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 \( N\left(0 ;-\frac{1}{2}\right) \) 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 \( g \). (1) 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 \( p \), \( q \), and \( a \)** From the domain \( x \in \mathbb{R}; x \neq -2 \), we can deduce that \( p = 2 \) because the function is undefined at \( x = -2 \). **Step 2: Use the \( x \)-intercept** The \( x \)-intercept is at \( K(1, 0) \). This means: \[ g(1) = 0 \] Substituting into the function: \[ 0 = \frac{a}{1 + 2} + q \implies 0 = \frac{a}{3} + q \implies q = -\frac{a}{3} \] **Step 3: Use the \( y \)-intercept** The \( y \)-intercept is at \( N(0, -\frac{1}{2}) \). This means: \[ g(0) = -\frac{1}{2} \] Substituting into the function: \[ -\frac{1}{2} = \frac{a}{0 + 2} + q \implies -\frac{1}{2} = \frac{a}{2} + q \] Substituting \( q = -\frac{a}{3} \): \[ -\frac{1}{2} = \frac{a}{2} - \frac{a}{3} \] To solve for \( a \), we need a common denominator: \[ -\frac{1}{2} = \frac{3a}{6} - \frac{2a}{6} = \frac{a}{6} \] Multiplying both sides by 6: \[ -3 = a \] **Step 4: Substitute \( a \) back to find \( q \)** \[ q = -\frac{-3}{3} = 1 \] **Step 5: Write the final equation** Now substituting \( a = -3 \), \( p = 2 \), and \( q = 1 \) into the function: \[ g(x) = \frac{-3}{x + 2} + 1 \] Thus, we have shown that: \[ g(x) = \frac{-3}{x + 2} + 1 \] ### 4.2 Write down the range of \( g \) The function \( g(x) = \frac{-3}{x + 2} + 1 \) has a vertical asymptote at \( x = -2 \). 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 \( y = 1 \). Thus, the range of \( g \) is: \[ (-\infty, 1) \cup (1, \infty) \] ### 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 that passes through the \( x \)-coordinate of the vertical asymptote. Since the vertical asymptote is at \( x = -2 \), the axis of symmetry is: \[ x = -2 \] To express this in the form \( y = mx + c \), we can rewrite it as: \[ y = 0 \cdot x + 0 \quad \text{(since it is a vertical line)} \] However, since we need \( m > 0 \), we can express the axis of symmetry as: \[ y = 0 \quad \text{(the x-axis)} \] ### 4.4 Write down the coordinates of \( K' \), the image of \( K \) reflected over \( h \) The point \( K(1, 0) \) lies on the axis of symmetry \( y = 0 \). When reflecting a point over the x-axis, the y-coordinate changes sign. Therefore, the coordinates of \( K' \) are: \[ K' = (1, 0) \] ### Summary of Answers 1. \( g(x) = \frac{-3}{x + 2} + 1 \) 2. Range of \( g \): \( (-\infty, 1) \cup (1, \infty) \) 3. Axis of symmetry \( h \): \( y = 0 \) 4. Coordinates of \( K' \): \( (1, 0) \)