If `f(x)=(x+2)/((x+2)(x^(2)-4))`, its graph will have

Options

A. one horizontal and three vertical asymptotes
B. one horizontal and two vertical asymototes
C. one horizontal and one vertical asymptote
D. zero horizontal and one vertical asymptote

Explanation

To determine the asymptotes of the function \( f(x) = \frac{x + 2}{(x + 2)(x^2 - 4)} \), we will follow these steps:

### Step 1: Simplify the Function
First, we can simplify the function by factoring the denominator.

The denominator \( x^2 - 4 \) can be factored as:
\[ x^2 - 4 = (x - 2)(x + 2) \]
Thus, we can rewrite \( f(x) \):
\[ f(x) = \frac{x + 2}{(x + 2)(x - 2)(x + 2)} \]
Now, we can cancel the \( (x + 2) \) term from the numerator and denominator (noting that \( x \neq -2 \)):
\[ f(x) = \frac{1}{(x - 2)(x + 2)} \]

### Step 2: Identify Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero (provided the numerator is not also zero).

Setting the denominator equal to zero:
\[ (x - 2)(x + 2) = 0 \]
This gives us:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
Thus, we have vertical asymptotes at \( x = 2 \) and \( x = -2 \).

### Step 3: Identify Horizontal Asymptotes
To find horizontal asymptotes, we look at the behavior of \( f(x) \) as \( x \) approaches infinity.

As \( x \) approaches infinity, the function simplifies to:
\[ f(x) \approx \frac{1}{x^2} \quad \text{(since the leading term dominates)} \]
As \( x \to \infty \) or \( x \to -\infty \), \( f(x) \to 0 \).

Thus, there is a horizontal asymptote at \( y = 0 \).

### Conclusion
In summary, the function \( f(x) \) has:
- **Vertical Asymptotes** at \( x = 2 \) and \( x = -2 \).
- **Horizontal Asymptote** at \( y = 0 \).

Therefore, the correct answer is **Option B: one horizontal and two vertical asymptotes**.