To determine the number of asymptotes for the function
\[ f(x) = \frac{x^2 - x - 6}{x^2 - 5x - 6}, \]
we need to investigate both vertical and horizontal asymptotes.
Step 1: Find Vertical Asymptotes
Vertical asymptotes occur where the function is undefined, which is where the denominator is equal to zero. We need to factor the denominator:
The denominator is
\[ x^2 - 5x - 6. \]
To factor it, we find two numbers that multiply to \(-6\) (the constant term) and add up to \(-5\) (the coefficient of \(x\)). These numbers are \(-6\) and \(1\). Thus, the factorization is:
\[ x^2 - 5x - 6 = (x - 6)(x + 1). \]
Setting the denominator to zero, we solve:
\[ (x - 6)(x + 1) = 0. \]
This gives us the vertical asymptotes:
\[ x - 6 = 0 \quad \Rightarrow \quad x = 6, \] \[ x + 1 = 0 \quad \Rightarrow \quad x = -1. \]
Step 2: Find Horizontal Asymptotes
To find horizontal asymptotes, we analyze the degrees of the polynomial in the numerator and the denominator.
The degrees of both the numerator and the denominator are 2 (the leading term of both is \(x^2\)). When the degrees are equal, the horizontal asymptote can be found by taking the ratio of the leading coefficients.
The leading coefficient of the numerator \(x^2 - x - 6\) is \(1\) and the leading coefficient of the denominator \(x^2 - 5x - 6\) is also \(1\). Thus, the horizontal asymptote is:
\[ y = \frac{1}{1} = 1. \]
Summary
We have found:
- Vertical asymptotes at \(x = 6\) and \(x = -1\) (2 vertical asymptotes).
- 1 horizontal asymptote at \(y = 1\).
Conclusion
Therefore, the function \(f(x)\) has a total of
\[ \boxed{3} \]
asymptotes (2 vertical and 1 horizontal).