Consider the graphs of \( g(x)=\frac{6}{x+3}+\frac{3}{2} \) and \( h(x)=\frac{6}{x-3}+2 \) 9.1 Write down the domain of \( g \). 9.3 If the graph of \( g \) is shifted so that it coincides with the graph of \( h \). 9.3.1 How many units must the graph be shifted horizontally? 9.3.2 How many units must the graph be shifted verticall \( y \) ? 9.4 Write down the equations of the asymptotes of \( g \). 9.5 Calculate the \( x \)-intercept of \( g \). 9.6 Sketch the graph of \( g \) in your ANSwER BOOK. Show clearly all the 9.7 Determine the value of \( k \) if \( h(x)=-x+k \) is an axis of symmetry of \( g \). 9.8 For which value( \( (s) \) of \( x \) will \( \frac{6}{x+3}-\frac{3}{2}>-x+k \) ? 9.9 The graph of \( g \) is reflected in the \( x \)-axis. Write down the new equation in the form \( y= \)...

Let's go through each part of the problem step by step. ### 9.1 Domain of \( g \) The function \( g(x) = \frac{6}{x+3} + \frac{3}{2} \) is undefined when the denominator is zero. Therefore, we set the denominator equal to zero: \[ x + 3 = 0 \implies x = -3 \] Thus, the domain of \( g \) is all real numbers except \( -3 \): \[ \text{Domain of } g: \{ x \in \mathbb{R} \mid x \neq -3 \} \] ### 9.3 Shifting the graph of \( g \) to coincide with \( h \) The functions are: \[ g(x) = \frac{6}{x+3} + \frac{3}{2} \] \[ h(x) = \frac{6}{x-3} + 2 \] #### 9.3.1 Horizontal Shift To find the horizontal shift, we need to determine how to transform \( g(x) \) into \( h(x) \). The term \( x+3 \) in \( g(x) \) must be changed to \( x-3 \) in \( h(x) \). To achieve this, we can set: \[ x + 3 = x' - 3 \implies x' = x + 6 \] Thus, the graph of \( g \) must be shifted **6 units to the left**. #### 9.3.2 Vertical Shift Next, we look at the vertical shift. The constant term in \( g(x) \) is \( \frac{3}{2} \) and in \( h(x) \) it is \( 2 \). To find the vertical shift, we calculate: \[ 2 - \frac{3}{2} = \frac{4}{2} - \frac{3}{2} = \frac{1}{2} \] Thus, the graph of \( g \) must be shifted **upward by \( \frac{1}{2} \)**. ### 9.4 Asymptotes of \( g \) The vertical asymptote occurs where the denominator is zero: \[ x + 3 = 0 \implies x = -3 \] The horizontal asymptote can be found by considering the behavior of \( g(x) \) as \( x \to \infty \): \[ \lim_{x \to \infty} g(x) = \lim_{x \to \infty} \left( \frac{6}{x+3} + \frac{3}{2} \right) = 0 + \frac{3}{2} = \frac{3}{2} \] Thus, the equations of the asymptotes of \( g \) are: - Vertical asymptote: \( x = -3 \) - Horizontal asymptote: \( y = \frac{3}{2} \) ### 9.5 \( x \)-intercept of \( g \) To find the \( x \)-intercept, we set \( g(x) = 0 \): \[ \frac{6}{x+3} + \frac{3}{2} = 0 \] Solving for \( x \): \[ \frac{6}{x+3} = -\frac{3}{2} \] Cross-multiplying gives: \[ 6 \cdot 2 = -3(x + 3) \implies 12 = -3x - 9 \implies 3x = -21 \implies x = -7 \] Thus, the \( x \)-intercept of \( g \) is \( (-7, 0) \). ### 9.6 Sketch of the graph of \( g \) To sketch the graph, plot the vertical asymptote at \( x = -3 \), the horizontal asymptote at \( y = \frac{3}{2} \), and the \( x \)-intercept at \( (-7, 0) \). The graph will approach the asymptotes and will have a hyperbolic shape. ### 9.7 Value of \( k \) for axis of symmetry The axis of symmetry \( h(x) = -x + k \) must satisfy the condition that \( g(x) \) is symmetric about this line. The line \( y = -x + k \) has a slope of -1, which means it is a diagonal line. To find \( k \), we need the midpoint of the points \( (x, g(x)) \