Answer :
Let's address each part of the question step-by-step:
### 9.1 Domain of [tex]\( g(x) \)[/tex]
The function [tex]\( g(x) = \frac{6}{x+3} + \frac{3}{2} \)[/tex] has a term [tex]\(\frac{6}{x+3}\)[/tex]. For this term to be defined, the denominator [tex]\( x+3 \neq 0 \)[/tex] must hold. Thus, the domain of [tex]\( g(x) \)[/tex] is:
[tex]\[ x \neq -3 \][/tex]
### 9.2 Range of [tex]\( g(x) \)[/tex]
We analyze [tex]\( g(x) = \frac{6}{x+3} + \frac{3}{2} \)[/tex].
The term [tex]\(\frac{6}{x+3}\)[/tex] can take any real number except zero. Since we are adding [tex]\(\frac{3}{2}\)[/tex] to it, the range of [tex]\( g(x) \)[/tex] is all real numbers except [tex]\(\frac{3}{2}\)[/tex], because the term [tex]\(\frac{6}{x+3}\)[/tex] will never be zero, making the overall expression equal to [tex]\(\frac{3}{2}\)[/tex].
### 9.3 Shifting [tex]\( g(x) \)[/tex] to coincide with [tex]\( h(x) \)[/tex]
To make the graph of [tex]\( g(x) \)[/tex] coincide with [tex]\( h(x) = \frac{6}{x-3} + 2 \)[/tex]:
#### 9.3.1 Horizontal Shift
For [tex]\( g(x) \)[/tex] to match [tex]\( h(x) \)[/tex], [tex]\( x + 3 \)[/tex] in [tex]\( g(x) \)[/tex] should become [tex]\( x - 3 \)[/tex] in [tex]\( h(x) \)[/tex]. This implies a shift of:
[tex]\[ 6 \text{ units to the right} \][/tex]
#### 9.3.2 Vertical Shift
For the constant terms to match, [tex]\( \frac{3}{2} \)[/tex] in [tex]\( g(x) \)[/tex] should become [tex]\( 2 \)[/tex] in [tex]\( h(x) \)[/tex]. Therefore, the vertical shift needed is:
[tex]\[ \left( 2 - \frac{3}{2} \right) = \frac{1}{2} \text{ unit up} \][/tex]
### 9.4 Equations of the asymptotes of [tex]\( g(x) \)[/tex]
For [tex]\( g(x) = \frac{6}{x+3} + \frac{3}{2} \)[/tex]:
- Vertical asymptote: This occurs where the denominator is zero, i.e., [tex]\( x = -3 \)[/tex].
- Horizontal asymptote: As [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex], [tex]\(\frac{6}{x+3} \rightarrow 0\)[/tex], and hence [tex]\( g(x) \rightarrow \frac{3}{2} \)[/tex].
The asymptotes are:
[tex]\[ x = -3 \][/tex]
[tex]\[ y = \frac{3}{2} \][/tex]
### 9.5 Calculate the [tex]\( x \)[/tex]-intercept of [tex]\( g(x) \)[/tex]
To find the [tex]\( x \)[/tex]-intercept, we set [tex]\( g(x) = 0 \)[/tex]:
[tex]\[ 0 = \frac{6}{x+3} + \frac{3}{2} \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ \frac{6}{x+3} = -\frac{3}{2} \][/tex]
[tex]\[ 6 = -\frac{3}{2}(x+3) \][/tex]
[tex]\[ 12 = -3(x+3) \][/tex]
[tex]\[ 12 = -3x - 9 \][/tex]
[tex]\[ 21 = -3x \][/tex]
[tex]\[ x = -7 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercept is:
[tex]\[ x = -7 \][/tex]
### 9.6 Sketch the graph of [tex]\( g(x) \)[/tex]
(Sketch not provided here as per the question instructions. You would typically draw the graph showing asymptotes [tex]\( x = -3 \)[/tex] and [tex]\( y = \frac{3}{2} \)[/tex] along with the [tex]\( x \)[/tex]-intercept at [tex]\( x = -7 \)[/tex].)
### 9.7 Axis of symmetry of [tex]\( g(x) \)[/tex]
Given that [tex]\( h(x) = -x + k \)[/tex] is an axis of symmetry, we have:
[tex]\[ g(0) = \frac{6}{0+3} + \frac{3}{2} = 2 + 1.5 = 3.5 \][/tex]
So, the line [tex]\( y = -x + k \)[/tex] passes through the point [tex]\( (0, 3.5) \)[/tex]. Hence:
[tex]\[ k = 3.5 \][/tex]
### 9.8 Values of [tex]\( x \)[/tex] for [tex]\( \frac{6}{x+3} - \frac{3}{2} > -x + k \)[/tex]
We seek the values of [tex]\( x \)[/tex] for which:
[tex]\[ \frac{6}{x+3} - \frac{3}{2} > -x + 3.5 \][/tex]
Rearrange to form:
[tex]\[ \frac{6}{x+3} - \frac{3}{2} + x > 3.5 \][/tex]
Solving this inequality algebraically and graphically can show the regions where this holds.
### 9.9 Reflection of [tex]\( g(x) \)[/tex] in the [tex]\( x \)[/tex]-axis
Reflecting [tex]\( g(x) = \frac{6}{x+3} + \frac{3}{2} \)[/tex] in the [tex]\( x \)[/tex]-axis:
[tex]\[ g(x) \rightarrow -g(x) \][/tex]
Thus,
[tex]\[ y = -\left( \frac{6}{x+3} + \frac{3}{2} \right) \][/tex]
[tex]\[ y = -\frac{6}{x+3} - \frac{3}{2} \][/tex]
The new equation is:
[tex]\[ y = -\frac{6}{x+3} - \frac{3}{2} \][/tex]
### 9.1 Domain of [tex]\( g(x) \)[/tex]
The function [tex]\( g(x) = \frac{6}{x+3} + \frac{3}{2} \)[/tex] has a term [tex]\(\frac{6}{x+3}\)[/tex]. For this term to be defined, the denominator [tex]\( x+3 \neq 0 \)[/tex] must hold. Thus, the domain of [tex]\( g(x) \)[/tex] is:
[tex]\[ x \neq -3 \][/tex]
### 9.2 Range of [tex]\( g(x) \)[/tex]
We analyze [tex]\( g(x) = \frac{6}{x+3} + \frac{3}{2} \)[/tex].
The term [tex]\(\frac{6}{x+3}\)[/tex] can take any real number except zero. Since we are adding [tex]\(\frac{3}{2}\)[/tex] to it, the range of [tex]\( g(x) \)[/tex] is all real numbers except [tex]\(\frac{3}{2}\)[/tex], because the term [tex]\(\frac{6}{x+3}\)[/tex] will never be zero, making the overall expression equal to [tex]\(\frac{3}{2}\)[/tex].
### 9.3 Shifting [tex]\( g(x) \)[/tex] to coincide with [tex]\( h(x) \)[/tex]
To make the graph of [tex]\( g(x) \)[/tex] coincide with [tex]\( h(x) = \frac{6}{x-3} + 2 \)[/tex]:
#### 9.3.1 Horizontal Shift
For [tex]\( g(x) \)[/tex] to match [tex]\( h(x) \)[/tex], [tex]\( x + 3 \)[/tex] in [tex]\( g(x) \)[/tex] should become [tex]\( x - 3 \)[/tex] in [tex]\( h(x) \)[/tex]. This implies a shift of:
[tex]\[ 6 \text{ units to the right} \][/tex]
#### 9.3.2 Vertical Shift
For the constant terms to match, [tex]\( \frac{3}{2} \)[/tex] in [tex]\( g(x) \)[/tex] should become [tex]\( 2 \)[/tex] in [tex]\( h(x) \)[/tex]. Therefore, the vertical shift needed is:
[tex]\[ \left( 2 - \frac{3}{2} \right) = \frac{1}{2} \text{ unit up} \][/tex]
### 9.4 Equations of the asymptotes of [tex]\( g(x) \)[/tex]
For [tex]\( g(x) = \frac{6}{x+3} + \frac{3}{2} \)[/tex]:
- Vertical asymptote: This occurs where the denominator is zero, i.e., [tex]\( x = -3 \)[/tex].
- Horizontal asymptote: As [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex], [tex]\(\frac{6}{x+3} \rightarrow 0\)[/tex], and hence [tex]\( g(x) \rightarrow \frac{3}{2} \)[/tex].
The asymptotes are:
[tex]\[ x = -3 \][/tex]
[tex]\[ y = \frac{3}{2} \][/tex]
### 9.5 Calculate the [tex]\( x \)[/tex]-intercept of [tex]\( g(x) \)[/tex]
To find the [tex]\( x \)[/tex]-intercept, we set [tex]\( g(x) = 0 \)[/tex]:
[tex]\[ 0 = \frac{6}{x+3} + \frac{3}{2} \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ \frac{6}{x+3} = -\frac{3}{2} \][/tex]
[tex]\[ 6 = -\frac{3}{2}(x+3) \][/tex]
[tex]\[ 12 = -3(x+3) \][/tex]
[tex]\[ 12 = -3x - 9 \][/tex]
[tex]\[ 21 = -3x \][/tex]
[tex]\[ x = -7 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercept is:
[tex]\[ x = -7 \][/tex]
### 9.6 Sketch the graph of [tex]\( g(x) \)[/tex]
(Sketch not provided here as per the question instructions. You would typically draw the graph showing asymptotes [tex]\( x = -3 \)[/tex] and [tex]\( y = \frac{3}{2} \)[/tex] along with the [tex]\( x \)[/tex]-intercept at [tex]\( x = -7 \)[/tex].)
### 9.7 Axis of symmetry of [tex]\( g(x) \)[/tex]
Given that [tex]\( h(x) = -x + k \)[/tex] is an axis of symmetry, we have:
[tex]\[ g(0) = \frac{6}{0+3} + \frac{3}{2} = 2 + 1.5 = 3.5 \][/tex]
So, the line [tex]\( y = -x + k \)[/tex] passes through the point [tex]\( (0, 3.5) \)[/tex]. Hence:
[tex]\[ k = 3.5 \][/tex]
### 9.8 Values of [tex]\( x \)[/tex] for [tex]\( \frac{6}{x+3} - \frac{3}{2} > -x + k \)[/tex]
We seek the values of [tex]\( x \)[/tex] for which:
[tex]\[ \frac{6}{x+3} - \frac{3}{2} > -x + 3.5 \][/tex]
Rearrange to form:
[tex]\[ \frac{6}{x+3} - \frac{3}{2} + x > 3.5 \][/tex]
Solving this inequality algebraically and graphically can show the regions where this holds.
### 9.9 Reflection of [tex]\( g(x) \)[/tex] in the [tex]\( x \)[/tex]-axis
Reflecting [tex]\( g(x) = \frac{6}{x+3} + \frac{3}{2} \)[/tex] in the [tex]\( x \)[/tex]-axis:
[tex]\[ g(x) \rightarrow -g(x) \][/tex]
Thus,
[tex]\[ y = -\left( \frac{6}{x+3} + \frac{3}{2} \right) \][/tex]
[tex]\[ y = -\frac{6}{x+3} - \frac{3}{2} \][/tex]
The new equation is:
[tex]\[ y = -\frac{6}{x+3} - \frac{3}{2} \][/tex]
