SOLUTION: R(s) = x^3 - 2x^2 - 7 / x^6 - 2x^5 = 6x^2 - 4; how many vertical asymptotes are possible? a. 3 b. 2 c. 6 d. 5

Algebra ->  Equations -> SOLUTION: R(s) = x^3 - 2x^2 - 7 / x^6 - 2x^5 = 6x^2 - 4; how many vertical asymptotes are possible? a. 3 b. 2 c. 6 d. 5      Log On


   

Question 147585: R(s) = x^3 - 2x^2 - 7 / x^6 - 2x^5 = 6x^2 - 4; how many vertical asymptotes are possible?
a. 3
b. 2
c. 6
d. 5
Answer by nabla(475) About Me  (Show Source):
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Seeing as the denominator could have up to 6 zeroes, we shall say (c) 6. Division by zeroes in rational functions results in vertical asymptotes.