The function \( f(x)=\frac{-2}{x+2}+2 \) is given. 11 White down the equations of the asymptotes of the function \( f \) 12 Sheth the gaph of the function Show the intercepts with the ares and the asymptotet clearly. \( \begin{array}{l}\text { Give the eqpation of the line of symmetry of the function with a negative gradient. } \\ \text { (3) } \\ \text { (2) } \\ \text { [7] }\end{array} \)

To identify the asymptotes of the function \( f(x)=\frac{-2}{x+2}+2 \), we observe that there is a vertical asymptote at \( x = -2 \) since the denominator becomes zero here. The horizontal asymptote can be determined by examining the behavior as \( x \) approaches infinity or negative infinity. As \( x \to \infty \) or \( x \to -\infty \), the term \( \frac{-2}{x+2} \) approaches zero, so the horizontal asymptote is at \( y = 2 \). Thus, the equations of the asymptotes are \( x = -2 \) and \( y = 2 \). Plotting the graph of \( f(x) \), you'll see that it curves around the vertical asymptote \( x = -2 \) and flattens out at the horizontal asymptote \( y = 2 \). The intercepts can be found by setting \( f(x) = 0 \): \( 0 = \frac{-2}{x+2} + 2 \), leading to \( x = -2 \) for the x-intercept, but this point is not valid as it coincides with the vertical asymptote. The y-intercept can be found by evaluating \( f(0) = \frac{-2}{2} + 2 = 1 \), yielding the point (0, 1). For the line of symmetry, since the function has a negative gradient and is asymptotic, a linear function such as \( y = -x + 1 \) could act as a line of symmetry, reflecting on the behavior of the function around the asymptotes. Consequently, it's crucial to understand geometric properties when dealing with rational functions.