Determine restrictions of rational functions

For real-valued rational functions, the domain consists of all real inputs except those that make the original denominator equal to zero; restrictions are identified by solving the denominator equation and expressed symbolically or in interval notation. Factoring and cancellation reveal whether an excluded value creates a removable discontinuity (a hole) or remains associated with a vertical asymptote, and a canceled factor does not restore the original input. The focus is on real polynomial quotients, not complex-domain or more advanced abstract treatments.

Detailed Explanation: 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.

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