I see that you've uploaded an image containing several exercises related to finding the domain of functions algebraically and supporting the answer graphically. Let's go through the steps to find the domain for each of the functions in Exercises 9-16.

Exercises 9-16: Finding the Domain Algebraically

9. $f(x) = -x^2 + 4$

• Domain: The function $f(x) = -x^2 + 4$ is a quadratic function, which is defined for all real numbers.
• Answer: Domain = $(-\infty, \infty)$

10. $h(x) = \frac{5}{x - 3}$

• Domain: This is a rational function. The function is undefined where the denominator is zero. So, set $x - 3 = 0$, which gives $x = 3$. Exclude this value from the domain.
• Answer: Domain = $(-\infty, 3) \cup (3, \infty)$

11. $f(x) = \frac{3x - 1}{(x + 3)(x - 1)}$

• Domain: The function is undefined where the denominator is zero. Set $(x + 3)(x - 1) = 0$, so $x = -3$ and $x = 1$. Exclude these values.
• Answer: Domain = $(-\infty, -3) \cup (-3, 1) \cup (1, \infty)$

12. $h(x) = \frac{5 + x}{x - 3}$

• Domain: This is another rational function. Set the denominator $x - 3 = 0$, which gives $x = 3$. Exclude this value.
• Answer: Domain = $(-\infty, 3) \cup (3, \infty)$

13. $f(x) = \frac{1}{x^2 - 4x}$

• Domain: The denominator is $x^2 - 4x = 0$. Factoring gives $x(x - 4) = 0$, so $x = 0$ and $x = 4$. Exclude these values.
• Answer: Domain = $(-\infty, 0) \cup (0, 4) \cup (4, \infty)$

14. $h(x) = \frac{\sqrt{x - 3}}{x - 1}$

• Domain: The square root requires that $x - 3 \geq 0$, which gives $x \geq 3$. Also, the denominator $x - 1 \neq 0$, so $x \neq 1$. Thus, the domain is $x \geq 3$, but exclude $x = 1$ if within that range.
• Answer: Domain = $[3, \infty)$, excluding $x = 1$.

15. $k(x) = \frac{\sqrt{x^2 - 5x}}{x - 1}$

• Domain: The square root requires $x^2 - 5x \geq 0$, and the denominator $x - 1 \neq 0$. First, solve the inequality for $x^2 - 5x \geq 0$. After factoring $x(x - 5) \geq 0$, the solution is $x \leq 0$ or $x \geq 5$. Also, exclude $x = 1$.
• Answer: Domain = $(-\infty, 0] \cup [5, \infty)$, excluding $x = 1$.

16. $f(x) = \sqrt{x^2 - 16x}$

• Domain: The square root requires that the expression inside the square root be non-negative: $x^2 - 16x \geq 0$. Factor to get $x(x - 16) \geq 0$. Solve this inequality to get $x \leq 0$ or $x \geq 16$.
• Answer: Domain = $(-\infty, 0] \cup [16, \infty)$

Would you like more detailed explanations for any of these problems or assistance with graphing the domains?

Here are five related questions:

1. How do you solve inequalities involving square roots and rational expressions?
2. What graphical techniques can help visualize the domain of rational functions?
3. How do you factor quadratic expressions to find restrictions on the domain?
4. What is the significance of excluding specific points from the domain of a function?
5. How does the concept of domain change for composite functions?

Tip: Always check for domain restrictions by analyzing the denominator, square roots, and logarithmic expressions. These often define boundaries for the allowable values of $x$.