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

1. amistre64

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

2. amistre64

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

3. amistre64

ha is y=0

4. amistre64

its like i was never wrong ;)

5. anonymous

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

6. 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

7. anonymous

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

8. amistre64

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

9. anonymous

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

10. 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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