Question

How many asymptotes does the function f(x)= x^2-x-6/x^2-5x-6 have

Answer 1

To find the asymptotes of the function \( f(x) = \frac{x^2 - x - 6}{x^2 - 5x - 6} \), we need to identify vertical and horizontal asymptotes.

### Step 1: Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (as long as the numerator is not zero at those points). We first need to factor the denominator:

1. **Factor the denominator**:
\[
x^2 - 5x - 6 = (x - 6)(x + 1)
\]

The denominator is zero when:
\[
(x - 6)(x + 1) = 0 \implies x = 6 \quad \text{or} \quad x = -1
\]

2. **Check the numerator at these points**:
Now we check if the numerator is zero at these points.
\[
\text{Numerator: } x^2 - x - 6 = (x - 3)(x + 2)
\]

- At \( x = 6 \):
\[
6^2 - 6 - 6 = 36 - 6 - 6 = 24 \quad (\text{not zero})
\]
- At \( x = -1 \):
\[
(-1)^2 - (-1) - 6 = 1 + 1 - 6 = -4 \quad (\text{not zero})
\]

Since the numerator is not zero at both points, we have vertical asymptotes at \( x = 6 \) and \( x = -1 \).

### Step 2: Horizontal Asymptotes

Horizontal asymptotes depend on the degrees of the polynomial in the numerator and denominator.

1. **Degree comparison**:
The degree of the numerator \( (x^2 - x - 6) \) is 2, and the degree of the denominator \( (x^2 - 5x - 6) \) is also 2.

2. **Calculate the horizontal asymptote**:
When the degrees are equal, the horizontal asymptote is given by the ratio of the leading coefficients. The leading coefficients of the numerator and denominator are both 1:
\[
y = \frac{1}{1} = 1
\]

### Conclusion

The function \( f(x) \) has:

- **Vertical asymptotes** at \( x = 6 \) and \( x = -1 \) (2 vertical asymptotes).
- **One horizontal asymptote** at \( y = 1 \).

In total, the function has **3 asymptotes** (2 vertical and 1 horizontal).