The equation provided is:

$f(x) = \frac{1}{(x+3)^2}$

To find the horizontal asymptote of the graph, we analyze the behavior of $f(x)$ as $x \to \infty$ or $x \to -\infty$.

1. When $x \to \infty$: $(x+3)^2 \to \infty \implies \frac{1}{(x+3)^2} \to 0$

2. When $x \to -\infty$: $(x+3)^2 \to \infty \implies \frac{1}{(x+3)^2} \to 0$

Thus, the horizontal asymptote of $f(x)$ is:

$y = 0$

The correct answer is:

$y = 0$

Would you like me to explain this further or clarify any points?

Here are 5 related questions for exploration:

1. What is the vertical asymptote of the function $f(x) = \frac{1}{(x+3)^2}$?
2. How would the horizontal asymptote change if the numerator of $f(x)$ was $2$ instead of $1$?
3. How does the graph of $f(x) = \frac{1}{(x+3)^2}$ differ from $f(x) = \frac{1}{x^2}$?
4. What is the domain of $f(x) = \frac{1}{(x+3)^2}$?
5. How can we identify horizontal asymptotes using limits in general?

Tip: Horizontal asymptotes represent the end behavior of a function as $x$ approaches infinity or negative infinity.