3.6 Rational Functions

3.6.1 Learning Objectives

By the end of this section, you should be able to:

determine the domain of a rational function;
distinguish holes from vertical asymptotes;
find vertical, horizontal, and slant asymptotes;
graph rational functions using intercepts, asymptotes, and sign analysis; and
construct and interpret rational-function models.

3.6.2 Rational Functions and Their Domains

A rational function is a quotient of polynomials:

\[ r(x)=\frac{P(x)}{Q(x)}, \]

where \(Q(x)\) is not the zero polynomial. Its domain includes all real numbers except those that make the original denominator zero.

Factor before simplifying. A cancelled factor creates a removable discontinuity, or hole, but its input is still excluded from the domain.

Example 3.28 Finding a Domain and a Hole

Analyze

\[ r(x)=\frac{x^2-9}{x^2-x-6}. \]

Solution

Factor:

\[ r(x)=\frac{(x-3)(x+3)}{(x-3)(x+2)}. \]

The original denominator is zero at \(x=3\) and \(x=-2\), so both values are excluded. Cancelling gives

\[ r(x)=\frac{x+3}{x+2},\qquad x\ne3,-2. \]

The cancelled factor produces a hole at \(x=3\). Its \(y\)-coordinate comes from the simplified expression:

\[ \frac{3+3}{3+2}=\frac65. \]

\[ \boxed{\text{Domain }(-\infty,-2)\cup(-2,3)\cup(3,\infty);\quad \text{hole }\left(3,\frac65\right)} \]

Exercise 3.28 Find the domain and any hole of \(r(x)=\dfrac{x^2-4}{x^2+x-6}\).

3.6.3 Reciprocal Transformations

The parent reciprocal function is

\[ f(x)=\frac1x. \]

Its vertical asymptote is \(x=0\), its horizontal asymptote is \(y=0\), and its domain and range both exclude \(0\).

For

\[ g(x)=\frac{a}{x-h}+k, \]

the vertical asymptote is \(x=h\), the horizontal asymptote is \(y=k\), and the graph is transformed according to \(a\), \(h\), and \(k\).

Example 3.29 Graphing a Transformed Reciprocal

Describe the graph of

\[ g(x)=-\frac{2}{x-3}+1. \]

Solution

The graph of \(1/x\) is reflected across the \(x\)-axis, stretched vertically by factor \(2\), shifted right \(3\), and shifted up \(1\).

\[ \boxed{\text{Vertical asymptote }x=3,\quad \text{horizontal asymptote }y=1} \]

Exercise 3.29 State the asymptotes of \(g(x)=\dfrac{4}{x+2}-5\).

3.6.4 Vertical Asymptotes

A vertical line \(x=a\) is a vertical asymptote if function values grow without bound as \(x\) approaches \(a\) from at least one side.

To find vertical asymptotes:

factor the numerator and denominator;
cancel common factors;
set the remaining denominator equal to zero.

Zeros from cancelled denominator factors are holes, not vertical asymptotes.

Example 3.30 Distinguishing a Hole from an Asymptote

Find the vertical asymptotes and holes of

\[ r(x)=\frac{(x-1)(x+4)}{(x-1)(x-2)(x+3)}. \]

Solution

The factor \(x-1\) cancels, so \(x=1\) produces a hole. The remaining denominator is zero at \(x=2\) and \(x=-3\), so these are vertical asymptotes.

Using the simplified function

\[ \frac{x+4}{(x-2)(x+3)}, \]

the hole’s \(y\)-coordinate is

\[ \frac{5}{(-1)(4)}=-\frac54. \]

\[ \boxed{\text{Vertical asymptotes }x=2,-3;\quad \text{hole }\left(1,-\frac54\right)} \]

Exercise 3.30 Find the vertical asymptotes and holes of \(r(x)=\dfrac{(x+2)(x-5)}{(x+2)(x-1)}\).

3.6.5 Horizontal Asymptotes

For

\[ r(x)=\frac{P(x)}{Q(x)}, \]

compare the degrees of \(P\) and \(Q\):

if \(\deg P<\deg Q\), the horizontal asymptote is \(y=0\);
if \(\deg P=\deg Q\), the horizontal asymptote is the ratio of leading coefficients;
if \(\deg P>\deg Q\), there is no horizontal asymptote.

A graph may cross a horizontal asymptote. The asymptote describes end behaviour, not a forbidden line.

Example 3.31 Finding a Horizontal Asymptote

Find the horizontal asymptote of

\[ r(x)=\frac{6x^3-x+2}{2x^3+5x^2-4}. \]

Solution

The numerator and denominator have equal degree. The ratio of leading coefficients is \(6/2=3\).

\[ \boxed{y=3} \]

Exercise 3.31 Find the horizontal asymptote of \(r(x)=\dfrac{4x^2+1}{x^5-2}\).

3.6.6 Slant Asymptotes and End Behaviour

If the numerator’s degree is exactly one more than the denominator’s degree, polynomial division gives a linear quotient. The line represented by that quotient is a slant asymptote.

More generally, division writes

\[ r(x)=Q(x)+\frac{R(x)}{D(x)}. \]

When the remainder term approaches \(0\) for large \(|x|\), the graph approaches \(y=Q(x)\).

Example 3.32 Finding a Slant Asymptote

Find the slant asymptote of

\[ r(x)=\frac{x^2+3x+5}{x+1}. \]

Solution

Polynomial division gives

\[ \frac{x^2+3x+5}{x+1}=x+2+\frac3{x+1}. \]

As \(|x|\) becomes large, \(3/(x+1)\) approaches \(0\). Therefore,

\[ \boxed{y=x+2} \]

Exercise 3.32 Find the slant asymptote of \(r(x)=\dfrac{2x^2-x+4}{x-2}\).

3.6.7 Graphing Rational Functions

To sketch a rational function:

factor and simplify while preserving domain restrictions;
identify holes;
find vertical asymptotes;
find horizontal or slant asymptotes;
find \(x\)- and \(y\)-intercepts;
use test points to determine signs and branch locations; and
use Desmos to verify the completed sketch.

Example 3.33 Building a Rational Graph

Describe the key features of

\[ r(x)=\frac{x+1}{x-2}. \]

Solution

There are no common factors. The denominator is zero at \(x=2\), giving a vertical asymptote. Equal degrees give horizontal asymptote \(y=1\).

The numerator is zero at \(x=-1\), so the \(x\)-intercept is \((-1,0)\). Also,

\[ r(0)=\frac1{-2}=-\frac12, \]

so the \(y\)-intercept is \((0,-1/2)\).

\[ \boxed{x=2,\quad y=1,\quad x\text{-intercept }(-1,0),\quad y\text{-intercept }\left(0,-\frac12\right)} \]

Exercise 3.33 Find the asymptotes and intercepts of \(r(x)=\dfrac{2x-6}{x+1}\).

3.6.8 Rational Models

Rational functions often model rates, averages, and combined quantities. The formula’s excluded inputs and asymptotic behaviour must be interpreted in context.

For two resistors with resistances \(R_1\) and \(R_2\) ohms connected in parallel, the equivalent resistance \(R\) satisfies

\[ \frac1R=\frac1{R_1}+\frac1{R_2}. \]

Example 3.34 Equivalent Resistance

A variable resistor of \(x\) ohms is connected in parallel with a \(12\)-ohm resistor. Write the equivalent resistance as a function of \(x\), and find it when \(x=6\).

Solution

\[ \frac1R=\frac1x+\frac1{12}=\frac{x+12}{12x}. \]

Therefore,

\[ R(x)=\frac{12x}{x+12},\qquad x>0. \]

At \(x=6\),

\[ R(6)=\frac{72}{18}=4. \]

\[ \boxed{R(x)=\frac{12x}{x+12},\quad R(6)=4\ \Omega} \]

Exercise 3.34 A resistor of \(x\) ohms is connected in parallel with an \(8\)-ohm resistor. Write the equivalent-resistance function.

3.6.9 Conceptual Takeaways

Domain restrictions come from the original denominator.
Cancelled factors produce holes; uncancelled denominator zeros produce vertical asymptotes.
Horizontal and slant asymptotes describe end behaviour.
Intercepts, asymptotes, holes, and signs work together to determine the graph.
A rational model must be interpreted using its physical domain and units.

3.6.10 Skills You Should Be Able to Do

Determine the domain of a rational function.
Identify holes and vertical asymptotes.
Find horizontal and slant asymptotes.
Sketch rational functions and verify them with Desmos.
Construct and interpret rational-function models.

3.6.11 Practice Problems with Solutions

Find the domain of \(r(x)=\dfrac{x+4}{x^2-9}\).

Show Solution

The denominator factors as \((x-3)(x+3)\), so exclude \(x=\pm3\):

\[ \boxed{(-\infty,-3)\cup(-3,3)\cup(3,\infty)}. \]
Find any holes and vertical asymptotes of \(r(x)=\dfrac{x^2-1}{x^2-3x+2}\).

Show Solution

Factor:

\[ r(x)=\frac{(x-1)(x+1)}{(x-1)(x-2)}. \]

The cancelled factor gives a hole at \(x=1\). The simplified value there is \(2/(-1)=-2\), so the hole is \((1,-2)\). The remaining denominator gives vertical asymptote \(x=2\).

\[ \boxed{\text{Hole }(1,-2);\quad \text{vertical asymptote }x=2} \]
Find the horizontal asymptote of \(r(x)=\dfrac{5x^4-2}{2x^4+7x}\).

Show Solution

The degrees are equal, so use the ratio of leading coefficients:

\[ \boxed{y=\frac52}. \]
Find the slant asymptote of \(r(x)=\dfrac{x^2-4x+7}{x-3}\).

Show Solution

Division gives

\[ \frac{x^2-4x+7}{x-3}=x-1+\frac4{x-3}. \]

Therefore, \(\boxed{y=x-1}\).
Find all asymptotes and intercepts of \(r(x)=\dfrac{x-4}{x+2}\).

Show Solution

The denominator gives vertical asymptote \(x=-2\). Equal degrees give horizontal asymptote \(y=1\). The numerator gives \(x\)-intercept \((4,0)\), and

\[ r(0)=\frac{-4}{2}=-2, \]

so the \(y\)-intercept is \((0,-2)\).

\[ \boxed{x=-2,\quad y=1,\quad (4,0),\quad (0,-2)} \]
Determine where \(r(x)=\dfrac{x+1}{(x-2)(x+3)}\) is positive.

Show Solution

Critical values are \(-3\), \(-1\), and \(2\). Testing the intervals gives signs:

\[ (-\infty,-3):-,\quad (-3,-1):+,\quad (-1,2):-,\quad (2,\infty):+. \]

The zero at \(-1\) is not included in a positive interval, and the asymptotes are excluded. Thus,

\[ \boxed{(-3,-1)\cup(2,\infty)}. \]
Use Desmos to graph \(r(x)=\dfrac{x^2-4}{x^2-x-2}\). Identify holes, asymptotes, and intercepts.

Show Solution

Factor:

\[ r(x)=\frac{(x-2)(x+2)}{(x-2)(x+1)}. \]

There is a hole at \(x=2\), with \(y=(2+2)/(2+1)=4/3\). The remaining denominator gives vertical asymptote \(x=-1\), and equal original degrees give horizontal asymptote \(y=1\). The uncancelled numerator gives \(x\)-intercept \((-2,0)\), and \(r(0)=2\).

\[ \boxed{\text{Hole }\left(2,\frac43\right),\quad x=-1,\quad y=1,\quad (-2,0),\quad (0,2)} \]
Two resistors of \(10\ \Omega\) and \(x\ \Omega\) are connected in parallel. Find the value of \(x\) that gives an equivalent resistance of \(4\ \Omega\).

Show Solution

The model is

\[ R(x)=\frac{10x}{x+10}. \]

Set \(R(x)=4\):

\[ \frac{10x}{x+10}=4. \]

Then

\[ 10x=4x+40,\qquad 6x=40,\qquad x=\frac{20}{3}. \]

\[ \boxed{x=\frac{20}{3}\ \Omega\approx6.67\ \Omega} \]