If,$$ f(x)=\frac{3 x-4}{x^{2}+4 x+4} $$

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To analyze the function $$ f(x) = \frac{3x - 4}{x^2 + 4x + 4} $$, we can start by simplifying the expression and examining its properties.

1. **Factor the denominator**:
   The denominator $$ x^2 + 4x + 4 $$ can be factored as:
   $$
   x^2 + 4x + 4 = (x + 2)^2
   $$
   Thus, we can rewrite the function as:
   $$
   f(x) = \frac{3x - 4}{(x + 2)^2}
   $$

2. **Identify the domain**:
   The function is defined for all $$ x $$ except where the denominator is zero. Setting the denominator equal to zero:
   $$
   (x + 2)^2 = 0 \implies x + 2 = 0 \implies x = -2
   $$
   Therefore, the domain of $$ f(x) $$ is $$ x \in \mathbb{R} \setminus \{-2\} $$.

3. **Find the vertical asymptote**:
   Since the denominator approaches zero as $$ x $$ approaches -2, there is a vertical asymptote at $$ x = -2 $$.

4. **Find the horizontal asymptote**:
   To find the horizontal asymptote, we can analyze the behavior of $$ f(x) $$ as $$ x $$ approaches infinity:
   $$
   f(x) = \frac{3x - 4}{(x + 2)^2} \approx \frac{3x}{x^2} = \frac{3}{x} \to 0 \quad \text{as } x \to \infty
   $$
   Thus, the horizontal asymptote is $$ y = 0 $$.

5. **Find the x-intercept**:
   The x-intercept occurs when $$ f(x) = 0 $$:
   $$
   \frac{3x - 4}{(x + 2)^2} = 0 \implies 3x - 4 = 0 \implies 3x = 4 \implies x = \frac{4}{3}
   $$
   Therefore, the x-intercept is at $$ \left( \frac{4}{3}, 0 \right) $$.

6. **Find the y-intercept**:
   The y-intercept occurs when $$ x = 0 $$:
   $$
   f(0) = \frac{3(0) - 4}{(0 + 2)^2} = \frac{-4}{4} = -1
   $$
   Therefore, the y-intercept is at $$ (0, -1) $$.

7. **Summary of properties**:
   - Domain: $$ x \in \mathbb{R} \setminus \{-2\} $$
   - Vertical asymptote: $$ x = -2 $$
   - Horizontal asymptote: $$ y = 0 $$
   - x-intercept: $$ \left( \frac{4}{3}, 0 \right) $$
   - y-intercept: $$ (0, -1) $$

If you have any specific questions or need further analysis on this function, feel free to ask!
The function $$ f(x) = \frac{3x - 4}{(x + 2)^2} $$ has a vertical asymptote at $$ x = -2 $$, a horizontal asymptote at $$ y = 0 $$, an x-intercept at $$ \left( \frac{4}{3}, 0 \right) $$, and a y-intercept at $$ (0, -1) $$.

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