Question 6 Given: \( g(x)=\frac{6}{x+2}-1 \) 6.1 Write down the asymptotes of \( g \). 6.2 Caicuiate the \( x \) and \( y \) intercepts of \( g \).

To find the asymptotes of \( g(x)=\frac{6}{x+2}-1 \), we get a vertical asymptote at \( x = -2 \) where the function is undefined. For the horizontal asymptote, as \( x \) approaches infinity, \( g(x) \) approaches \( -1 \). So, the asymptotes are \( x = -2 \) (vertical) and \( y = -1 \) (horizontal). For the intercepts, the \( y \)-intercept occurs when \( x = 0 \). Plugging this value into \( g(x) \), we get \( g(0) = \frac{6}{0+2} - 1 = 3 - 1 = 2 \). Thus, the \( y \)-intercept is \( (0, 2) \). For the \( x \)-intercept, set \( g(x) = 0 \): \( \frac{6}{x+2} - 1 = 0 \) leads to \( \frac{6}{x+2} = 1 \), giving \( x + 2 = 6 \) or \( x = 4 \). The \( x \)-intercept is \( (4, 0) \).