Certainly! To construct a function that satisfies all the given conditions, we can use piecewise definitions and appropriate functions that fit the behavior described. Here’s a formula for a function 𝑓 ( 𝑥 ) f(x) that meets all the conditions:

Vertical asymptote at 𝑥

− 2 x=−2 with a limit approaching 1. Value at 𝑥

0 x=0 should be 2. Value at 𝑥

3 x=3 should be 0. As 𝑥 x approaches 0 − 0 − , 𝑓 ( 𝑥 ) f(x) approaches 2. As 𝑥 x approaches 0 + 0 + , 𝑓 ( 𝑥 ) f(x) approaches infinity. Suggested Formula: \frac{1}{x + 2} + 1 & \text{for } x < 0 \text{ (approaching } -2 \text{ from the left side)} \ \frac{2}{x} + 2 & \text{for } x > 0 \text{ (approaching } 0 \text{ from the right side)} \ 0 & \text{for } x = 3 \text{ (discrete point)} \ 2 & \text{for } x = 0 \text{ (discrete point)} \end{cases} ] ### Explanation: 1. For ( x < -2 ): - Use ( \frac{1}{x + 2} + 1 ). This term approaches 1 as ( x ) approaches -2 from the left side. 2. For ( x > 0 ): - Use ( \frac{2}{x} + 2 ). As ( x ) approaches 0 from the right, this term goes to infinity, matching the behavior described. 3. At ( x = 0 ): - Set the function explicitly to 2 to satisfy the condition ( f(0) = 2 ). 4. At ( x = 3 ): - Set the function explicitly to 0 to meet the condition ( f(3) = 0 ). ### Piecewise Construction: - For ( x \to -2 ) from the left, ( \frac{1}{x + 2} + 1 ) approaches 1. - For ( x \to 0^- ), the function ( \frac{2}{x} + 2 ) approaches 2, satisfying ( \lim_{x \to 0^-} f(x) = 2 ). - For ( x \to 0^+ ), the function ( \frac{2}{x} + 2 ) goes to infinity, satisfying ( \lim_{x \to 0^+} f(x) = \infty ).