Skill: Determine restrictions of rational functions

For a rational function, the domain includes every real number except values that make the original denominator equal to zero.

Consider

$$f(x)=\frac{x^2-4}{x^2-x-6}.$$

1. Set the original denominator equal to zero

The denominator is $x^2-x-6$, so solve

$$x^2-x-6=0.$$

Factor:

$$(x-3)(x+2)=0.$$

Therefore,

$$x=3 \quad \text{or} \quad x=-2.$$

These values are restrictions because they make the original denominator zero.

2. State the domain

The domain is all real numbers except $-2$ and $3$:

$$\boxed{x\ne -2,\; x\ne 3}$$

In interval notation, this is

$$\boxed{(-\infty,-2)\cup(-2,3)\cup(3,\infty)}.$$

3. Factor the numerator to identify what happens at each restriction

Factor the numerator:

$$x^2-4=(x-2)(x+2).$$

So the function becomes

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

The factor $x+2$ cancels:

$$f(x)=\frac{x-2}{x-3},$$

but the original function is still undefined at $x=-2$. Cancellation does not restore an excluded value.

• $x=-2$ corresponds to a removable discontinuity, or hole, because the factor $(x+2)$ canceled.
• $x=3$ corresponds to a vertical asymptote because the factor $(x-3)$ did not cancel.

Thus, the restrictions are found from the original denominator, and factoring tells you whether each restriction creates a hole or a vertical asymptote.