QUESTION 4 Consider: \( g(x)=\frac{a}{x+p}+q \) The following information of \( g \) is given: - Domain: \( x \in \mathrm{R} ; x \neq-2 \) - \( x \)-intercept at \( \mathrm{K}(1 ; 0) \) - \( y \)-intercept at \( \mathrm{N}\left(0 ;-\frac{1}{2}\right) \) 4.1 Show that the equation of \( g \) is given by: \( g(x)=\frac{-3}{x+2}+1 \) 4.2 Write down the range of \( g \). 4.3 Determine the equation of \( h \), the axis of symmetry of \( g \), in the form \( y=m x+c \), where \( m>0 \). 4.4 Write down the coordinates of \( K^{\prime} \), the image of \( K \) reflected over \( h \). (2) [12]

To solve the problem step by step, we will analyze the function \( g(x) = \frac{a}{x+p} + q \) using the given conditions. ### 4.1 Show that the equation of \( g \) is given by: \( g(x) = \frac{-3}{x+2} + 1 \) **Step 1: Identify the parameters using the intercepts.** 1. **\( x \)-intercept at \( K(1, 0) \)**: \[ g(1) = 0 \implies \frac{a}{1+p} + q = 0 \] 2. **\( y \)-intercept at \( N(0, -\frac{1}{2}) \)**: \[ g(0) = -\frac{1}{2} \implies \frac{a}{0+p} + q = -\frac{1}{2} \] **Step 2: Substitute \( q \) from the first equation into the second.** From the first equation: \[ q = -\frac{a}{1+p} \] Substituting into the second equation: \[ \frac{a}{p} - \frac{a}{1+p} = -\frac{1}{2} \] **Step 3: Solve for \( a \) and \( p \).** Multiply through by \( p(1+p) \): \[ a(1+p) - a p = -\frac{1}{2} p(1+p) \] \[ a = -\frac{1}{2} \frac{p(1+p)}{1} \] **Step 4: Use the domain condition.** The domain condition \( x \neq -2 \) implies \( p = 2 \). Substituting \( p = 2 \): \[ a = -\frac{1}{2} \cdot 2(1+2) = -\frac{1}{2} \cdot 6 = -3 \] **Step 5: Substitute \( a \) and \( p \) back into \( g(x) \).** Now we have: \[ g(x) = \frac{-3}{x+2} + q \] To find \( q \): Using \( q = -\frac{a}{1+p} = -\frac{-3}{1+2} = 1 \). Thus, the equation of \( g \) is: \[ g(x) = \frac{-3}{x+2} + 1 \] ### 4.2 Write down the range of \( g \). The function \( g(x) = \frac{-3}{x+2} + 1 \) has a vertical asymptote at \( x = -2 \) and approaches \( y = 1 \) as \( x \to \infty \) or \( x \to -\infty \). The function can take all values except \( y = 1 \). **Range**: \( y \in \mathbb{R}, y \neq 1 \). ### 4.3 Determine the equation of \( h \), the axis of symmetry of \( g \). The axis of symmetry for a rational function of the form \( g(x) = \frac{a}{x+p} + q \) is a vertical line at \( x = -p \). Here, \( p = 2 \), so the axis of symmetry is: \[ x = -2 \] To express it in the form \( y = mx + c \), we can write: \[ y = 0 \cdot x + 0 \quad \text{(since it is a vertical line)} \] ### 4.4 Write down the coordinates of \( K' \), the image of \( K \) reflected over \( h \). **Step 1: Find the coordinates of \( K(1, 0) \).** **Step 2: Reflect over the line \( x = -2 \).** The distance from \( K(1, 0) \) to the line \( x = -2 \) is: \[ 1 - (-2) = 3 \] Thus, the reflected point \( K' \) will be: \[ -2 - 3 = -5 \] So, the coordinates of \( K' \) are: \[ K'(-5, 0) \] ### Summary of Answers: - **4.1**: \( g(x) = \frac{-3}{x+2} + 1 \) - **4.2**: Range: \( y \in \mathbb{R}, y \neq 1 \) - **4.3**: Axis of symmetry: \( y = 0 \cdot x + 0 \) - **4.4**: Coordinates of \( K' \): \( K'(-5, 0) \)