Determine whether the function has a vertical asymptote or a removable discontinuity. Classify the discontinuity. h(x)= (x^3+1)/(x+1)

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Determine whether the function has a vertical asymptote or a removable discontinuity. Classify the discontinuity. h(x)= (x^3+1)/(x+1)

Calculus1
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in this case set the denominator to zero x+1=0, then solve for x. X = -1 if you could factor the numerator and cancel something out you would find a hole or removable discontinuity. In this case, I don't see this as possible. Therefore, the vertical asymptote is x= -1 Hope this helps
The numerator can be factored since it's a convenient sum of cubes: \[x^3+1^3=(x+1)(x^2-x+1)\]

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