To find the holes of a rational function, we need to identify any values of $x$ that make both the numerator and the denominator zero simultaneously. Here’s the process:

Given the function: $f(x) = \frac{2x - 6}{3x - 9}$

1. Factorize the Numerator and Denominator:

   The numerator, $2x - 6$, can be factored as: $2x - 6 = 2(x - 3)$

   The denominator, $3x - 9$, can be factored as: $3x - 9 = 3(x - 3)$

   So, we can rewrite $f(x)$ as: $f(x) = \frac{2(x - 3)}{3(x - 3)}$

2. Identify the Common Factor:

   Both the numerator and the denominator have a common factor of $(x - 3)$. When this factor is canceled out, the function becomes: $f(x) = \frac{2}{3}$

3. Determine the Hole:

   Since $x = 3$ was canceled from both the numerator and the denominator, it indicates a hole at $x = 3$.

4. Locate the Coordinates of the Hole:

   To find the y-coordinate of the hole, substitute $x = 3$ into the simplified form $f(x) = \frac{2}{3}$. The hole is therefore at: $\text{Hole at } (3, \frac{2}{3})$

In summary, the hole of the function $f(x) = \frac{2x - 6}{3x - 9}$ is at the point $(3, \frac{2}{3})$.

Would you like further explanation on any step or have additional questions?

Related Questions

1. How do you identify vertical asymptotes in a rational function?
2. What differentiates a hole from a vertical asymptote?
3. How would you find the domain of $f(x) = \frac{2x - 6}{3x - 9}$?
4. Can a function have more than one hole? If so, how?
5. What impact do holes have on the graph of a function?

Tip:

When factoring, always look for common terms that may cancel out, as they often indicate holes in rational functions.