Skill: Identify vertical and horizontal asymptotes

To identify asymptotes of a rational function:

1. Factor the numerator and denominator.
2. Find vertical asymptotes:
   • Set the remaining denominator equal to zero.
   • A factor that cancels creates a hole, not a vertical asymptote.
3. Find the horizontal asymptote by comparing the degrees and leading coefficients:
   • Same degree: divide the leading coefficients.
   • Numerator degree smaller: $y=0$.
   • Numerator degree larger: there is no horizontal asymptote.

Consider the function

$$f(x)=\frac{x^2-x-2}{x^2+x-6}.$$

Step 1: Factor

Factor the numerator and denominator:

$$f(x)=\frac{(x-2)(x+1)}{(x-2)(x+3)}.$$

The original denominator is zero when $x=2$ or $x=-3$, so these values are excluded from the domain.

Step 2: Identify a hole and a vertical asymptote

Cancel the common factor $(x-2)$:

$$f(x)=\frac{x+1}{x+3}, \qquad x\ne 2.$$

Because $(x-2)$ canceled, $x=2$ creates a removable hole, not a vertical asymptote.

The remaining denominator is $x+3$. Set it equal to zero:

$$x+3=0 \quad\Rightarrow\quad x=-3.$$

Therefore, the vertical asymptote is

$$\boxed{x=-3}.$$

As $x$ gets close to $-3$, the function values increase or decrease without bound.

Step 3: Identify the horizontal asymptote

Use the simplified function:

$$\frac{x+1}{x+3}.$$

The numerator and denominator have the same degree. Compare their leading coefficients:

$$\frac{1}{1}=1.$$

Therefore, the horizontal asymptote is

$$\boxed{y=1}.$$

So the function has:

• Vertical asymptote: $\boxed{x=-3}$
• Horizontal asymptote: $\boxed{y=1}$
• Hole: at $x=2$; its corresponding $y$-value is $\frac{2+1}{2+3}=\frac35$, so the hole is $\left(2,\frac35\right)$.