The function given in the image is:

$R(x) = \frac{x - 1}{3x^2 + 3x}$

We are asked to find the following: a) Domain
b) x-intercepts
c) y-intercepts
d) Zeros
e) Vertical asymptote
f) Horizontal asymptote
g) Hole

Let’s go step by step:

a) Domain:

The domain of a rational function is all real numbers except where the denominator is equal to zero.

The denominator is $3x^2 + 3x = 3x(x + 1)$.

Setting the denominator equal to zero and solving for $x$: $3x(x + 1) = 0$ $x = 0 \quad \text{or} \quad x = -1$

Thus, the domain is all real numbers except $x = 0$ and $x = -1$.

Domain: $(-\infty, -1) \cup (-1, 0) \cup (0, \infty)$

b) x-intercepts:

The x-intercepts occur when $R(x) = 0$, which means the numerator is equal to zero.

The numerator is $x - 1$. Set it equal to zero: $x - 1 = 0 \quad \Rightarrow \quad x = 1$

So, the x-intercept is at $x = 1$.

c) y-intercepts:

The y-intercept occurs when $x = 0$.

However, in this case, $x = 0$ is not in the domain of the function, so there is no y-intercept.

y-intercept: None

d) Zeros:

The zeros of the function are the same as the x-intercepts. Therefore, the zero is at $x = 1$.

Zero: $x = 1$

e) Vertical Asymptote:

Vertical asymptotes occur where the denominator equals zero, provided the numerator is non-zero.

From the domain calculation, the denominator is zero at $x = 0$ and $x = -1$. Let's check if the numerator is non-zero at these points:

• At $x = 0$, the numerator $x - 1 = -1$, so there is a vertical asymptote at $x = 0$.
• At $x = -1$, the numerator $x - 1 = -2$, so there is a vertical asymptote at $x = -1$.

Vertical asymptotes: $x = 0$ and $x = -1$

f) Horizontal Asymptote:

For horizontal asymptotes, we compare the degrees of the numerator and the denominator.

• The degree of the numerator is 1.
• The degree of the denominator is 2.

Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $y = 0$.

Horizontal asymptote: $y = 0$

g) Hole:

A hole occurs where a factor cancels from both the numerator and the denominator.

In this case, there are no common factors between the numerator and denominator, so there is no hole.

Hole: None

Would you like further details or clarifications?
Here are five related questions you can explore:

1. How do you determine the slant asymptote for a rational function?
2. What happens to the graph of a function at a vertical asymptote?
3. How can you determine end behavior for different rational functions?
4. What is the relationship between a function's degree and its asymptotes?
5. How do you calculate the limits approaching vertical asymptotes?

Tip: Always check if common factors can cancel when finding vertical asymptotes and holes in rational functions.