The function in the image is given as:

$f(x) = \frac{6}{-2x + 1}$

This is a rational function. Let’s break down key aspects of the function:

1. Domain: The denominator cannot be zero, as division by zero is undefined. To find the domain, set the denominator equal to zero and solve for $x$: $-2x + 1 = 0 \implies x = \frac{1}{2}$ So, the function is undefined at $x = \frac{1}{2}$. The domain of $f(x)$ is all real numbers except $x = \frac{1}{2}$, i.e., $x \in (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$.

2. Vertical Asymptote: A vertical asymptote occurs where the denominator is zero. As calculated above, there is a vertical asymptote at $x = \frac{1}{2}$.

3. Horizontal Asymptote: To find the horizontal asymptote, consider the degree of the numerator and the denominator. Since both have degree 1 (degree of $x$), we look at the ratio of the leading coefficients. In this case, the leading coefficient of the denominator is $-2$ and the numerator is 6, so the horizontal asymptote is: $y = \frac{6}{-2} = -3$

Would you like to explore the graph or any specific behavior of the function? Here are some related questions:

1. How can we compute the intercepts (x- and y-intercepts) of the function?
2. Can you explain why the function has a vertical asymptote at $x = \frac{1}{2}$?
3. How does the function behave near the vertical asymptote $x = \frac{1}{2}$?
4. What is the significance of the horizontal asymptote $y = -3$?
5. How would you find the inverse of the function $f(x)$?

Tip: For rational functions like this, vertical asymptotes indicate where the function tends towards infinity or negative infinity, while horizontal asymptotes describe the function's behavior as $x$ approaches large positive or negative values.