A rational function is a function that can be written as the quotient of two polynomials:
\[
R(x)=\frac{P(x)}{Q(x)},\qquad Q(x)\ne0.
\]
The denominator is especially important because every zero of \(Q(x)\) is excluded from the
domain. Some excluded values create vertical asymptotes, while others create removable
discontinuities, also called holes.
1. Domain restrictions
The domain of a rational function includes all real numbers except values that make the
denominator zero.
\[
Q(x)=0 \quad\Rightarrow\quad x\text{ is excluded from the domain.}
\]
For example, if
\[
R(x)=\frac{x^2-4}{x^2-3x+2},
\]
then the denominator factors as
\[
x^2-3x+2=(x-1)(x-2).
\]
Therefore, \(x=1\) and \(x=2\) are excluded from the domain.
2. Factoring and cancellation
Factoring the numerator and denominator is the key step. In the example above,
\[
x^2-4=(x-2)(x+2).
\]
So the rational function becomes
\[
\frac{x^2-4}{x^2-3x+2}
=
\frac{(x-2)(x+2)}{(x-1)(x-2)}.
\]
The factor \(x-2\) appears in both numerator and denominator, so it cancels algebraically.
However, the original function is still undefined at \(x=2\). This creates a hole.
\[
R(x)=\frac{x+2}{x-1},\qquad x\ne1,\ x\ne2.
\]
3. Holes
A hole occurs when a denominator factor cancels completely. To find the coordinates of the
hole, substitute the excluded \(x\)-value into the simplified function.
In the example, the canceled value is \(x=2\). The simplified expression is
\[
\frac{x+2}{x-1}.
\]
Now substitute \(x=2\):
\[
y=\frac{2+2}{2-1}=4.
\]
Therefore, the hole is at
\[
(2,4).
\]
4. Vertical asymptotes
A vertical asymptote comes from a denominator factor that does not cancel. In the example,
after cancellation, the denominator still contains \(x-1\). Therefore,
\[
x-1=0\quad\Rightarrow\quad x=1.
\]
So the vertical asymptote is
\[
x=1.
\]
5. Horizontal asymptotes
Horizontal asymptotes describe end behavior. For
\[
R(x)=\frac{P(x)}{Q(x)},
\]
compare the degree of the numerator with the degree of the denominator.
6. Oblique asymptotes
If the numerator degree is exactly one greater than the denominator degree, the function has an
oblique, or slant, asymptote. To find it, divide the numerator by the denominator.
For example,
\[
R(x)=\frac{x^2+1}{x-1}.
\]
Polynomial long division gives
\[
\frac{x^2+1}{x-1}=x+1+\frac{2}{x-1}.
\]
Since the remainder term \(\frac{2}{x-1}\) approaches \(0\) as \(x\) becomes very large in
magnitude, the oblique asymptote is
\[
y=x+1.
\]
7. Intercepts
\(x\)-intercepts occur where the rational function equals zero. After cancellation, solve the
simplified numerator equal to zero, but avoid excluded values.
\[
R(x)=0 \quad\Rightarrow\quad \text{simplified numerator}=0.
\]
The \(y\)-intercept is found by evaluating the original function at \(x=0\), if \(x=0\) is in
the domain:
\[
y=R(0).
\]
8. Complete example
Analyze
\[
R(x)=\frac{x^2-4}{x^2-3x+2}.
\]
Factor:
\[
R(x)=\frac{(x-2)(x+2)}{(x-1)(x-2)}.
\]
Cancel the common factor:
\[
R(x)=\frac{x+2}{x-1},\qquad x\ne1,\ x\ne2.
\]
The canceled value \(x=2\) gives a hole:
\[
\left(2,\frac{2+2}{2-1}\right)=(2,4).
\]
The uncanceled denominator zero gives a vertical asymptote:
\[
x=1.
\]
The simplified numerator gives the \(x\)-intercept:
\[
x+2=0\Rightarrow x=-2.
\]
The \(y\)-intercept is
\[
R(0)=\frac{-4}{2}=-2.
\]
Since the simplified numerator and denominator have the same degree, the horizontal asymptote is
the ratio of leading coefficients:
\[
y=1.
\]
9. Final checklist
- Write the rational function as \(\frac{P(x)}{Q(x)}\).
- Factor both \(P(x)\) and \(Q(x)\).
- Set \(Q(x)=0\) to find excluded values.
- Cancel common factors.
- Canceled denominator zeros become holes if the simplified denominator is nonzero there.
- Uncanceled denominator zeros become vertical asymptotes.
- Use degree comparison or long division to find horizontal, oblique, or polynomial asymptotes.
- Find intercepts from the simplified function while respecting the original domain.
