Question

consider the following function:
$f(x) = \frac{25 - x^2}{x^2 - 4x - 5}$
which of the following are correct? choose three correct answers.

• $y = -1$ is the horizontal asymptote.
• there is only one vertical asymptote.
• $m = n$
• $m \

eq n$

Explanation:

Step1: Analyze Horizontal Asymptote

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), if the degrees of \( N(x) \) and \( D(x) \) are equal (let the degree be \( n \) for numerator and \( m \) for denominator, here \( m = n = 2 \)), the horizontal asymptote is \( y=\frac{\text{leading coefficient of } N(x)}{\text{leading coefficient of } D(x)} \). The numerator \( 25 - x^2=-x^2 + 25 \) has leading coefficient - 1, denominator \( x^2-4x - 5 \) has leading coefficient 1. So \( y=\frac{-1}{1}=-1 \), so " \( y = - 1 \) is the horizontal asymptote" is correct.

Step2: Analyze Vertical Asymptotes

First, factor numerator and denominator. Numerator: \( 25 - x^2=(5 - x)(5 + x)=-(x - 5)(x + 5) \). Denominator: \( x^2-4x - 5=(x - 5)(x+1) \). Then, cancel common factor \( (x - 5) \) (note \( x
eq5 \) to avoid division by zero). The simplified function is \( f(x)=\frac{-(x + 5)}{x + 1},x
eq5 \). The vertical asymptote occurs where \( D(x)=0 \) and not canceled, so \( x+1 = 0\Rightarrow x=-1 \). The value \( x = 5 \) is a hole (removable discontinuity), not a vertical asymptote. So there is only one vertical asymptote (\( x=-1 \)), so "There is only one vertical asymptote" is correct.

Step3: Analyze Degrees of Numerator and Denominator

The numerator \( 25 - x^2 \) is a quadratic (degree \( n = 2 \)), denominator \( x^2-4x - 5 \) is also a quadratic (degree \( m = 2 \)). So \( m=n \), so " \( m = n \)" is correct, and " \( m
eq n \)" is incorrect.

Answer:

1. \( y = - 1 \) is the horizontal asymptote.
2. There is only one vertical asymptote.
3. \( m = n \)