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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]
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