To find the x-intercept of a function, we need to solve for $x$ when $f(x) = 0$. In the given equation:

$f(x) = \frac{x^2 + 6x + 9}{x^2 - 9}$

The x-intercept occurs when the numerator of the rational function equals zero, since a fraction is zero only when the numerator is zero and the denominator is non-zero.

Step 1: Set the numerator equal to zero.

The numerator is: $x^2 + 6x + 9$ Set it equal to zero and solve for $x$: $x^2 + 6x + 9 = 0$

Step 2: Factor the quadratic equation.

This can be factored as: $(x + 3)^2 = 0$

Step 3: Solve for $x$.

$x + 3 = 0 \quad \Rightarrow \quad x = -3$

So, the x-intercept is $x = -3$.

Step 4: Check the denominator.

We must ensure the denominator does not become zero at $x = -3$. The denominator is: $x^2 - 9 = (x - 3)(x + 3)$

At $x = -3$, the denominator becomes zero, which creates a hole in the graph at this point. This means the function is undefined at $x = -3$, so there is no x-intercept for this function because the function does not exist at $x = -3$.

Conclusion:

There is no x-intercept because the point $x = -3$ is not in the domain of the function due to a zero in the denominator.

Would you like further details or explanations?

Here are 5 related questions for practice:

1. What is the y-intercept of this function?
2. How would you find the vertical asymptotes of this function?
3. What are the horizontal asymptotes of rational functions in general?
4. How can you determine holes in the graph of a rational function?
5. How do you factor the difference of squares in rational functions?

Tip: When dealing with rational functions, always check both the numerator and denominator before concluding about x-intercepts.