Let's analyze the rational function provided in the image:

Given Function

$f(x) = \frac{x - 6}{x^2 - 36}$ This function has a numerator of $x - 6$ and a denominator of $x^2 - 36$, which can be factored further.

Step 1: Factor the Denominator

The denominator $x^2 - 36$ is a difference of squares, which factors as: $x^2 - 36 = (x + 6)(x - 6)$ So, the function becomes: $f(x) = \frac{x - 6}{(x + 6)(x - 6)}$

Step 2: Simplify the Function

We see that $x - 6$ is present in both the numerator and denominator, which allows us to cancel out $x - 6$, resulting in: $f(x) = \frac{1}{x + 6}, \quad x \neq 6$ The restriction $x \neq 6$ arises because setting $x = 6$ in the original function would lead to a zero in the denominator, indicating a hole at $x = 6$.

Analyzing the Questions

X-Intercept(s)

To find x-intercepts, we set the numerator equal to zero: $x - 6 = 0$ This gives $x = 6$. However, $x = 6$ results in a hole in the simplified function (not an x-intercept), so there is no x-intercept.

Therefore, the answer is:

• B. There is no x-intercept.

Vertical Asymptote(s) and Hole(s)

Vertical asymptotes occur where the denominator of the simplified function is zero (after removing any common factors). The simplified function is: $f(x) = \frac{1}{x + 6}$ Setting $x + 6 = 0$ gives $x = -6$, which is the location of a vertical asymptote.

Additionally, since $x = 6$ was canceled out in the simplification, it represents a hole in the graph.

Thus, the answer is:

• B. The equation(s) of the vertical asymptote(s) is/are $x = -6$, and the graph has hole(s) at $x = 6$.

Horizontal Asymptote(s)

To determine the horizontal asymptote, we look at the degrees of the numerator and denominator in the original function $f(x) = \frac{x - 6}{x^2 - 36}$:

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

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

Thus, the answer is:

• A. The equation(s) of the horizontal asymptote(s) is/are $y = 0$.

Would you like further details on these steps, or have any questions?

Follow-up Questions

1. What is a hole in the context of a rational function, and how is it different from an asymptote?
2. How do we determine the horizontal asymptote for rational functions with different degrees in the numerator and denominator?
3. Why does canceling terms in the numerator and denominator create a hole instead of an asymptote?
4. Can a function have both a hole and an x-intercept at the same value?
5. What would change in the analysis if the numerator and denominator were of the same degree?

Tip

For rational functions, always start by factoring the numerator and denominator to identify potential holes and simplify the function if possible. This will make it easier to analyze asymptotes and intercepts.