Let's analyze the given function:
[tex]\[ f(x) = \frac{25 - x^2}{x^2 - 4x - 5} \][/tex]
### Step-by-Step Analysis
1. Identify the degrees of the numerator and denominator:
[tex]\[ 25 - x^2 \][/tex]
This is a quadratic polynomial with the highest power of [tex]\(x\)[/tex] being 2 (degree 2).
[tex]\[ x^2 - 4x - 5 \][/tex]
This is also a quadratic polynomial with the highest power of [tex]\(x\)[/tex] being 2 (degree 2).
Conclusion: The degrees of the numerator (m) and the denominator (n) are both 2.
[tex]\[ m = n \][/tex]
2. Vertical asymptotes:
Vertical asymptotes occur where the denominator is zero (since division by zero is undefined), except where the numerator is also zero at the same point (if it cancels out).
We find the roots of the denominator:
[tex]\[ x^2 - 4x - 5 = 0 \][/tex]
Factoring the quadratic equation:
[tex]\[ (x - 5)(x + 1) = 0 \][/tex]
So, [tex]\(x = 5\)[/tex] and [tex]\(x = -1\)[/tex] are the points where the denominator is zero.
Next, we check if these points also make the numerator zero:
[tex]\[ 25 - x^2 = 0 \][/tex]
[tex]\[ x^2 = 25 \][/tex]
[tex]\[ x = \pm 5 \][/tex]
From this, [tex]\(x = 5\)[/tex] zeros out both the numerator and the denominator, which means [tex]\(x = 5\)[/tex] is a hole, not a vertical asymptote.
For [tex]\(x = -1\)[/tex]:
[tex]\[ 25 - (-1)^2 = 25 - 1 = 24 \neq 0 \][/tex]
At [tex]\(x = -1\)[/tex], the numerator is not zero.
Conclusion: The vertical asymptote is only at [tex]\(x = -1\)[/tex].
3. Horizontal asymptote:
Horizontal asymptotes are determined by the degrees of the numerator and the denominator.
- If [tex]\(m < n\)[/tex], the horizontal asymptote is [tex]\(y = 0\)[/tex].
- If [tex]\(m > n\)[/tex], there is no horizontal asymptote (oblique asymptote might exist).
- If [tex]\(m = n\)[/tex], the horizontal asymptote is [tex]\(y = \frac{a_m}{b_n}\)[/tex], where [tex]\(a_m\)[/tex] and [tex]\(b_n\)[/tex] are the leading coefficients of the numerator and denominator respectively.
Here, [tex]\(m = n = 2\)[/tex]:
The leading coefficients are both -1 (numerator) and 1 (denominator).
[tex]\[ y = \frac{-1}{1} = -1 \][/tex]
Conclusion: The horizontal asymptote is [tex]\(y = -1\)[/tex].
### Summary of Results
- [tex]\(m \neq n\)[/tex]: False ([tex]\(m = n\)[/tex]).
- [tex]\(m = n\)[/tex]: True.
- There is only one vertical asymptote: True ([tex]\(x = -1\)[/tex]).
- [tex]\(y = -1\)[/tex] is the horizontal asymptote: True.
So, the correct boxes to check are:
- [tex]\(m = n\)[/tex]
- There is only one vertical asymptote.
- [tex]\(y = -1\)[/tex] is the horizontal asymptote.
