Consider the rational function and answer the questions below: f(x)=x+2 OVER x^2-1 1. Hole(s) = 2. Vertical asymptote(s) = 3. Horizontal asymptote(s)=

\text{1. Holes}
There are no common factors in the numerator \( x + 2 \) and the denominator \( (x - 1)(x + 1) \), so there are no holes:
\text{Hole(s)} = \text{None}

\text{2. Vertical Asymptotes}
Vertical asymptotes occur where the denominator is zero. Solving \( (x - 1)(x + 1) = 0 \), we get:
x = 1 \text{ and } x = -1
Therefore:
\text{Vertical asymptote(s)} = x = 1 \text{ and } x = -1

\text{3. Horizontal Asymptotes}
The degree of the numerator is 1 and the degree of the denominator is 2. Since the degree of the numerator is less than the degree of the denominator:
\text{Horizontal asymptote(s)} = y = 0

\text{Hole(s)} = \text{None}
\text{Vertical asymptote(s)} = x = 1 \text{ and } x = -1
\text{Horizontal asymptote(s)} = y = 0