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Mathematics 24 Online
OpenStudy (anonymous):

Find the horizontal asymptote of y=1/(x-2) -3

OpenStudy (anonymous):

y=-3

OpenStudy (anonymous):

thats nice, but how do i find it on my own?

OpenStudy (anonymous):

your function is a hyperbola. For it to have a horizontal asymptote,it means that if x goes to +- infinity y should aprouch some kind of value: so take the limit:\[\lim_{x \rightarrow \infty} 1/(x-2) -3 = -3\] that's it

OpenStudy (anonymous):

so as x gets larger it pull 1/x-2 to 0

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

leaving me -3

OpenStudy (anonymous):

right

OpenStudy (anonymous):

thanks a ton

OpenStudy (anonymous):

you wellcome

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