Beskou: \( g(x)=\frac{a}{x+p}+q \) Die volgende inligting van \( g \) word gegee: - Definisieversameling: \( x \in \mathrm{R} ; x \neq-2 \) - \( x \)-afsnit by \( \mathrm{K}(1 ; 0) \) - \( y \)-afsnit by \( \mathrm{N}\left(0 ;-\frac{1}{2}\right) \) 5.1 Toon dat die vergelyking van \( g \) geges word deur: \( g(x)=\frac{-3}{x+2}+1 \) 5.2 Skryf die waardeversameling van \( g \) neer. 5.3 Bepaal die vergelyking van \( h \), die simmetrie-as van \( g \), in die vorm \( y=m x+c \), waar \( m>0 \). 5.4 Skryf die koördinate van \( \mathrm{K}^{\prime} \), die beeld van K gereflekteer oor \( h \), neer.

Let's solve the problem step by step. ### 5.1 Show that the equation of \( g \) is given by: \( g(x)=\frac{-3}{x+2}+1 \) We know the general form of the function is: \[ g(x) = \frac{a}{x+p} + q \] From the information provided: - The \( x \)-intercept is at \( K(1, 0) \). This means when \( g(1) = 0 \). - The \( y \)-intercept is at \( N(0, -\frac{1}{2}) \). This means when \( g(0) = -\frac{1}{2} \). - The domain is \( x \in \mathbb{R} \) and \( x \neq -2 \), which indicates \( p = 2 \). Substituting \( p = 2 \) into the function gives: \[ g(x) = \frac{a}{x+2} + q \] Now, we will use the intercepts to find \( a \) and \( q \). 1. **Using the \( x \)-intercept \( K(1, 0) \)**: \[ g(1) = 0 \implies \frac{a}{1+2} + q = 0 \implies \frac{a}{3} + q = 0 \quad (1) \] 2. **Using the \( y \)-intercept \( N(0, -\frac{1}{2}) \)**: \[ g(0) = -\frac{1}{2} \implies \frac{a}{0+2} + q = -\frac{1}{2} \implies \frac{a}{2} + q = -\frac{1}{2} \quad (2) \] Now we have a system of equations: From equation (1): \[ q = -\frac{a}{3} \] Substituting \( q \) into equation (2): \[ \frac{a}{2} - \frac{a}{3} = -\frac{1}{2} \] To solve this, we need a common denominator, which is 6: \[ \frac{3a}{6} - \frac{2a}{6} = -\frac{1}{2} \] This simplifies to: \[ \frac{a}{6} = -\frac{1}{2} \implies a = -3 \] Now substituting \( a \) back into equation (1) to find \( q \): \[ q = -\frac{-3}{3} = 1 \] Thus, we have: \[ g(x) = \frac{-3}{x+2} + 1 \] ### 5.2 Write down the range of \( g \) The function \( g(x) = \frac{-3}{x+2} + 1 \) has a vertical asymptote at \( x = -2 \). As \( x \) approaches \( -2 \) from the left, \( g(x) \) approaches \( -\infty \), and as \( x \) approaches \( -2 \) from the right, \( g(x) \) approaches \( +\infty \). The horizontal asymptote is \( y = 1 \) (as \( x \to \infty \) or \( x \to -\infty \)). Therefore, the range of \( g \) is: \[ \text{Range of } g: y \in \mathbb{R}, y \neq 1 \] ### 5.3 Determine the equation of \( h \), the axis of symmetry of \( g \) The function \( g(x) \) is a rational function and does not have a traditional axis of symmetry like a parabola. However, we can find the line of symmetry by reflecting points across the horizontal asymptote \( y = 1 \). The equation of the line of symmetry \( h \) is: \[ y = 1 \] ### 5.4 Write down the coordinates of \( K' \), the image of \( K \) reflected over \( h \) The point \( K(1, 0) \) is reflected over the line \( y = 1 \). The distance from \( K \) to the line \( y = 1 \) is: \[ 1 - 0 = 1 \] Thus, the reflected point \( K' \) will be: \[ K'(1, 1 + 1) = K'(1, 2) \] ### Summary of Answers - **5.1**: \( g(x) = \frac{-3}{x+2} + 1 \) - **5.2**: Range of \( g \): \( y \in \mathbb{R}, y \neq 1 \) - **5.3**: Equation of \( h \): \( y = 1 \) - **5.4**: Coordinates of \( K' \): \( (1, 2) \)