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Sagot :
To determine which function among the given options has a vertical asymptote at \( x = -2 \), a horizontal asymptote at \( f(x) = -1 \), and a root at \( x = 2 \), we need to check each of the functions based on these criteria. Let's analyze each function step-by-step:
### Function A: \( f(x) = \frac{4}{x+2} - 1 \)
1. Vertical Asymptote: A vertical asymptote occurs where the denominator is zero. For \( f(x) = \frac{4}{x+2} - 1 \), the denominator is zero at \( x = -2 \). Therefore, it has a vertical asymptote at \( x = -2 \).
2. Horizontal Asymptote: Horizontal asymptotes for rational functions of the form \( \frac{a}{x+b} + c \) occur at \( y = c \) as \( x \) approaches infinity. Here, as \( x \rightarrow \infty \), \( f(x) \rightarrow -1 \).
3. Root: A root of the function occurs where \( f(x) = 0 \). Set the function equal to zero:
[tex]\[ \frac{4}{x+2} - 1 = 0 \implies \frac{4}{x+2} = 1 \implies 4 = x + 2 \implies x = 2 \][/tex]
Therefore, \( x = 2 \) is a root.
Since function A meets all criteria, it is a candidate. Let's verify the other functions to ensure there is only one correct answer.
### Function B: \( f(x) = \frac{4}{x-2} - 1 \)
1. Vertical Asymptote: The vertical asymptote occurs where the denominator is zero. Here, \( f(x) = \frac{4}{x-2} - 1 \) has a denominator of zero at \( x = 2 \), not \( x = -2 \). This function does not meet the first criterion.
### Function C: \( f(x) = \frac{-4}{x+2} + 1 \)
1. Vertical Asymptote: The denominator is zero at \( x = -2 \). This function has a vertical asymptote at \( x = -2 \).
2. Horizontal Asymptote: As \( x \rightarrow \infty \), the function approaches \( 1 \), not \(-1\). This function does not meet the second criterion.
### Function D: \( f(x) = \frac{-4}{x+2} - 1 \)
1. Vertical Asymptote: The vertical asymptote occurs where the denominator is zero, which is at \( x = -2 \). This function has a vertical asymptote at \( x = -2 \).
2. Horizontal Asymptote: As \( x \rightarrow \infty \), the function approaches \(-1\), which meets the second criterion.
3. Root: Set the function to zero:
[tex]\[ \frac{-4}{x+2} - 1 = 0 \implies \frac{-4}{x+2} = 1 \implies -4 = x + 2 \implies x = -6 \][/tex]
Therefore, \( x = -6 \) but we need a root at \( x = 2 \). This function does not meet the third criterion.
After evaluating all options, we find that:
- Function A: \( f(x) = \frac{4}{x+2} - 1 \) meets all criteria.
Hence, the correct answer is:
[tex]\[ \boxed{\text{A}} \][/tex]
### Function A: \( f(x) = \frac{4}{x+2} - 1 \)
1. Vertical Asymptote: A vertical asymptote occurs where the denominator is zero. For \( f(x) = \frac{4}{x+2} - 1 \), the denominator is zero at \( x = -2 \). Therefore, it has a vertical asymptote at \( x = -2 \).
2. Horizontal Asymptote: Horizontal asymptotes for rational functions of the form \( \frac{a}{x+b} + c \) occur at \( y = c \) as \( x \) approaches infinity. Here, as \( x \rightarrow \infty \), \( f(x) \rightarrow -1 \).
3. Root: A root of the function occurs where \( f(x) = 0 \). Set the function equal to zero:
[tex]\[ \frac{4}{x+2} - 1 = 0 \implies \frac{4}{x+2} = 1 \implies 4 = x + 2 \implies x = 2 \][/tex]
Therefore, \( x = 2 \) is a root.
Since function A meets all criteria, it is a candidate. Let's verify the other functions to ensure there is only one correct answer.
### Function B: \( f(x) = \frac{4}{x-2} - 1 \)
1. Vertical Asymptote: The vertical asymptote occurs where the denominator is zero. Here, \( f(x) = \frac{4}{x-2} - 1 \) has a denominator of zero at \( x = 2 \), not \( x = -2 \). This function does not meet the first criterion.
### Function C: \( f(x) = \frac{-4}{x+2} + 1 \)
1. Vertical Asymptote: The denominator is zero at \( x = -2 \). This function has a vertical asymptote at \( x = -2 \).
2. Horizontal Asymptote: As \( x \rightarrow \infty \), the function approaches \( 1 \), not \(-1\). This function does not meet the second criterion.
### Function D: \( f(x) = \frac{-4}{x+2} - 1 \)
1. Vertical Asymptote: The vertical asymptote occurs where the denominator is zero, which is at \( x = -2 \). This function has a vertical asymptote at \( x = -2 \).
2. Horizontal Asymptote: As \( x \rightarrow \infty \), the function approaches \(-1\), which meets the second criterion.
3. Root: Set the function to zero:
[tex]\[ \frac{-4}{x+2} - 1 = 0 \implies \frac{-4}{x+2} = 1 \implies -4 = x + 2 \implies x = -6 \][/tex]
Therefore, \( x = -6 \) but we need a root at \( x = 2 \). This function does not meet the third criterion.
After evaluating all options, we find that:
- Function A: \( f(x) = \frac{4}{x+2} - 1 \) meets all criteria.
Hence, the correct answer is:
[tex]\[ \boxed{\text{A}} \][/tex]
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