The function you're asking about is:

$f(x) = \frac{x^2 - 4x - 12}{x + 2}$

Domain:

The domain of a function consists of all the input values (x-values) for which the function is defined. In this case, the function is a rational function, and rational functions are undefined where the denominator is equal to zero.

Let's find when the denominator is zero:

$x + 2 = 0$

Solving for $x$:

$x = -2$

Thus, the function is undefined at $x = -2$. Therefore, the domain of the function is:

$\boxed{(-\infty, -2) \cup (-2, \infty)}$

Range:

To determine the range of the function, we need to analyze its behavior. We can begin by factoring the numerator and see if there are any common factors with the denominator that can be canceled out.

The numerator is:

$x^2 - 4x - 12$

Let's factor this expression:

$x^2 - 4x - 12 = (x - 6)(x + 2)$

Now, the function becomes:

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

For all $x \neq -2$, we can cancel out the $(x + 2)$ term:

$f(x) = x - 6, \quad \text{for} \quad x \neq -2$

This is a linear function, which is defined for all $x \neq -2$. The function $f(x) = x - 6$ is a straight line with a slope of 1 and a y-intercept of -6.

Since the function is a straight line (except at $x = -2$), it can take any value for $y$ except the value it would take if $x = -2$, which is:

$f(-2) = -2 - 6 = -8$

Thus, the function never reaches $y = -8$, so the range is all real numbers except $-8$:

$\boxed{(-\infty, -8) \cup (-8, \infty)}$

Final Answer:

• Domain: $(-\infty, -2) \cup (-2, \infty)$
• Range: $(-\infty, -8) \cup (-8, \infty)$

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Would you like a more detailed explanation of how to find the domain and range? Or do you have any specific questions on this?

Here are 5 related questions for you:

1. How do we find the vertical asymptotes of rational functions?
2. What are the horizontal asymptotes of a rational function?
3. How can you determine the end behavior of a function like this one?
4. How does the factorization of the numerator and denominator help simplify the function?
5. What would happen if the denominator had a factor of $(x - 6)$ instead of $(x + 2)$?

Tip: Always remember to check for undefined values in the denominator when working with rational functions to accurately find the domain.