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

Algebra ->  Finance -> 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      Log On


   

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) About Me  (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)
= %28x-2%29%2A%28x%2B2%29%2A%28x-1%29%2A%28x%2B1%29%2F%28%28x%2B1%29%2A%28x-2%29%5E2%29
= %28x%2B2%29%2A%28x-1%29%2F%28x-2%29
Only the x = 2 remains in the DEN