anonymous
  • anonymous
determine the vertical asymptotes and the end behavior of the following function. 2x^2+3x+1/((x^2-2)(5-x^2))
Mathematics
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anonymous
  • anonymous
determine the vertical asymptotes and the end behavior of the following function. 2x^2+3x+1/((x^2-2)(5-x^2))
Mathematics
chestercat
  • chestercat
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amistre64
  • amistre64
i dont see it canceling... got 2 linears up top and 2 irreduciles on the bottom
amistre64
  • amistre64
x = +- sqrt(2) and +-sqrt(5) are your vas then
amistre64
  • amistre64
ha is y=0

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amistre64
  • amistre64
its like i was never wrong ;)
anonymous
  • anonymous
haha how did you get y=0? is it from the denominator ?
amistre64
  • amistre64
it from the shortcut to finding has.... if the bottom degree is greater than the top degree; at all goes to 0 in the end
anonymous
  • anonymous
so you expand the denomiator out first. right youll get 2X^2/x^4
amistre64
  • amistre64
pretty much; the coeffs are meangingless in this since the x^4 overpowers everything else
anonymous
  • anonymous
cool that was easy.. why does it look soo hard. LOL
anonymous
  • anonymous
they make it seem that way to annoy you. big denominator means small number (by which i mean close to zero). think about \[\frac{(10^6)^2}{(10^6)^4}=\frac{10^{12}}{10^{24}}=\frac{1}{10^{12}}\] a number very close to zero

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