1. **State the problem:** Identify vertical asymptotes, horizontal asymptotes, and holes of the function $$f(x) = \frac{(2x - 6)(x + 1)}{(x + 1)(x - 1)}.$$\n\n2. **Simplify the function:** Notice that the factor $(x + 1)$ appears in both numerator and denominator.\n\n$$f(x) = \frac{(2x - 6)\cancel{(x + 1)}}{\cancel{(x + 1)}(x - 1)} = \frac{2x - 6}{x - 1}, \quad x \neq -1$$\n\nThe restriction $x \neq -1$ is because the original function is undefined there due to division by zero.\n\n3. **Identify holes:** A hole occurs where a factor cancels out. Since $(x + 1)$ cancels, there is a hole at $x = -1$. To find the $y$-value of the hole, substitute $x = -1$ into the simplified function:\n\n$$f(-1) = \frac{2(-1) - 6}{-1 - 1} = \frac{-2 - 6}{-2} = \frac{-8}{-2} = 4.$$\n\nSo, there is a hole at $(-1, 4)$.\n\n4. **Identify vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the factor does not cancel. The denominator is zero at $x = 1$ and $x = -1$, but $x = -1$ is a hole, so the vertical asymptote is at $x = 1$.\n\n5. **Identify horizontal asymptotes:** Compare degrees of numerator and denominator in simplified function $\frac{2x - 6}{x - 1}$. Both numerator and denominator are degree 1.\n\nThe horizontal asymptote is given by the ratio of leading coefficients:\n\n$$y = \frac{2}{1} = 2.$$\n\n**Final answers:**\n- Hole at $(-1, 4)$\n- Vertical asymptote at $x = 1$\n- Horizontal asymptote at $y = 2$