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anonymous
 5 years ago
determine the vertical asymptotes and the end behavior of the following function.
2x^2+3x+1/((x^22)(5x^2))
anonymous
 5 years ago
determine the vertical asymptotes and the end behavior of the following function. 2x^2+3x+1/((x^22)(5x^2))

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i dont see it canceling... got 2 linears up top and 2 irreduciles on the bottom

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x = + sqrt(2) and +sqrt(5) are your vas then

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1its like i was never wrong ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha how did you get y=0? is it from the denominator ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1it 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
 5 years ago
Best ResponseYou've already chosen the best response.0so you expand the denomiator out first. right youll get 2X^2/x^4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1pretty much; the coeffs are meangingless in this since the x^4 overpowers everything else

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool that was easy.. why does it look soo hard. LOL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they 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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