Answer :
Let's start by analyzing the function [tex]\( f(x) = \frac{x^2 - 81}{x^2 - 11x + 18} \)[/tex]. To understand the most appropriate viewing window for graphing this function and to determine the domain and range, follow these steps:
1. Factor the numerator and the denominator to identify any points of discontinuity:
The numerator [tex]\( x^2 - 81 \)[/tex] can be factored as:
[tex]\[ x^2 - 81 = (x - 9)(x + 9) \][/tex]
The denominator [tex]\( x^2 - 11x + 18 \)[/tex] can be factored as:
[tex]\[ x^2 - 11x + 18 = (x - 2)(x - 9) \][/tex]
2. Identify any restrictions on the domain by finding the values of [tex]\( x \)[/tex] that make the denominator zero:
Setting the denominator equal to zero:
[tex]\[ (x - 2)(x - 9) = 0 \][/tex]
gives:
[tex]\[ x = 2 \quad \text{and} \quad x = 9 \][/tex]
Therefore, the function is undefined at [tex]\( x = 2 \)[/tex] and [tex]\( x = 9 \)[/tex].
3. Determine the asymptotes:
- Vertical Asymptotes: These occur where the denominator is zero. Based on our factorization:
[tex]\[ x = 2 \quad \text{and} \quad x = 9 \][/tex]
- Horizontal Asymptote: Since the degrees of the numerator and the denominator are the same (both quadratic), we look at the leading coefficients:
[tex]\[ \text{Horizontal Asymptote:} \quad y = \frac{1}{1} = 1 \][/tex]
4. Determine the simplified form of the function:
Notice that [tex]\( (x - 9) \)[/tex] is a common factor in both the numerator and the denominator:
[tex]\[ f(x) = \frac{(x - 9)(x + 9)}{(x - 2)(x - 9)} = \frac{x + 9}{x - 2} \quad \text{for} \quad x \neq 9 \][/tex]
At [tex]\( x = 9 \)[/tex], the function has a removable discontinuity, so we need to consider the limit as [tex]\( x \)[/tex] approaches 9.
5. Choosing an appropriate viewing window:
To capture the essential features of the function:
- We need a wide enough x-range to capture the behavior of the function around the vertical asymptotes [tex]\( x = 2 \)[/tex] and [tex]\( x = 9 \)[/tex] and to see the horizontal asymptote.
- Ensuring the y-range captures the behavior around the horizontal asymptote and the excursions near the vertical asymptotes.
Given the factors and analysis, the most appropriate viewing window would ensure we catch the vertical asymptotes and observe the overall behavior of the function:
[tex]\[ \text{Xmin: } -10, \quad \text{Xmax: } 10, \quad \text{Ymin: } -10, \quad \text{Ymax: } 10 \][/tex]
This large viewing window allows for a comprehensive view including the discontinuities, asymptotic behavior, and the general trend of the function. Thus, the correct answer is:
[tex]\[ \text{Xmin: } -10, \quad \text{Xmax: } 10, \quad \text{Ymin: } -10, \quad \text{Ymax: } 10 \][/tex]
1. Factor the numerator and the denominator to identify any points of discontinuity:
The numerator [tex]\( x^2 - 81 \)[/tex] can be factored as:
[tex]\[ x^2 - 81 = (x - 9)(x + 9) \][/tex]
The denominator [tex]\( x^2 - 11x + 18 \)[/tex] can be factored as:
[tex]\[ x^2 - 11x + 18 = (x - 2)(x - 9) \][/tex]
2. Identify any restrictions on the domain by finding the values of [tex]\( x \)[/tex] that make the denominator zero:
Setting the denominator equal to zero:
[tex]\[ (x - 2)(x - 9) = 0 \][/tex]
gives:
[tex]\[ x = 2 \quad \text{and} \quad x = 9 \][/tex]
Therefore, the function is undefined at [tex]\( x = 2 \)[/tex] and [tex]\( x = 9 \)[/tex].
3. Determine the asymptotes:
- Vertical Asymptotes: These occur where the denominator is zero. Based on our factorization:
[tex]\[ x = 2 \quad \text{and} \quad x = 9 \][/tex]
- Horizontal Asymptote: Since the degrees of the numerator and the denominator are the same (both quadratic), we look at the leading coefficients:
[tex]\[ \text{Horizontal Asymptote:} \quad y = \frac{1}{1} = 1 \][/tex]
4. Determine the simplified form of the function:
Notice that [tex]\( (x - 9) \)[/tex] is a common factor in both the numerator and the denominator:
[tex]\[ f(x) = \frac{(x - 9)(x + 9)}{(x - 2)(x - 9)} = \frac{x + 9}{x - 2} \quad \text{for} \quad x \neq 9 \][/tex]
At [tex]\( x = 9 \)[/tex], the function has a removable discontinuity, so we need to consider the limit as [tex]\( x \)[/tex] approaches 9.
5. Choosing an appropriate viewing window:
To capture the essential features of the function:
- We need a wide enough x-range to capture the behavior of the function around the vertical asymptotes [tex]\( x = 2 \)[/tex] and [tex]\( x = 9 \)[/tex] and to see the horizontal asymptote.
- Ensuring the y-range captures the behavior around the horizontal asymptote and the excursions near the vertical asymptotes.
Given the factors and analysis, the most appropriate viewing window would ensure we catch the vertical asymptotes and observe the overall behavior of the function:
[tex]\[ \text{Xmin: } -10, \quad \text{Xmax: } 10, \quad \text{Ymin: } -10, \quad \text{Ymax: } 10 \][/tex]
This large viewing window allows for a comprehensive view including the discontinuities, asymptotic behavior, and the general trend of the function. Thus, the correct answer is:
[tex]\[ \text{Xmin: } -10, \quad \text{Xmax: } 10, \quad \text{Ymin: } -10, \quad \text{Ymax: } 10 \][/tex]
