SOLUTION: 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 Thank you

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
Thank you
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
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)
=
=
Only the x = 2 remains in the DEN
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