The given function in the image is:

$f(x) = \frac{x}{x - 3}$

Let’s address each part of the question.

A. Find the following:

1. Domain: The domain of a function consists of all the input values (x) for which the function is defined. The only point where the function is undefined is when the denominator equals zero: $x - 3 = 0 \implies x = 3$ Therefore, the domain is all real numbers except $x = 3$: $\boxed{\text{Domain}: (-\infty, 3) \cup (3, \infty)}$

2. Range: The range consists of all possible output values (y). To determine the range, notice that the function is undefined for the value of $f(x)$ when $x = 3$. The function's horizontal asymptote (discussed below) is $y = 1$, meaning that the function can approach but never equal this value. Thus, the range is: $\boxed{\text{Range}: (-\infty, 1) \cup (1, \infty)}$

3. Vertical Asymptote: A vertical asymptote occurs when the denominator of the function goes to zero. For this function: $x - 3 = 0 \implies x = 3$ So, the vertical asymptote is: $\boxed{x = 3}$

4. Horizontal Asymptote: To find the horizontal asymptote, we examine the end behavior of the function as $x$ approaches infinity. For large values of $x$, the highest degree terms in the numerator and denominator dominate: $\lim_{x \to \infty} \frac{x}{x - 3} = \lim_{x \to \infty} \frac{1}{1 - \frac{3}{x}} = 1$ Therefore, the horizontal asymptote is: $\boxed{y = 1}$

5. x-intercept (where $f(x) = 0$): The x-intercept occurs when the numerator of the function equals zero: $x = 0 \implies \boxed{(0, 0)}$

6. y-intercept (where $x = 0$): Substitute $x = 0$ into the function: $f(0) = \frac{0}{0 - 3} = 0$ So, the y-intercept is: $\boxed{(0, 0)}$

7. Table of Values: Here’s a table of values for several different values of $x$:

   \hline x & f(x) = \frac{x}{x - 3} \\ \hline -2 & \frac{-2}{-2 - 3} = \frac{-2}{-5} = 0.4 \\ 0 & \frac{0}{0 - 3} = 0 \\ 1 & \frac{1}{1 - 3} = \frac{1}{-2} = -0.5 \\ 2 & \frac{2}{2 - 3} = \frac{2}{-1} = -2 \\ 4 & \frac{4}{4 - 3} = \frac{4}{1} = 4 \\ 5 & \frac{5}{5 - 3} = \frac{5}{2} = 2.5 \\ \hline \end{array}$$

B. Sketch the graph:

The graph should include:

• A vertical asymptote at $x = 3$.
• A horizontal asymptote at $y = 1$.
• Points such as $(0, 0)$, $(2, -2)$, and $(4, 4)$. The function will tend to infinity as it approaches $x = 3$ from either side and flatten toward $y = 1$ as $x$ moves toward infinity.

Would you like more details on any step, or help sketching the graph?

Here are 5 related questions:

1. How does changing the denominator of the function affect the vertical asymptote?
2. What happens to the range if the function was $f(x) = \frac{x + 1}{x - 3}$?
3. Can you describe what would happen if the numerator had a higher degree than the denominator?
4. How do you find the oblique asymptote in rational functions?
5. How does the behavior of the function change when you modify the horizontal asymptote?

Tip: For rational functions, always start by examining the asymptotes, as they shape the graph's structure.