1. The problem is to analyze the function $f(x) = \log_2(x) + 2$. 2. The logarithmic function $\log_2(x)$ is defined only for $x > 0$. 3. The function $f(x) = \log_2(x) + 2$ shifts the graph of $\log_2(x)$ vertically upward by 2 units. 4. The domain of $f(x)$ remains $x > 0$ because the logarithm is undefined for $x \leq 0$. 5. The vertical asymptote is at $x = 0$ since $\log_2(x)$ approaches $-\infty$ as $x$ approaches 0 from the right. 6. The $y$-intercept does not exist because the function is undefined at $x=0$ and for $x<0$. 7. To find the $x$-intercept, set $f(x) = 0$: $$\log_2(x) + 2 = 0$$ $$\log_2(x) = -2$$ $$x = 2^{-2} = \frac{1}{4}$$ 8. So the $x$-intercept is at $\left(\frac{1}{4}, 0\right)$. 9. The function is increasing because the logarithm base 2 is an increasing function. 10. Summary: The graph of $f(x) = \log_2(x) + 2$ is the graph of $\log_2(x)$ shifted up by 2 units, with domain $x > 0$, vertical asymptote at $x=0$, and $x$-intercept at $\left(\frac{1}{4}, 0\right)$.