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Vertical asymptotes of (x^2-4)/(x^3+2x^2+x+2)
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set denom equal to 0 and those are your vertical asymptotes. THese are the values which x can NOT be.
I dont think i did it right
because 0 doesnt make the denominator equal to 0
It has three solutions \[(x+2)(x^2 + 1)\] solutions are: \[x=-2, x=\pm i\]
yeah that looks right.. i was getting imaginary numbers too.
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