Answer :
To graph the rational function [tex]\( f(x) = \frac{-x + 5}{-2x + 1} \)[/tex], we'll start by identifying key characteristics such as vertical and horizontal asymptotes, intercepts, and overall behavior of the function.
### 1. Vertical Asymptote
The vertical asymptote occurs where the denominator is zero because the function will be undefined at that point. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -2x + 1 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -2x = -1 \][/tex]
[tex]\[ x = \frac{1}{2} \][/tex]
Thus, there is a vertical asymptote at [tex]\( x = \frac{1}{2} \)[/tex].
### 2. Horizontal Asymptote
The horizontal asymptote is determined by the degrees of the numerator and the denominator. Since both are linear functions (degree 1), we compare the leading coefficients. The leading coefficient of the numerator is [tex]\(-1\)[/tex] and the leading coefficient of the denominator is [tex]\(-2\)[/tex]. The horizontal asymptote is:
[tex]\[ y = \frac{-1}{-2} = \frac{1}{2} \][/tex]
Thus, the horizontal asymptote is [tex]\( y = \frac{1}{2} \)[/tex].
### 3. Intercepts
#### x-intercept
The [tex]\( x \)[/tex]-intercept occurs where the numerator is zero. Set the numerator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -x + 5 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 5 \][/tex]
So, the [tex]\( x \)[/tex]-intercept is at [tex]\( (5, 0) \)[/tex].
#### y-intercept
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = \frac{-(0) + 5}{-2(0) + 1} = \frac{5}{1} = 5 \][/tex]
So, the [tex]\( y \)[/tex]-intercept is at [tex]\( (0, 5) \)[/tex].
### 4. Behavior Near Asymptotes
We need to determine the behavior of the function as [tex]\( x \)[/tex] approaches the vertical asymptote from both sides and as [tex]\( x \)[/tex] approaches infinity.
- As [tex]\( x \to \frac{1}{2}^- \)[/tex] (approaching from the left), the values of [tex]\( f(x) \)[/tex] tend to [tex]\( -\infty \)[/tex].
- As [tex]\( x \to \frac{1}{2}^+ \)[/tex] (approaching from the right), the values of [tex]\( f(x) \)[/tex] tend to [tex]\( \infty \)[/tex].
As for the horizontal asymptote:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \frac{1}{2} \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to \frac{1}{2} \)[/tex].
### 5. Plotting the Graph
Using the information gathered, we can sketch the graph. Here are the steps:
1. Draw the vertical asymptote line at [tex]\( x = \frac{1}{2} \)[/tex].
2. Draw the horizontal asymptote line at [tex]\( y = \frac{1}{2} \)[/tex].
3. Plot the intercept points [tex]\( (5, 0) \)[/tex] and [tex]\( (0, 5) \)[/tex].
4. Sketch the curve approaching the vertical asymptote: as [tex]\( x \to \frac{1}{2}^- \)[/tex], the function goes to [tex]\( -\infty \)[/tex]; as [tex]\( x \to \frac{1}{2}^+ \)[/tex], the function goes to [tex]\( \infty \)[/tex].
5. Make sure that as [tex]\( x \)[/tex] moves towards [tex]\( \infty \)[/tex] or [tex]\( -\infty \)[/tex], the curve approaches [tex]\( y = \frac{1}{2} \)[/tex].
By following these steps, you should have a clear graph of the rational function [tex]\( f(x) = \frac{-x + 5}{-2x + 1} \)[/tex] with all key features accurately represented.
### 1. Vertical Asymptote
The vertical asymptote occurs where the denominator is zero because the function will be undefined at that point. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -2x + 1 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -2x = -1 \][/tex]
[tex]\[ x = \frac{1}{2} \][/tex]
Thus, there is a vertical asymptote at [tex]\( x = \frac{1}{2} \)[/tex].
### 2. Horizontal Asymptote
The horizontal asymptote is determined by the degrees of the numerator and the denominator. Since both are linear functions (degree 1), we compare the leading coefficients. The leading coefficient of the numerator is [tex]\(-1\)[/tex] and the leading coefficient of the denominator is [tex]\(-2\)[/tex]. The horizontal asymptote is:
[tex]\[ y = \frac{-1}{-2} = \frac{1}{2} \][/tex]
Thus, the horizontal asymptote is [tex]\( y = \frac{1}{2} \)[/tex].
### 3. Intercepts
#### x-intercept
The [tex]\( x \)[/tex]-intercept occurs where the numerator is zero. Set the numerator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -x + 5 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 5 \][/tex]
So, the [tex]\( x \)[/tex]-intercept is at [tex]\( (5, 0) \)[/tex].
#### y-intercept
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = \frac{-(0) + 5}{-2(0) + 1} = \frac{5}{1} = 5 \][/tex]
So, the [tex]\( y \)[/tex]-intercept is at [tex]\( (0, 5) \)[/tex].
### 4. Behavior Near Asymptotes
We need to determine the behavior of the function as [tex]\( x \)[/tex] approaches the vertical asymptote from both sides and as [tex]\( x \)[/tex] approaches infinity.
- As [tex]\( x \to \frac{1}{2}^- \)[/tex] (approaching from the left), the values of [tex]\( f(x) \)[/tex] tend to [tex]\( -\infty \)[/tex].
- As [tex]\( x \to \frac{1}{2}^+ \)[/tex] (approaching from the right), the values of [tex]\( f(x) \)[/tex] tend to [tex]\( \infty \)[/tex].
As for the horizontal asymptote:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \frac{1}{2} \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to \frac{1}{2} \)[/tex].
### 5. Plotting the Graph
Using the information gathered, we can sketch the graph. Here are the steps:
1. Draw the vertical asymptote line at [tex]\( x = \frac{1}{2} \)[/tex].
2. Draw the horizontal asymptote line at [tex]\( y = \frac{1}{2} \)[/tex].
3. Plot the intercept points [tex]\( (5, 0) \)[/tex] and [tex]\( (0, 5) \)[/tex].
4. Sketch the curve approaching the vertical asymptote: as [tex]\( x \to \frac{1}{2}^- \)[/tex], the function goes to [tex]\( -\infty \)[/tex]; as [tex]\( x \to \frac{1}{2}^+ \)[/tex], the function goes to [tex]\( \infty \)[/tex].
5. Make sure that as [tex]\( x \)[/tex] moves towards [tex]\( \infty \)[/tex] or [tex]\( -\infty \)[/tex], the curve approaches [tex]\( y = \frac{1}{2} \)[/tex].
By following these steps, you should have a clear graph of the rational function [tex]\( f(x) = \frac{-x + 5}{-2x + 1} \)[/tex] with all key features accurately represented.
