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Given f(x) =(x^3+1)/(x^2+2x), find any vertical asymptotes?
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- tubz84Lv 410 years ago
Vertical asymptotes exist when the denominator (bottom of the fraction) is equal to zero
so x^2 + 2x = 0
which leads to x(x+2) = 0
x(x+2) is equal to 0 when either x or x+2 = 0
which leads to x = 0 and x+2 = 0, i.e. x = -2
so f(x) will have vertical asymptotes at x = 0, x = -2
- Anonymous10 years ago
use computers do it
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