1. We are given the function $f(x) = \frac{\sqrt{x-1}}{x+3}$. We need to analyze it. 2. First, consider the domain. The expression inside the square root, $x-1$, must be non-negative, so $x-1 \geq 0 \Rightarrow x \geq 1$. 3. Also, the denominator $x+3$ must be non-zero, so $x+3 \neq 0 \Rightarrow x \neq -3$. Since the domain must satisfy both, the domain is $[1, \infty)$ because $x \geq 1$ already excludes $-3$. 4. Let's identify the intercepts. - To find the $x$-intercept(s), set $f(x) = 0$, so $\frac{\sqrt{x-1}}{x+3} = 0 \Rightarrow \sqrt{x-1} = 0 \Rightarrow x-1 = 0 \Rightarrow x=1$. - The $y$-intercept is $f(0)$, but since $0$ is not in the domain, there is no $y$-intercept. 5. To find extrema, we would differentiate and analyze critical points, but since this is not requested explicitly, we omit this. Final answer: The domain of $f(x)$ is $[1, \infty)$ and the $x$-intercept is at $x=1$. There is no $y$-intercept.