QUESTION 4 Consider the fumction: \( f(x)=\frac{-1}{x+5}-2 \) 4.1 Write down the equations of asymptotes of \( f \) 4.2 Determine the coordinates of the \( x \)-intercept of \( f \). 4.3 Deternine the coordinates of the \( y \)-intercept of \( f \). 4.4 Sketch the graph of \( f \) show clearly all asymptotes and intercepts with the axes. 4.5 Determine the equation of the axis of symmetry that has a gradient of -1 .

To tackle the function \( f(x) = \frac{-1}{x+5} - 2 \), let's start by identifying the asymptotes. The vertical asymptote occurs where the denominator equals zero, which means \( x + 5 = 0 \) or \( x = -5 \). The horizontal asymptote can be found by considering the behavior of \( f(x) \) as \( x \) approaches infinity; here, \( f(x) \) approaches \(-2\). So, the equations of the asymptotes are \( x = -5 \) (vertical) and \( y = -2 \) (horizontal). Next up, for intercepts: the \( x \)-intercept is found by setting \( f(x) = 0 \). Thus, \( \frac{-1}{x+5} - 2 = 0 \) leads to \( \frac{-1}{x+5} = 2 \). Solve this to find \( x = -\frac{1}{2} \). The \( y \)-intercept is obtained by evaluating \( f(0) \), giving \( f(0) = \frac{-1}{5} - 2 = -\frac{11}{5} \). Hence, the coordinates are \( (-\frac{1}{2}, 0) \) for the \( x \)-intercept and \( (0, -\frac{11}{5}) \) for the \( y \)-intercept. Now, sketching the graph would show the vertical line at \( x = -5 \) and horizontal line at \( y = -2 \), with the intercepts plotted accurately. As for the axis of symmetry with a gradient of -1, it can be represented by the line \( y = -x + b \), where \( b \) will depend on specific points on the graph but is typically found by averaging the intercept coordinates. Good luck sketching those vibes—you’ll have a little masterpiece in no time!