Given the function:

$f(x) = \frac{x - 2}{(x - 2)(x + 4)(x - 5)}$

Let's analyze this function in detail by finding its domain, range, intercepts, asymptotes, and a table of values.

1. Domain

The domain of the function is determined by finding the values of $x$ that make the denominator zero because division by zero is undefined.

The denominator is:

$(x - 2)(x + 4)(x - 5)$

Set each factor to zero:

• $x - 2 = 0 \implies x = 2$
• $x + 4 = 0 \implies x = -4$
• $x - 5 = 0 \implies x = 5$

Thus, the domain of $f(x)$ is:

$\text{Domain} = \{x \in \mathbb{R} \mid x \neq -4, 2, 5\}$

2. Simplification

The expression $x - 2$ cancels out in the numerator and denominator (except at $x = 2$), simplifying the function to:

$f(x) = \frac{1}{(x + 4)(x - 5)} \quad \text{for } x \neq 2$

However, at $x = 2$, the function is undefined because of the original cancellation, creating a hole at $x = 2$.

3. Range

Since the simplified function $f(x) = \frac{1}{(x + 4)(x - 5)}$ is a rational function, it can take on any real number except the value it approaches as asymptotes. However, specific exclusions may depend on further behavior analysis.

4. Asymptotes

• Vertical Asymptotes: These occur where the denominator is zero and the numerator is not zero:

  $x = -4 \quad \text{and} \quad x = 5$

• Horizontal Asymptote: Since the degrees of the numerator and denominator are the same after simplification, the horizontal asymptote is found by dividing the leading coefficients:

  $y = 0$

• Hole: A removable discontinuity at $x = 2$.

5. Intercepts

• x-intercept: There is no x-intercept because the numerator (after simplification) is a constant 1, which never equals zero.

• y-intercept: Find the value of $f(x)$ when $x = 0$:

  $f(0) = \frac{1}{(0 + 4)(0 - 5)} = \frac{1}{-20} = -\frac{1}{20}$

6. Table of Values

Here is a sample table of values around key points:

$x$	$f(x)$
$-5$	$\frac{1}{-9}$
$-3$	$\frac{1}{1}$
$0$	$-\frac{1}{20}$
$1$	$-\frac{1}{12}$
$3$	$\frac{1}{7}$
$4$	$\frac{1}{36}$

Summary

• Domain: $x \neq -4, 2, 5$
• Range: All real values except the asymptotic approach.
• Vertical Asymptotes: $x = -4$, $x = 5$
• Horizontal Asymptote: $y = 0$
• Hole: At $x = 2$
• Intercepts: No x-intercept, y-intercept at $-\frac{1}{20}$.

Would you like further details, or do you have any questions? Here are some related questions:

1. How do vertical asymptotes affect the graph's behavior?
2. Why does the hole occur at $x = 2$?
3. How can you find slant asymptotes in more complex functions?
4. How would changes to the numerator affect the intercepts?
5. What does the behavior near the asymptotes tell us about the function?

Tip: Always carefully simplify functions to reveal holes and asymptotes properly!