganeshie8):

the numerator ate the factor x+1

ganeshie8 (ganeshie8):

\[\large f(x) = \dfrac{x+1}{x^2+x} = \dfrac{x+1}{x(x+1)} = \dfrac{1}{x}~~, x\ne -1\]

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