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Sagot :
To determine the values of [tex]\( a \)[/tex] and [tex]\( m \)[/tex] for which the function [tex]\( f(x) \)[/tex] has a horizontal asymptote at [tex]\( y=2 \)[/tex] and a vertical asymptote at [tex]\( x=1 \)[/tex], let's analyze each option in detail.
We start by considering the general form of a rational function:
[tex]\[ f(x) = \frac{ax + b}{x - m} \][/tex]
### Conditions to Satisfy:
1. Horizontal Asymptote at [tex]\( y=2 \)[/tex]:
For a rational function, the horizontal asymptote is determined by the ratio of the leading coefficients of the numerator and the denominator as [tex]\( x \)[/tex] approaches infinity. Thus, we need:
[tex]\[ \lim_{x \to \infty} \frac{ax + b}{x - m} = 2 \][/tex]
This simplifies to:
[tex]\[ \frac{a}{1} = 2 \Rightarrow a = 2 \][/tex]
2. Vertical Asymptote at [tex]\( x=1 \)[/tex]:
A vertical asymptote occurs where the denominator equals zero but the numerator does not. Thus, we need:
[tex]\[ x - m = 0 \Rightarrow x = m \Rightarrow m = 1 \][/tex]
### Analyzing the Given Options:
Now let's consider the given options to find the correct combination:
- Option 1: [tex]\( a = -1, m = 0 \)[/tex]
[tex]\[ f(x) = \frac{-1x + b}{x - 0} = \frac{-x + b}{x} \][/tex]
This function does not have a vertical asymptote at [tex]\( x=1 \)[/tex].
- Option 2: [tex]\( a = 1, m = 0 \)[/tex]
[tex]\[ f(x) = \frac{1x + b}{x - 0} = \frac{x + b}{x} \][/tex]
This function also does not have a vertical asymptote at [tex]\( x=1 \)[/tex].
- Option 3: [tex]\( a = -1, m = 1 \)[/tex]
[tex]\[ f(x) = \frac{-1x + b}{x - 1} = \frac{-x + b}{x - 1} \][/tex]
This function has a vertical asymptote at [tex]\( x=1 \)[/tex], but the horizontal asymptote would be:
[tex]\[ \lim_{x \to \infty} \frac{-x + b}{x - 1} = -1 \quad \text{(leading coefficient ratio)} \][/tex]
So, the horizontal asymptote is [tex]\( y=-1 \)[/tex], not [tex]\( y=2 \)[/tex].
- Option 4: [tex]\( a = 1, m = 1 \)[/tex]
[tex]\[ f(x) = \frac{1x + b}{x - 1} = \frac{x + b}{x - 1} \][/tex]
This function has a vertical asymptote at [tex]\( x=1 \)[/tex], but the horizontal asymptote would be:
[tex]\[ \lim_{x \to \infty} \frac{x + b}{x - 1} = 1 \quad \text{(leading coefficient ratio)} \][/tex]
So, the horizontal asymptote is [tex]\( y=1 \)[/tex], not [tex]\( y=2 \)[/tex].
After carefully reviewing each option, none of the provided options of [tex]\( a \)[/tex] and [tex]\( m \)[/tex] produce a function that has both the required horizontal asymptote at [tex]\( y=2 \)[/tex] and the vertical asymptote at [tex]\( x=1 \)[/tex]. Hence, we find that none of the given options are correct.
We start by considering the general form of a rational function:
[tex]\[ f(x) = \frac{ax + b}{x - m} \][/tex]
### Conditions to Satisfy:
1. Horizontal Asymptote at [tex]\( y=2 \)[/tex]:
For a rational function, the horizontal asymptote is determined by the ratio of the leading coefficients of the numerator and the denominator as [tex]\( x \)[/tex] approaches infinity. Thus, we need:
[tex]\[ \lim_{x \to \infty} \frac{ax + b}{x - m} = 2 \][/tex]
This simplifies to:
[tex]\[ \frac{a}{1} = 2 \Rightarrow a = 2 \][/tex]
2. Vertical Asymptote at [tex]\( x=1 \)[/tex]:
A vertical asymptote occurs where the denominator equals zero but the numerator does not. Thus, we need:
[tex]\[ x - m = 0 \Rightarrow x = m \Rightarrow m = 1 \][/tex]
### Analyzing the Given Options:
Now let's consider the given options to find the correct combination:
- Option 1: [tex]\( a = -1, m = 0 \)[/tex]
[tex]\[ f(x) = \frac{-1x + b}{x - 0} = \frac{-x + b}{x} \][/tex]
This function does not have a vertical asymptote at [tex]\( x=1 \)[/tex].
- Option 2: [tex]\( a = 1, m = 0 \)[/tex]
[tex]\[ f(x) = \frac{1x + b}{x - 0} = \frac{x + b}{x} \][/tex]
This function also does not have a vertical asymptote at [tex]\( x=1 \)[/tex].
- Option 3: [tex]\( a = -1, m = 1 \)[/tex]
[tex]\[ f(x) = \frac{-1x + b}{x - 1} = \frac{-x + b}{x - 1} \][/tex]
This function has a vertical asymptote at [tex]\( x=1 \)[/tex], but the horizontal asymptote would be:
[tex]\[ \lim_{x \to \infty} \frac{-x + b}{x - 1} = -1 \quad \text{(leading coefficient ratio)} \][/tex]
So, the horizontal asymptote is [tex]\( y=-1 \)[/tex], not [tex]\( y=2 \)[/tex].
- Option 4: [tex]\( a = 1, m = 1 \)[/tex]
[tex]\[ f(x) = \frac{1x + b}{x - 1} = \frac{x + b}{x - 1} \][/tex]
This function has a vertical asymptote at [tex]\( x=1 \)[/tex], but the horizontal asymptote would be:
[tex]\[ \lim_{x \to \infty} \frac{x + b}{x - 1} = 1 \quad \text{(leading coefficient ratio)} \][/tex]
So, the horizontal asymptote is [tex]\( y=1 \)[/tex], not [tex]\( y=2 \)[/tex].
After carefully reviewing each option, none of the provided options of [tex]\( a \)[/tex] and [tex]\( m \)[/tex] produce a function that has both the required horizontal asymptote at [tex]\( y=2 \)[/tex] and the vertical asymptote at [tex]\( x=1 \)[/tex]. Hence, we find that none of the given options are correct.
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