anonymous
  • anonymous
Find the horizontal asymptote of y=1/(x-2) -3
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
y=-3
anonymous
  • anonymous
thats nice, but how do i find it on my own?
anonymous
  • 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

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anonymous
  • anonymous
so as x gets larger it pull 1/x-2 to 0
anonymous
  • anonymous
yes
anonymous
  • anonymous
leaving me -3
anonymous
  • anonymous
right
anonymous
  • anonymous
thanks a ton
anonymous
  • anonymous
you wellcome

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