Questions: Problem 2 (6 Points) You are given f(x)=(x^10-x-2)/(x+1) and f(x) is defined for all values except x ≠ -1. Using the equation of f(x) and WITHOUT graphing, check whether or not x=-1 is a vertical asymptote or a removable discontinuity. You will need to reference what must be true about f(x) for there to be a removable discontinuity at x=-1 and show all algebraic work!

Determine if \( x = -1 \) is a vertical asymptote or a removable discontinuity for the function \( f(x) = \frac{x^{10} - x - 2}{x + 1} \).

Check if \( x + 1 \) is a factor of the numerator \( x^{10} - x - 2 \).

To check if \( x + 1 \) is a factor, we can perform polynomial long division or synthetic division. If the result is a polynomial without a remainder, then \( x + 1 \) is a factor. In this case, we find that \( \frac{x^{10} - x - 2}{x + 1} \) does not simplify to a polynomial, indicating that \( x + 1 \) is not a factor.

Determine the type of discontinuity at \( x = -1 \).

Since \( x + 1 \) is not a factor of the numerator, \( x = -1 \) is not a removable discontinuity. Therefore, it must be a vertical asymptote.

The answer is that \( x = -1 \) is a vertical asymptote.

The answer is that \( x = -1 \) is a vertical asymptote.
\\(\boxed{\text{vertical asymptote}}\\)