The function you've provided is:

$f(x) = \frac{x^2 + x - 6}{x^2 - 4}$

To analyze this function, we can do the following:

Step 1: Factor the numerator and the denominator

Let's factor both the numerator and denominator to simplify the function.

Numerator: $x^2 + x - 6$

We need two numbers that multiply to $-6$ and add up to $1$ (the coefficient of $x$). The numbers are $3$ and $-2$, so:

$x^2 + x - 6 = (x + 3)(x - 2)$

Denominator: $x^2 - 4$

This is a difference of squares, which factors as:

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

Step 2: Simplify the function

Now, the function becomes:

$f(x) = \frac{(x + 3)(x - 2)}{(x - 2)(x + 2)}$

We can cancel out the common factor $(x - 2)$ in the numerator and denominator, but we must note that $x \neq 2$, as this would make the denominator zero and cause the function to be undefined at $x = 2$.

The simplified function is:

$f(x) = \frac{x + 3}{x + 2}, \quad x \neq 2$

Step 3: Identify any restrictions

• The function is undefined at $x = 2$ and $x = -2$ because the original denominator becomes zero at those points.
• There is a vertical asymptote at $x = -2$, where the denominator equals zero but no cancellation occurs.
• At $x = 2$, there is a hole in the graph because the factor $x - 2$ was canceled out.

Summary:

The simplified function is:

$f(x) = \frac{x + 3}{x + 2}, \quad x \neq 2, x \neq -2$

• Vertical asymptote at $x = -2$
• Hole at $x = 2$

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Would you like more details on how these concepts (asymptotes, holes) are derived?

Related Questions:

1. How do you find horizontal asymptotes of rational functions?
2. How do you graph rational functions with vertical asymptotes and holes?
3. What happens to the function near $x = -2$ and $x = 2$?
4. How does the function behave as $x \to \infty$?
5. Can rational functions have more than one vertical asymptote?

Tip:

When simplifying rational functions, always check for points where the original function is undefined, even after canceling terms.