\( F(x)=\frac{(x+2)(x+4)}{(x-4)(x+2)} \) V.A \( = \) H.A \( = \) Hules \( =x=-2 \) x-intercept \( = \) x- intercept

Let's analyze the function: \[ F(x) = \frac{(x + 2)(x + 4)}{(x - 4)(x + 2)} \] ### Simplifying the Function First, notice that the \((x + 2)\) terms in the numerator and denominator cancel out, except at \(x = -2\), where the function is undefined. This results in: \[ F(x) = \frac{x + 4}{x - 4}, \quad \text{with a hole at } x = -2 \] ### 1. **Vertical Asymptote (V.A)** Vertical asymptotes occur where the denominator is zero (and the numerator is not zero at those points). \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] **Vertical Asymptote:** \( x = 4 \) --- ### 2. **Horizontal Asymptote (H.A)** For rational functions, if the degrees of the numerator and the denominator are the same, the horizontal asymptote is the ratio of the leading coefficients. Both the numerator and the denominator are degree 1 (linear), and the leading coefficients are both 1. \[ H.A = \frac{1}{1} = 1 \] **Horizontal Asymptote:** \( y = 1 \) --- ### 3. **Hole (Hules)** A hole occurs at values of \(x\) that make both the numerator and denominator zero. From the canceled factor: \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] **Hole:** \( x = -2 \) --- ### 4. **x-Intercepts** x-intercepts occur where the numerator is zero. After simplifying the function: \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \] **x-Intercept:** \( (-4, 0) \) --- ### Summary - **Vertical Asymptote (V.A):** \( x = 4 \) - **Horizontal Asymptote (H.A):** \( y = 1 \) - **Hole:** \( x = -2 \) - **x-Intercept:** \( (-4, 0) \) --- ### Additional Information (Optional) **y-Intercept:** To find the y-intercept, evaluate \( F(0) \): \[ F(0) = \frac{0 + 4}{0 - 4} = -1 \] **y-Intercept:** \( (0, -1) \)