Answer :
Sure, let's go through each part of the question step-by-step.
### 2.1 Asymptotes
#### Vertical Asymptote:
To determine the vertical asymptote, we need to consider the values which make the denominator zero. The given function is:
[tex]\[ f(x) = \frac{-3}{x-2} + 1 \][/tex]
The vertical asymptote occurs when the denominator is zero:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
So, the vertical asymptote is:
[tex]\[ x = 2 \][/tex]
#### Horizontal Asymptote:
To find the horizontal asymptote, we need to analyze the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\(\pm\infty\)[/tex].
As [tex]\( x \to \pm\infty \)[/tex], [tex]\(\frac{-3}{x-2} \to 0\)[/tex]. Therefore,
[tex]\[ f(x) \approx 0 + 1 = 1 \][/tex]
So, the horizontal asymptote is:
[tex]\[ y = 1 \][/tex]
### 2.2 Intercepts
#### [tex]\(x\)[/tex]-Intercept:
The [tex]\(x\)[/tex]-intercept occurs when [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \frac{-3}{x-2} + 1 = 0 \][/tex]
Solving this equation:
[tex]\[ \frac{-3}{x-2} = -1 \][/tex]
[tex]\[ \frac{3}{x-2} = 1 \][/tex]
[tex]\[ 3 = x - 2 \][/tex]
[tex]\[ x = 5 \][/tex]
So, the [tex]\(x\)[/tex]-intercept is:
[tex]\[ (5, 0) \][/tex]
#### [tex]\(y\)[/tex]-Intercept:
The [tex]\(y\)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{-3}{0-2} + 1 \][/tex]
[tex]\[ f(0) = \frac{-3}{-2} + 1 \][/tex]
[tex]\[ f(0) = 1.5 + 1 \][/tex]
[tex]\[ f(0) = 2.5 \][/tex]
So, the [tex]\(y\)[/tex]-intercept is:
[tex]\[ (0, 2.5) \][/tex]
### 2.3 Sketching the Graph
To sketch the graph of [tex]\( f(x) \)[/tex], you should include the following:
- The vertical asymptote at [tex]\( x = 2 \)[/tex]
- The horizontal asymptote at [tex]\( y = 1 \)[/tex]
- The [tex]\( x \)[/tex]-intercept at [tex]\( (5, 0) \)[/tex]
- The [tex]\( y \)[/tex]-intercept at [tex]\( (0, 2.5) \)[/tex]
The graph is a hyperbola that approaches these asymptotes.
### 2.4 Range of [tex]\( y = -f(x) \)[/tex]
Given [tex]\( f(x) = \frac{-3}{x-2} + 1 \)[/tex], the transformation [tex]\( -f(x) \)[/tex] reverses the sign of [tex]\( f(x) \)[/tex]:
[tex]\[ -f(x) = -\left(\frac{-3}{x-2} + 1\right) \][/tex]
[tex]\[ -f(x) = \frac{3}{x-2} - 1 \][/tex]
As [tex]\( x \to \pm\infty \)[/tex], [tex]\(\frac{3}{x-2} \to 0\)[/tex] and thus:
[tex]\[ y = -1 \][/tex]
So, the horizontal asymptote of [tex]\( y = -f(x) \)[/tex] is [tex]\( y = -1 \)[/tex].
The function [tex]\( \frac{3}{x-2} - 1 \)[/tex] can take any real value except [tex]\(-1\)[/tex]. Therefore, the range of [tex]\( y = -f(x) \)[/tex] is:
[tex]\[ (-\infty, -1) \cup (-1, \infty) \][/tex]
### 2.5 Axes of Symmetry
The function [tex]\( f(x) = \frac{-3}{x-2} + 1 \)[/tex] is symmetric about the line [tex]\( x = 2 \)[/tex] because it is centered around the vertical asymptote.
If you consider horizontal symmetry, you can reflect the graph over the horizontal asymptote [tex]\( y = 1 \)[/tex]. Therefore, the equations of the axes of symmetry are:
[tex]\[ x = 2 \][/tex]
[tex]\[ y = 1 \][/tex]
### 2.6 Solving Inequality [tex]\( \frac{-3}{x-2} \leq -1 \)[/tex]
First solve the equality:
[tex]\[ \frac{-3}{x-2} = -1 \][/tex]
[tex]\[ \frac{3}{x-2} = 1 \][/tex]
[tex]\[ 3 = x - 2 \][/tex]
[tex]\[ x = 5 \][/tex]
To solve the inequality [tex]\( \frac{-3}{x-2} \leq -1 \)[/tex]:
Rewriting:
[tex]\[ \frac{3}{x-2} \geq 1 \][/tex]
This inequality holds when:
[tex]\[ 3 \geq x - 2 \][/tex]
[tex]\[ x \leq 5 \][/tex]
Thus, the solution set for the inequality is:
[tex]\[ (-\infty, 5] \][/tex]
### 2.7 Translating [tex]\( f \)[/tex] 3 units to the right and 4 units down
To translate the function [tex]\( f(x) \)[/tex] 3 units to the right, substitute [tex]\( x \)[/tex] with [tex]\( x-3 \)[/tex]:
[tex]\[ f(x-3) = \frac{-3}{(x-3)-2} + 1 = \frac{-3}{x-5} + 1 \][/tex]
Then to translate it 4 units down, subtract 4 from the result:
[tex]\[ h(x) = \frac{-3}{x-5} + 1 - 4 \][/tex]
[tex]\[ h(x) = \frac{-3}{x-5} - 3 \][/tex]
So, the equation of [tex]\( h \)[/tex] is:
[tex]\[ h(x) = \frac{-3}{x-5} - 3 \][/tex]
### 2.1 Asymptotes
#### Vertical Asymptote:
To determine the vertical asymptote, we need to consider the values which make the denominator zero. The given function is:
[tex]\[ f(x) = \frac{-3}{x-2} + 1 \][/tex]
The vertical asymptote occurs when the denominator is zero:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
So, the vertical asymptote is:
[tex]\[ x = 2 \][/tex]
#### Horizontal Asymptote:
To find the horizontal asymptote, we need to analyze the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\(\pm\infty\)[/tex].
As [tex]\( x \to \pm\infty \)[/tex], [tex]\(\frac{-3}{x-2} \to 0\)[/tex]. Therefore,
[tex]\[ f(x) \approx 0 + 1 = 1 \][/tex]
So, the horizontal asymptote is:
[tex]\[ y = 1 \][/tex]
### 2.2 Intercepts
#### [tex]\(x\)[/tex]-Intercept:
The [tex]\(x\)[/tex]-intercept occurs when [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \frac{-3}{x-2} + 1 = 0 \][/tex]
Solving this equation:
[tex]\[ \frac{-3}{x-2} = -1 \][/tex]
[tex]\[ \frac{3}{x-2} = 1 \][/tex]
[tex]\[ 3 = x - 2 \][/tex]
[tex]\[ x = 5 \][/tex]
So, the [tex]\(x\)[/tex]-intercept is:
[tex]\[ (5, 0) \][/tex]
#### [tex]\(y\)[/tex]-Intercept:
The [tex]\(y\)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{-3}{0-2} + 1 \][/tex]
[tex]\[ f(0) = \frac{-3}{-2} + 1 \][/tex]
[tex]\[ f(0) = 1.5 + 1 \][/tex]
[tex]\[ f(0) = 2.5 \][/tex]
So, the [tex]\(y\)[/tex]-intercept is:
[tex]\[ (0, 2.5) \][/tex]
### 2.3 Sketching the Graph
To sketch the graph of [tex]\( f(x) \)[/tex], you should include the following:
- The vertical asymptote at [tex]\( x = 2 \)[/tex]
- The horizontal asymptote at [tex]\( y = 1 \)[/tex]
- The [tex]\( x \)[/tex]-intercept at [tex]\( (5, 0) \)[/tex]
- The [tex]\( y \)[/tex]-intercept at [tex]\( (0, 2.5) \)[/tex]
The graph is a hyperbola that approaches these asymptotes.
### 2.4 Range of [tex]\( y = -f(x) \)[/tex]
Given [tex]\( f(x) = \frac{-3}{x-2} + 1 \)[/tex], the transformation [tex]\( -f(x) \)[/tex] reverses the sign of [tex]\( f(x) \)[/tex]:
[tex]\[ -f(x) = -\left(\frac{-3}{x-2} + 1\right) \][/tex]
[tex]\[ -f(x) = \frac{3}{x-2} - 1 \][/tex]
As [tex]\( x \to \pm\infty \)[/tex], [tex]\(\frac{3}{x-2} \to 0\)[/tex] and thus:
[tex]\[ y = -1 \][/tex]
So, the horizontal asymptote of [tex]\( y = -f(x) \)[/tex] is [tex]\( y = -1 \)[/tex].
The function [tex]\( \frac{3}{x-2} - 1 \)[/tex] can take any real value except [tex]\(-1\)[/tex]. Therefore, the range of [tex]\( y = -f(x) \)[/tex] is:
[tex]\[ (-\infty, -1) \cup (-1, \infty) \][/tex]
### 2.5 Axes of Symmetry
The function [tex]\( f(x) = \frac{-3}{x-2} + 1 \)[/tex] is symmetric about the line [tex]\( x = 2 \)[/tex] because it is centered around the vertical asymptote.
If you consider horizontal symmetry, you can reflect the graph over the horizontal asymptote [tex]\( y = 1 \)[/tex]. Therefore, the equations of the axes of symmetry are:
[tex]\[ x = 2 \][/tex]
[tex]\[ y = 1 \][/tex]
### 2.6 Solving Inequality [tex]\( \frac{-3}{x-2} \leq -1 \)[/tex]
First solve the equality:
[tex]\[ \frac{-3}{x-2} = -1 \][/tex]
[tex]\[ \frac{3}{x-2} = 1 \][/tex]
[tex]\[ 3 = x - 2 \][/tex]
[tex]\[ x = 5 \][/tex]
To solve the inequality [tex]\( \frac{-3}{x-2} \leq -1 \)[/tex]:
Rewriting:
[tex]\[ \frac{3}{x-2} \geq 1 \][/tex]
This inequality holds when:
[tex]\[ 3 \geq x - 2 \][/tex]
[tex]\[ x \leq 5 \][/tex]
Thus, the solution set for the inequality is:
[tex]\[ (-\infty, 5] \][/tex]
### 2.7 Translating [tex]\( f \)[/tex] 3 units to the right and 4 units down
To translate the function [tex]\( f(x) \)[/tex] 3 units to the right, substitute [tex]\( x \)[/tex] with [tex]\( x-3 \)[/tex]:
[tex]\[ f(x-3) = \frac{-3}{(x-3)-2} + 1 = \frac{-3}{x-5} + 1 \][/tex]
Then to translate it 4 units down, subtract 4 from the result:
[tex]\[ h(x) = \frac{-3}{x-5} + 1 - 4 \][/tex]
[tex]\[ h(x) = \frac{-3}{x-5} - 3 \][/tex]
So, the equation of [tex]\( h \)[/tex] is:
[tex]\[ h(x) = \frac{-3}{x-5} - 3 \][/tex]
