Question

How many asymptotes does the function LaTeX: f\left(x\right)=\frac{x^2-x-6}{x^2-5x-6} have?

Answer 1

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).