Given: \( f(x)=\frac{1}{x+2}+3 \) 8.1 Determine the equation of the asymptotes of \( f \). 8.2 Write down the \( y \)-intercept of \( f \). 8.3 Calculate the \( x \)-intercept of \( f \). 8.4 Sketch the graph of \( f \). Clearly label ALL intercepts with the axes and any asym

The function \( f(x) = \frac{1}{x+2} + 3 \) has a vertical asymptote when the denominator is zero, which occurs at \( x + 2 = 0 \), or \( x = -2 \). The horizontal asymptote is determined by analyzing the behavior as \( x \) approaches infinity or negative infinity; because the term \( \frac{1}{x+2} \) approaches \( 0 \), the horizontal asymptote is \( y = 3 \). The \( y \)-intercept occurs when \( x = 0 \): \[ f(0) = \frac{1}{0+2} + 3 = \frac{1}{2} + 3 = \frac{7}{2} \] Thus, the \( y \)-intercept is \( (0, \frac{7}{2}) \). To find the \( x \)-intercept, we set \( f(x) = 0 \): \[ 0 = \frac{1}{x+2} + 3 \Rightarrow \frac{1}{x+2} = -3 \] This simplifies to: \[ 1 = -3(x+2) \Rightarrow 1 = -3x - 6 \Rightarrow 3x = -7 \Rightarrow x = -\frac{7}{3} \] Therefore, the \( x \)-intercept is \( (-\frac{7}{3}, 0) \). To sketch the graph, start by plotting the vertical asymptote at \( x = -2 \) and the horizontal asymptote at \( y = 3 \). Mark the \( y \)-intercept at \( (0, \frac{7}{2}) \) and the \( x \)-intercept at \( (-\frac{7}{3}, 0) \). Connect these points smoothly, noting that as \( x \) approaches -2 from the left, \( f(x) \) heads towards negative infinity, and as it approaches from the right, \( f(x) \) shoots up towards positive infinity. The graph should approach the horizontal asymptote as \( x \) heads to either infinity. So the final features you should label are: - Vertical Asymptote: \( x = -2 \) - Horizontal Asymptote: \( y = 3 \) - \( y \)-Intercept: \( (0, \frac{7}{2}) \) - \( x \)-Intercept: \( (-\frac{7}{3}, 0) \)