Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
To analyze the function [tex]\( f(x) = \frac{x^2 - 3x + 2}{x - 2} \)[/tex], we will determine its domain, vertical and horizontal asymptotes, intercepts, and any removable discontinuities.
### 1. Domain
The domain of a function is the set of all possible input values (x-values) that do not cause the function to be undefined. For the given function, it will be undefined wherever the denominator is zero.
The denominator is [tex]\( x - 2 \)[/tex]. Setting this equal to zero, we find:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
Hence, the function is undefined at [tex]\( x = 2 \)[/tex]. Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers except [tex]\( x = 2 \)[/tex], which can be written in interval notation as:
[tex]\[ (-\infty, 2) \cup (2, \infty) \][/tex]
### 2. Vertical Asymptotes
Vertical asymptotes occur where the function approaches infinity as [tex]\( x \)[/tex] approaches a certain value within the domain. Typically, these are found at values that make the denominator zero but are not cancelled out by the numerator.
For [tex]\( f(x) \)[/tex], the problematic value is [tex]\( x = 2 \)[/tex]. However, for this specific function, there is no vertical asymptote at [tex]\( x = 2 \)[/tex]. Instead, we need to check for removable discontinuities in the steps below. Therefore, there are no vertical asymptotes in this function.
### 3. Horizontal Asymptotes
Horizontal asymptotes are determined by comparing the degrees of the numerator and the denominator.
- Degree of the numerator [tex]\( x^2 - 3x + 2 \)[/tex] is 2.
- Degree of the denominator [tex]\( x - 2 \)[/tex] is 1.
Since the degree of the numerator (2) is greater than the degree of the denominator (1), there are no horizontal asymptotes for this function.
### 4. Simplifying the Function (Removable Discontinuities)
To identify any removable discontinuities, we simplify the function algebraically. The numerator [tex]\( x^2 - 3x + 2 \)[/tex] can be factored as:
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
Thus, the function can be rewritten as:
[tex]\[ f(x) = \frac{(x - 1)(x - 2)}{x - 2} \][/tex]
For [tex]\( x \neq 2 \)[/tex], the [tex]\( (x - 2) \)[/tex] factors cancel out, giving:
[tex]\[ f(x) = x - 1 \][/tex]
So, the simplified function is:
[tex]\[ f(x) = x - 1, \quad \text{for } x \neq 2 \][/tex]
Here, [tex]\( x = 2 \)[/tex] represents a removable discontinuity. At [tex]\( x = 2 \)[/tex], the function would have been:
[tex]\[ f(2) = 2 - 1 = 1 \][/tex]
### 5. Intercepts
#### Y-Intercept
The y-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{0^2 - 3 \cdot 0 + 2}{0 - 2} = \frac{2}{-2} = -1 \][/tex]
So, the y-intercept is at [tex]\( (0, -1) \)[/tex].
#### X-Intercepts
To find the x-intercepts, set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \frac{x^2 - 3x + 2}{x - 2} = 0 \][/tex]
[tex]\[ x^2 - 3x + 2 = 0 \][/tex]
[tex]\[ (x - 1)(x - 2) = 0 \][/tex]
Solving these factors:
[tex]\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \][/tex]
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex] (but this is a removable discontinuity and not relevant as an intercept)
Thus, the x-intercept is at [tex]\( (1, 0) \)[/tex].
### Summary
- Domain: [tex]\( (-\infty, 2) \cup (2, \infty) \)[/tex]
- Vertical Asymptotes: None
- Horizontal Asymptotes: None
- Y-Intercept: [tex]\( (0, -1) \)[/tex]
- X-Intercepts: [tex]\( (1, 0) \)[/tex]
- Removable Discontinuity: At [tex]\( x = 2 \)[/tex]
### 1. Domain
The domain of a function is the set of all possible input values (x-values) that do not cause the function to be undefined. For the given function, it will be undefined wherever the denominator is zero.
The denominator is [tex]\( x - 2 \)[/tex]. Setting this equal to zero, we find:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
Hence, the function is undefined at [tex]\( x = 2 \)[/tex]. Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers except [tex]\( x = 2 \)[/tex], which can be written in interval notation as:
[tex]\[ (-\infty, 2) \cup (2, \infty) \][/tex]
### 2. Vertical Asymptotes
Vertical asymptotes occur where the function approaches infinity as [tex]\( x \)[/tex] approaches a certain value within the domain. Typically, these are found at values that make the denominator zero but are not cancelled out by the numerator.
For [tex]\( f(x) \)[/tex], the problematic value is [tex]\( x = 2 \)[/tex]. However, for this specific function, there is no vertical asymptote at [tex]\( x = 2 \)[/tex]. Instead, we need to check for removable discontinuities in the steps below. Therefore, there are no vertical asymptotes in this function.
### 3. Horizontal Asymptotes
Horizontal asymptotes are determined by comparing the degrees of the numerator and the denominator.
- Degree of the numerator [tex]\( x^2 - 3x + 2 \)[/tex] is 2.
- Degree of the denominator [tex]\( x - 2 \)[/tex] is 1.
Since the degree of the numerator (2) is greater than the degree of the denominator (1), there are no horizontal asymptotes for this function.
### 4. Simplifying the Function (Removable Discontinuities)
To identify any removable discontinuities, we simplify the function algebraically. The numerator [tex]\( x^2 - 3x + 2 \)[/tex] can be factored as:
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
Thus, the function can be rewritten as:
[tex]\[ f(x) = \frac{(x - 1)(x - 2)}{x - 2} \][/tex]
For [tex]\( x \neq 2 \)[/tex], the [tex]\( (x - 2) \)[/tex] factors cancel out, giving:
[tex]\[ f(x) = x - 1 \][/tex]
So, the simplified function is:
[tex]\[ f(x) = x - 1, \quad \text{for } x \neq 2 \][/tex]
Here, [tex]\( x = 2 \)[/tex] represents a removable discontinuity. At [tex]\( x = 2 \)[/tex], the function would have been:
[tex]\[ f(2) = 2 - 1 = 1 \][/tex]
### 5. Intercepts
#### Y-Intercept
The y-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{0^2 - 3 \cdot 0 + 2}{0 - 2} = \frac{2}{-2} = -1 \][/tex]
So, the y-intercept is at [tex]\( (0, -1) \)[/tex].
#### X-Intercepts
To find the x-intercepts, set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \frac{x^2 - 3x + 2}{x - 2} = 0 \][/tex]
[tex]\[ x^2 - 3x + 2 = 0 \][/tex]
[tex]\[ (x - 1)(x - 2) = 0 \][/tex]
Solving these factors:
[tex]\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \][/tex]
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex] (but this is a removable discontinuity and not relevant as an intercept)
Thus, the x-intercept is at [tex]\( (1, 0) \)[/tex].
### Summary
- Domain: [tex]\( (-\infty, 2) \cup (2, \infty) \)[/tex]
- Vertical Asymptotes: None
- Horizontal Asymptotes: None
- Y-Intercept: [tex]\( (0, -1) \)[/tex]
- X-Intercepts: [tex]\( (1, 0) \)[/tex]
- Removable Discontinuity: At [tex]\( x = 2 \)[/tex]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
