Skill: Identify holes and asymptotes of rational functions

To identify holes and asymptotes, first factor the numerator and denominator. Then look for common factors.

Consider

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

1. Factor completely

Factor the numerator and denominator:

$$x^2-1=(x-1)(x+1)$$

and

$$x^2-3x+2=(x-1)(x-2).$$

So,

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

The original denominator is zero when $x=1$ or $x=2$, so both values are excluded from the domain.

2. Find the hole

The common factor $(x-1)$ cancels:

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

Because $(x-1)$ was canceled, $x=1$ produces a hole.

To find the hole’s $y$-coordinate, substitute $x=1$ into the reduced function:

$$y=\frac{1+1}{1-2}=\frac{2}{-1}=-2.$$

Therefore, the hole is

$$\boxed{(1,-2)}.$$

3. Find the vertical asymptote

The denominator that remains after cancellation is $x-2$. Set it equal to zero:

$$x-2=0$$

so the vertical asymptote is

$$\boxed{x=2}.$$

The canceled factor $(x-1)$ does not give a vertical asymptote; it gives the hole.

4. Find the horizontal asymptote

In the original function, the numerator and denominator both have degree $2$. When the degrees are equal, divide the leading coefficients:

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

Thus, the horizontal asymptote is

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

This also describes the end behavior: as $x$ becomes very large positive or very large negative, the graph approaches $y=1$.

For this function:

• Hole: $\boxed{(1,-2)}$
• Vertical asymptote: $\boxed{x=2}$
• Horizontal asymptote: $\boxed{y=1}$