\( \begin{array}{ll}\text { Given: } f(x)=\frac{2}{x+2}+2 \\ \text { 4.2.1 } & \begin{array}{l}\text { Draw a sketch graph of } f \text { clearly showing the intercepts and } \\ \text { the esymptotes. }\end{array} \\ \begin{array}{lll}4.2 .2 & \text { Deternine for which values of } x \text { will: } \frac{2}{x+2} \geq-2 & \text { (4) } \\ 4.2 .3 & \begin{array}{l}\text { Determine the equation of the axes of symmetry for which the } \\ \text { gradient is negative. }\end{array} & \text { (2) }\end{array}\end{array} \)

To sketch the graph of \( f(x) = \frac{2}{x+2} + 2 \), we first note the vertical asymptote at \( x = -2 \) where the function is undefined. The horizontal asymptote, as \( x \to \infty \) or \( x \to -\infty \), is \( y = 2 \). The y-intercept can be found by substituting \( x = 0 \), yielding \( f(0) = 1 \). The x-intercept occurs when \( f(x) = 0 \), which gives us the point \( x = -2 \). The graph will exhibit a hyperbolic shape approaching the asymptotes. To solve \( \frac{2}{x+2} \geq -2 \), we first rewrite it as \( \frac{2}{x+2} + 2 \geq 0 \), leading to \( \frac{2 + 2(x+2)}{x+2} \geq 0 \), or \( \frac{2x + 6}{x+2} \geq 0 \). Setting the numerator equal to zero gives \( x = -3 \). The critical points are \( x = -3 \) and \( x = -2 \). Testing intervals around these points will show that the inequality holds for \( x \in (-\infty, -3] \cup (-2, \infty) \). To find the axes of symmetry where the gradient is negative, we analyze the function's derivative. The derivative of \( f(x) \) is found by differentiating \( \frac{2}{x+2} + 2 \), resulting in \( f'(x) = -\frac{2}{(x+2)^2} \). Since this derivative is always negative (except at the asymptote), the function is decreasing everywhere except for the undefined point at \( x = -2 \). Thus, the line \( x = -2 \) is the axis of symmetry, highlighting how the function behaves across this critical division.