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

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

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Question 147678: R(x) = x^3 - 2x^2 - 7 / x^6 - 2x^5 + 6x^2 - 4, how many vertical asymptotes are possible?
a. 5
b. 6
c. 2
d. 3
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Since the denominator is of degree 6, the maximum number of zeroes
is 6; so there are at most 6 vertical asymptotes.
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Cheers,
Stan H.
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