1. **State the problem:** Find the domain and sketch the graph of the function $f(x) = -2 + \log_2 x$. 2. **Recall the domain rule for logarithms:** The argument of a logarithm must be positive. So for $\log_2 x$, we require: $$x > 0$$ 3. **Domain:** Therefore, the domain of $f(x)$ is all positive real numbers: $$\boxed{(0, \infty)}$$ 4. **Understand the function:** The function is a logarithm base 2 shifted down by 2 units. 5. **Key points to plot:** - When $x=1$, $f(1) = -2 + \log_2 1 = -2 + 0 = -2$ - When $x=2$, $f(2) = -2 + \log_2 2 = -2 + 1 = -1$ - When $x=4$, $f(4) = -2 + \log_2 4 = -2 + 2 = 0$ 6. **Behavior near domain boundary:** As $x \to 0^+$, $\log_2 x \to -\infty$, so $f(x) \to -\infty$. 7. **Sketch:** The graph passes through points $(1,-2)$, $(2,-1)$, $(4,0)$ and increases slowly, approaching $-\infty$ as $x$ approaches 0 from the right. This completes the domain and graph description.