Answer :
Certainly! Let's analyze the rational function [tex]\( f(x) = \frac{2x + 9}{4x - 1} \)[/tex] step-by-step and identify its key features for plotting.
### 1. Vertical Asymptote
The vertical asymptote is found where the denominator is zero. Setting the denominator equal to zero and solving for [tex]\( x \)[/tex], we get:
[tex]\[ 4x - 1 = 0 \implies x = \frac{1}{4} \][/tex]
So there is a vertical asymptote at [tex]\( x = \frac{1}{4} \)[/tex].
### 2. Horizontal Asymptote
For horizontal asymptotes in rational functions, we compare the degrees of the numerator and the denominator. Here, both numerator and denominator have the same degree (1). The horizontal asymptote is given by the ratio of the leading coefficients:
[tex]\[ \lim_{x \to \infty} \frac{2x + 9}{4x - 1} = \frac{2}{4} = \frac{1}{2} \][/tex]
So the horizontal asymptote is at [tex]\( y = \frac{1}{2} \)[/tex].
### 3. [tex]\( x \)[/tex]-Intercept
The [tex]\( x \)[/tex]-intercept is found where the numerator is zero:
[tex]\[ 2x + 9 = 0 \implies x = -\frac{9}{2} \][/tex]
So the [tex]\( x \)[/tex]-intercept is at [tex]\( x = -\frac{9}{2} \)[/tex].
### 4. [tex]\( y \)[/tex]-Intercept
The [tex]\( y \)[/tex]-intercept is found by evaluating the function when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{2(0) + 9}{4(0) - 1} = \frac{9}{-1} = -9 \][/tex]
So the [tex]\( y \)[/tex]-intercept is at [tex]\( y = -9 \)[/tex].
### 5. Hole
There is no hole in this function since the numerator and the denominator do not have any common factors that can be canceled out.
### Plotting the Key Features
With the features calculated, let's summarize:
- Vertical Asymptote: [tex]\( x = \frac{1}{4} \)[/tex]
- Horizontal Asymptote: [tex]\( y = \frac{1}{2} \)[/tex]
- [tex]\( x \)[/tex]-Intercept: [tex]\( x = -\frac{9}{2} \)[/tex]
- [tex]\( y \)[/tex]-Intercept: [tex]\( y = -9 \)[/tex]
- Hole: None
Now you can plot these features on a graph of the function [tex]\( f(x) \)[/tex]:
1. Draw a vertical dashed line at [tex]\( x = \frac{1}{4} \)[/tex].
2. Draw a horizontal dashed line at [tex]\( y = \frac{1}{2} \)[/tex].
3. Mark the [tex]\( x \)[/tex]-intercept at [tex]\( (-\frac{9}{2}, 0) \)[/tex].
4. Mark the [tex]\( y \)[/tex]-intercept at [tex]\( (0, -9) \)[/tex].
The function should approach these asymptotes as [tex]\( x \)[/tex] goes to [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex] and it will pass through the intercepts calculated.
Ensure these points and lines are correctly translated onto your graph for a clear and accurate representation of the rational function [tex]\( f(x) = \frac{2x + 9}{4x - 1} \)[/tex].
### 1. Vertical Asymptote
The vertical asymptote is found where the denominator is zero. Setting the denominator equal to zero and solving for [tex]\( x \)[/tex], we get:
[tex]\[ 4x - 1 = 0 \implies x = \frac{1}{4} \][/tex]
So there is a vertical asymptote at [tex]\( x = \frac{1}{4} \)[/tex].
### 2. Horizontal Asymptote
For horizontal asymptotes in rational functions, we compare the degrees of the numerator and the denominator. Here, both numerator and denominator have the same degree (1). The horizontal asymptote is given by the ratio of the leading coefficients:
[tex]\[ \lim_{x \to \infty} \frac{2x + 9}{4x - 1} = \frac{2}{4} = \frac{1}{2} \][/tex]
So the horizontal asymptote is at [tex]\( y = \frac{1}{2} \)[/tex].
### 3. [tex]\( x \)[/tex]-Intercept
The [tex]\( x \)[/tex]-intercept is found where the numerator is zero:
[tex]\[ 2x + 9 = 0 \implies x = -\frac{9}{2} \][/tex]
So the [tex]\( x \)[/tex]-intercept is at [tex]\( x = -\frac{9}{2} \)[/tex].
### 4. [tex]\( y \)[/tex]-Intercept
The [tex]\( y \)[/tex]-intercept is found by evaluating the function when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{2(0) + 9}{4(0) - 1} = \frac{9}{-1} = -9 \][/tex]
So the [tex]\( y \)[/tex]-intercept is at [tex]\( y = -9 \)[/tex].
### 5. Hole
There is no hole in this function since the numerator and the denominator do not have any common factors that can be canceled out.
### Plotting the Key Features
With the features calculated, let's summarize:
- Vertical Asymptote: [tex]\( x = \frac{1}{4} \)[/tex]
- Horizontal Asymptote: [tex]\( y = \frac{1}{2} \)[/tex]
- [tex]\( x \)[/tex]-Intercept: [tex]\( x = -\frac{9}{2} \)[/tex]
- [tex]\( y \)[/tex]-Intercept: [tex]\( y = -9 \)[/tex]
- Hole: None
Now you can plot these features on a graph of the function [tex]\( f(x) \)[/tex]:
1. Draw a vertical dashed line at [tex]\( x = \frac{1}{4} \)[/tex].
2. Draw a horizontal dashed line at [tex]\( y = \frac{1}{2} \)[/tex].
3. Mark the [tex]\( x \)[/tex]-intercept at [tex]\( (-\frac{9}{2}, 0) \)[/tex].
4. Mark the [tex]\( y \)[/tex]-intercept at [tex]\( (0, -9) \)[/tex].
The function should approach these asymptotes as [tex]\( x \)[/tex] goes to [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex] and it will pass through the intercepts calculated.
Ensure these points and lines are correctly translated onto your graph for a clear and accurate representation of the rational function [tex]\( f(x) = \frac{2x + 9}{4x - 1} \)[/tex].
