You have the function $f(x) = \frac{x - 3}{9x + 4}$.

Let's break down some key properties of this function:

1. Domain:

The domain of the function includes all values of $x$ except where the denominator is zero, since division by zero is undefined.

Set the denominator $9x + 4 = 0$ and solve for $x$: $9x + 4 = 0 \quad \Rightarrow \quad 9x = -4 \quad \Rightarrow \quad x = -\frac{4}{9}$ Therefore, the domain is: $x \in (-\infty, -\frac{4}{9}) \cup (-\frac{4}{9}, \infty)$

2. Vertical Asymptote:

A vertical asymptote occurs where the denominator is zero. From the domain, we know there is a vertical asymptote at $x = -\frac{4}{9}$.

3. Horizontal Asymptote:

For rational functions like this one, we can determine the horizontal asymptote by comparing the degrees of the numerator and denominator.

• The degree of the numerator $x - 3$ is 1.
• The degree of the denominator $9x + 4$ is also 1.

When the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients. In this case: $\text{Horizontal Asymptote: } y = \frac{1}{9}$

4. Intercepts:

• x-intercept: Set $f(x) = 0$, which happens when the numerator is zero: $x - 3 = 0 \quad \Rightarrow \quad x = 3$ So, the x-intercept is $(3, 0)$.
• y-intercept: Set $x = 0$ and evaluate $f(0)$: $f(0) = \frac{0 - 3}{9(0) + 4} = \frac{-3}{4}$ So, the y-intercept is $\left( 0, -\frac{3}{4} \right)$.

Would you like further clarification or additional details on any part of this explanation?

Here are 5 questions to deepen your understanding:

1. How would the function change if the numerator were $x + 3$?
2. What would the domain and asymptotes be for $f(x) = \frac{2x - 5}{5x + 1}$?
3. How does the horizontal asymptote affect the behavior of $f(x)$ as $x \to \infty$?
4. Can you find the inverse of this function?
5. What happens to the function's behavior near the vertical asymptote?

Tip: Always check for points where the denominator equals zero to avoid undefined values.