1. **State the problem:** We are given the function $$y = \log_2 |x - 2|$$ and asked to analyze it. 2. **Understand the function:** This is a logarithmic function with base 2, where the argument is the absolute value of $$x - 2$$. The absolute value ensures the input to the logarithm is always positive, which is necessary because the logarithm is undefined for non-positive values. 3. **Domain:** Since $$|x - 2| > 0$$ for all $$x \neq 2$$, the domain is all real numbers except $$x = 2$$. 4. **Intercepts:** - **x-intercept:** Set $$y = 0$$: $$0 = \log_2 |x - 2| \implies |x - 2| = 1$$ So, $$x - 2 = 1$$ or $$x - 2 = -1$$, giving $$x = 3$$ or $$x = 1$$. Thus, x-intercepts are at $$(1, 0)$$ and $$(3, 0)$$. - **y-intercept:** Set $$x = 0$$: $$y = \log_2 |0 - 2| = \log_2 2 = 1$$ So, y-intercept is $$(0, 1)$$. 5. **Behavior near the vertical asymptote:** At $$x = 2$$, the argument of the logarithm approaches zero, so $$y \to -\infty$$ from both sides. 6. **Graph features:** The graph has a vertical asymptote at $$x = 2$$, passes through $$(1, 0)$$ and $$(3, 0)$$, and is symmetric about $$x = 2$$ due to the absolute value. **Final answer:** The function $$y = \log_2 |x - 2|$$ has domain $$x \in (-\infty, 2) \cup (2, \infty)$$, vertical asymptote at $$x = 2$$, x-intercepts at $$(1, 0)$$ and $$(3, 0)$$, and y-intercept at $$(0, 1)$$.