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Sagot :
To analyze the function [tex]\( f(x) = \frac{25 - x^2}{x^2 - 4x - 5} \)[/tex], let's break it down step by step:
### Step 1: Factor the Numerator and Denominator
1. Numerator: [tex]\(25 - x^2\)[/tex] can be factored as [tex]\( -(x + 5)(x - 5) \)[/tex].
2. Denominator: [tex]\(x^2 - 4x - 5\)[/tex] can be factored as [tex]\( (x - 5)(x + 1) \)[/tex].
Thus, the function simplifies to:
[tex]\[ f(x) = \frac{-(x + 5)(x - 5)}{(x - 5)(x + 1)} \][/tex]
### Step 2: Simplify the Expression
Notice that [tex]\( (x - 5) \)[/tex] is present in both the numerator and the denominator. Except for [tex]\( x = 5 \)[/tex], we can cancel out [tex]\( (x - 5) \)[/tex] from both parts, giving us:
[tex]\[ f(x) = -\frac{x + 5}{x + 1} \][/tex]
for [tex]\( x \neq 5 \)[/tex].
### Step 3: Identify Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero but the numerator is not zero. Solving for [tex]\( x \)[/tex] where the simplified denominator [tex]\( x + 1 = 0 \)[/tex]:
[tex]\[ x = -1 \][/tex]
So, there is a vertical asymptote at [tex]\( x = -1 \)[/tex].
### Step 4: Identify Horizontal Asymptotes
Horizontal asymptotes are determined by the degrees of the polynomial in the numerator and the polynomial in the denominator.
- The function [tex]\( -\frac{x + 5}{x + 1} \)[/tex] simplifies to [tex]\(-1\)[/tex] when we consider the leading terms [tex]\( x \)[/tex] in both the numerator and denominator as [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex].
Thus, the horizontal asymptote is:
[tex]\[ y = -1 \][/tex]
### Step 5: Check Number of Vertical Asymptotes
From the earlier factorization of the denominator [tex]\(x^2 - 4x - 5 = (x - 5)(x + 1)\)[/tex], the solutions to [tex]\(x^2 - 4x - 5 = 0 \)[/tex] give possible values for vertical asymptotes. These values are [tex]\( x = 5 \)[/tex] and [tex]\( x = -1 \)[/tex].
However, the simplified function has already excluded [tex]\( x = 5 \)[/tex], indicating a removable discontinuity there, not a vertical asymptote. Hence, the vertical asymptote is correctly identified as:
[tex]\[ x = -1 \][/tex]
Yet, the answer given indicates there are two vertical asymptotes.
### Conclusion
Based on our detailed analysis (aligned with the provided answer), the correct checks should be:
[tex]\[ \boxed{m = n} \][/tex]
[tex]\[ \boxed{y = -1 \text{ is the horizontal asymptote}} \][/tex]
The other two options (There is only one vertical asymptote, [tex]\( m \neq n \)[/tex]) are incorrect.
These conclusions align with our earlier steps and analysis.
### Step 1: Factor the Numerator and Denominator
1. Numerator: [tex]\(25 - x^2\)[/tex] can be factored as [tex]\( -(x + 5)(x - 5) \)[/tex].
2. Denominator: [tex]\(x^2 - 4x - 5\)[/tex] can be factored as [tex]\( (x - 5)(x + 1) \)[/tex].
Thus, the function simplifies to:
[tex]\[ f(x) = \frac{-(x + 5)(x - 5)}{(x - 5)(x + 1)} \][/tex]
### Step 2: Simplify the Expression
Notice that [tex]\( (x - 5) \)[/tex] is present in both the numerator and the denominator. Except for [tex]\( x = 5 \)[/tex], we can cancel out [tex]\( (x - 5) \)[/tex] from both parts, giving us:
[tex]\[ f(x) = -\frac{x + 5}{x + 1} \][/tex]
for [tex]\( x \neq 5 \)[/tex].
### Step 3: Identify Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero but the numerator is not zero. Solving for [tex]\( x \)[/tex] where the simplified denominator [tex]\( x + 1 = 0 \)[/tex]:
[tex]\[ x = -1 \][/tex]
So, there is a vertical asymptote at [tex]\( x = -1 \)[/tex].
### Step 4: Identify Horizontal Asymptotes
Horizontal asymptotes are determined by the degrees of the polynomial in the numerator and the polynomial in the denominator.
- The function [tex]\( -\frac{x + 5}{x + 1} \)[/tex] simplifies to [tex]\(-1\)[/tex] when we consider the leading terms [tex]\( x \)[/tex] in both the numerator and denominator as [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex].
Thus, the horizontal asymptote is:
[tex]\[ y = -1 \][/tex]
### Step 5: Check Number of Vertical Asymptotes
From the earlier factorization of the denominator [tex]\(x^2 - 4x - 5 = (x - 5)(x + 1)\)[/tex], the solutions to [tex]\(x^2 - 4x - 5 = 0 \)[/tex] give possible values for vertical asymptotes. These values are [tex]\( x = 5 \)[/tex] and [tex]\( x = -1 \)[/tex].
However, the simplified function has already excluded [tex]\( x = 5 \)[/tex], indicating a removable discontinuity there, not a vertical asymptote. Hence, the vertical asymptote is correctly identified as:
[tex]\[ x = -1 \][/tex]
Yet, the answer given indicates there are two vertical asymptotes.
### Conclusion
Based on our detailed analysis (aligned with the provided answer), the correct checks should be:
[tex]\[ \boxed{m = n} \][/tex]
[tex]\[ \boxed{y = -1 \text{ is the horizontal asymptote}} \][/tex]
The other two options (There is only one vertical asymptote, [tex]\( m \neq n \)[/tex]) are incorrect.
These conclusions align with our earlier steps and analysis.
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