determine the vertical asymptotes and the end behavior of the following function.

2x^2+3x+1/((x^2-2)(5-x^2))

Question

determine the vertical asymptotes and the end behavior of the following function.

2x^2+3x+1/((x^2-2)(5-x^2))

amistre64 · Answer

i dont see it canceling... got 2 linears up top and 2 irreduciles on the bottom

amistre64 · Answer

x = +- sqrt(2) and +-sqrt(5) are your vas then

amistre64 · Answer

ha is y=0

amistre64 · Answer

its like i was never wrong  ;)

anonymous · Answer

haha how did you get y=0? is it from the denominator
?

amistre64 · Answer

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 · Answer

so you expand the denomiator out first. right youll get 2X^2/x^4

amistre64 · Answer

pretty much; the coeffs are meangingless in this since the x^4 overpowers everything else

anonymous · Answer

cool that was easy.. why does it look soo hard. LOL

anonymous · Answer

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