Question 1005658
f(x) =((x^2-4)(x^2-1))/((x+1)(x-2)^2) why is the vertical asymptote only x=2  and not x = -1 too? I thought vertical asymptotes were the zeros of the denominator
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f(x) =((x^2-4)(x^2-1))/((x+1)(x-2)^2)

= {{{(x-2)*(x+2)*(x-1)*(x+1)/((x+1)*(x-2)^2)}}}
= {{{(x+2)*(x-1)/(x-2)}}}
Only the x = 2 remains in the DEN