SL 2.8 Analyse reciprocal and linear-over-linear rational functions | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Functions

IB Mathematics: Analysis and Approaches SL — study topic SL 2.8

A rational function is a quotient of polynomials. This lesson focuses on reciprocal functions and functions with a linear numerator and a linear denominator. The denominator is a useful starting point: any input that makes it zero is excluded from the domain. Depending on the function, the graph may have a vertical asymptote at that input, or a missing point if a common factor cancels. Algebra identifies these features; a table and graphing technology help you see and check the graph. The function’s domain may also be restricted by its context.

What you will learn

Identify the domain, intercepts, and asymptotes of reciprocal and linear-over-linear functions.
Rewrite a linear-over-linear function to reveal its graph features.
Use algebra, a value table, and a graphing calculator to analyse and check a function.
Explain how domain restrictions and asymptotes affect a graph and a simple context.

1. Start with the denominator

A reciprocal function can be written as

f(x)=\frac{a}{x-h}+k

, where

a

,

h

, and

k

are constants and

a\ne 0

. Its denominator is zero at

x=h

, so that input is excluded. The line

x=h

is a vertical asymptote: as the input approaches

h

, the function’s values increase or decrease without bound.

As

x

becomes very large positive or negative, the fraction approaches zero. Therefore, the graph approaches the horizontal line

y=k

. The coefficient

a

affects the branches’ direction and steepness. If

a>0

, the branches lie above-right and below-left of the point

(h,k)

; if

a<0

, they lie above-left and below-right.

A linear-over-linear function has the form

f(x)=\frac{mx+n}{px+q}

, with

p\ne 0

. Solve

px+q=0

to find the input excluded by the original denominator. If the numerator is also zero at that input, check for a common factor. Cancelling it may show that the graph is a line with one missing point, rather than a graph with a vertical asymptote. The original domain restriction still applies.

f(x)=\frac{a}{x-h}+k

Exclude every input that makes the original denominator zero.
For $f(x)=\frac{a}{x-h}+k$ , the asymptotes are $x=h$ and $y=k$ .
A cancelled factor does not restore an input to the original function’s domain.

2. Rewrite and sketch

To analyse a linear-over-linear function, rewrite its numerator as a multiple of its denominator plus a remainder. This puts the function into reciprocal form and reveals its asymptotes. For example, if the rewrite gives

f(x)=2+\frac{5}{x-3}

, the vertical asymptote is

x=3

and the horizontal asymptote is

y=2

.

For the

x

-intercept, set the numerator equal to zero and check that the resulting input is in the domain. For the

y

-intercept, substitute

x=0

, provided zero is in the domain. These intercepts, together with the asymptotes, help you sketch the graph. If a factor cancels, the point omitted from the original graph must still be shown as a missing point.

A value table helps show how the function behaves on both sides of a vertical asymptote. Use inputs on each side; values from only one side can give an incomplete picture. For a real-world model, distinguish the algebraic domain from realistic inputs. For example, if

t(x)=\frac{a}{x}

describes time per worker for a fixed task,

x

may need to represent a positive whole number of workers, even though the algebraic function is defined for other inputs too.

\frac{mx+n}{px+q}=\frac{m}{p}+\frac{np-mq}{p(px+q)}

Rewrite a linear-over-linear function to expose its reciprocal structure.
Find intercepts algebraically, checking that each input belongs to the domain.
Use values on both sides of a vertical asymptote when building a table or sketch.

3. Connect the algebra, table, and graph

A graphing calculator can check the overall shape, help estimate intercepts, and show whether a viewing window hides part of a branch. Enter the numerator and denominator with clear grouping. Choose a window that shows inputs on both sides of an excluded value and use the algebraic asymptotes to guide your scale.

Technology supports, but does not replace, the reasoning. Confirm restrictions by solving the original denominator equation. For a linear-over-linear function, confirm asymptotes by rewriting the expression. A graph that seems to cross a vertical asymptote may be displaying a scale or window issue; the excluded input is not part of the graph.

For an exam-style analysis, give the domain restriction, show the rewrite when it is useful, state the asymptotes, and calculate intercepts. Use a sketch or calculator graph to connect these results. In a context, state any units and realistic restrictions on the input.

Use the calculator to check a reasoned sketch, not to establish features by itself.
Choose a window that displays both sides of an excluded input.
State contextual restrictions and units when the function models a situation.

Worked example

Reciprocal form and graph features

Analyse

f(x)=\frac{6}{x+2}-1

. State its domain, asymptotes, intercepts, and the location of its branches.

Find the restriction
The denominator is zero when $x=-2$ . The function is undefined at that input, so exclude it from the domain.
$x\ne -2$
Read the asymptotes
The denominator shift gives the vertical asymptote. The constant outside the fraction gives the horizontal asymptote.
$x=-2,\qquad y=-1$
Calculate the intercepts
Set the function equal to zero to find the $x$ -intercept. Substitute zero for $x$ to find the $y$ -intercept.
$x=4,\qquad f(0)=2$
Describe the branches
The reciprocal coefficient is positive, so the branches lie above-right and below-left relative to the centre $(-2,-1)$ . The intercepts provide points for the sketch.
$(-2,-1)$

Answer: The domain is all real numbers except

-2

. The asymptotes are

x=-2

and

y=-1

. The intercepts are

(4,0)

and

(0,2)

.

Check: Substitution gives

f(4)=6/6-1=0

and

f(0)=6/2-1=2

.

Worked example

Rewrite a linear-over-linear function

Analyse

g(x)=\frac{3x+7}{x-2}

. Find its domain, asymptotes, and intercepts.

Rewrite the numerator
Write the numerator as three times the denominator plus a remainder. This reveals the reciprocal part of the function.
$g(x)=3+\frac{13}{x-2}$
State the domain and asymptotes
The original denominator is zero at $x=2$ , so exclude that input. The rewritten form shows the vertical and horizontal asymptotes.
$x\ne 2,\qquad x=2,\qquad y=3$
Find the intercepts
Set the numerator equal to zero for the $x$ -intercept. Substitute $x=0$ in the original function for the $y$ -intercept.
$x=-\frac{7}{3},\qquad g(0)=-\frac{7}{2}$

Answer: The domain excludes

2

. The asymptotes are

x=2

and

y=3

. The intercepts are

(-7/3,0)

and

(0,-7/2)

.

Check: The rewrite gives

g(0)=3+13/(-2)=-7/2

, which matches direct substitution.

Worked example

A cancellation and a missing point

Analyse

h(x)=\frac{2x-2}{x-1}

. State its domain and describe its graph.

Factor and simplify
The numerator is twice the denominator. Cancelling the common factor simplifies the rule, but the original denominator still excludes $x=1$ .
$h(x)=2,\qquad x\ne 1$
Identify the missing point
The simplified rule has the constant value $2$ , but the original function is undefined at $x=1$ . Its graph is therefore the horizontal line with the point $(1,2)$ omitted.
$(1,2)$

Answer: The domain is all real numbers except

1

. The graph is the line

y=2

with a missing point at

(1,2)

; there is no vertical asymptote.

Check: The original denominator is zero at

1

, while the simplified rule would give

2

there. This confirms that the point is missing.

Common mistakes and how to avoid them

Including an input that makes the denominator zero in the domain.

Correction: Solve the original denominator equation and exclude its solution, even if a factor later cancels.

Calling a cancelled factor’s excluded input a vertical asymptote.

Correction: If cancellation leaves a finite value, the graph has a missing point at that input, not a vertical asymptote.

Assuming a graph can never cross a horizontal asymptote.

Correction: An asymptote describes behaviour as the input becomes very large or approaches an excluded value; check the function before making claims about crossings.

Using a calculator graph as the only evidence for asymptotes.

Correction: Use the denominator and a reciprocal-form rewrite to establish the features analytically, then use the graph as a check.

Lesson summary

The original denominator determines which inputs are excluded.
Reciprocal form reveals the vertical and horizontal asymptotes.
Rewriting a linear-over-linear function makes its reciprocal structure visible.
Intercepts, selected table values, and asymptotes support a clear sketch.
When a common factor cancels, retain the original restriction and show the missing point.

Check your understanding

Question 1

For

r(x)=\frac{4}{x-5}+2

, which pair gives the vertical and horizontal asymptotes?

$x=5$ and $y=2$
$x=-5$ and $y=2$
$x=5$ and $y=-2$
$x=2$ and $y=5$

Show answer and explanation

x=5

and

y=2

The denominator is zero at

x=5

, and the outside constant gives the horizontal asymptote

y=2

.

Question 2

What is the domain of

s(x)=\frac{2x+1}{x+4}

?

All real numbers except $-4$
All real numbers except $4$
All real numbers
All real numbers except $-1/2$

Show answer and explanation

All real numbers except

-4

The denominator is zero when

x=-4

, so that input is excluded.

Question 3

The function

u(x)=\frac{2x-2}{x-1}

is defined by its original form only when

x\ne 1

. What happens at

x=1

on its graph?

There is a vertical asymptote at $x=1$ .
There is a missing point at $(1,2)$ .
The graph crosses the $x$ -axis at $(1,0)$ .
The graph has a horizontal asymptote $y=1$ .

Show answer and explanation

There is a missing point at

(1,2)

.

Cancelling the common factor gives the constant rule

2

for

x\ne 1

. The original function omits the point

(1,2)

.

Key terms

Rational function: A function expressed as one polynomial divided by another.
Domain: The set of input values for which a function is defined.
Vertical asymptote: A vertical line that the graph approaches as the input approaches an excluded value.
Horizontal asymptote: A horizontal line that the graph approaches as the input becomes very large positive or negative.
Missing point: A point omitted from the graph because the original expression is undefined there, even though a simplified rule has a finite value.

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SL 2.8 · Analyse reciprocal and linear-over-linear rational functions

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

IB Mathematics: Analysis and Approaches SL — study topic SL 2.8

What you will learn

1. Start with the denominator

2. Rewrite and sketch

3. Connect the algebra, table, and graph

Reciprocal form and graph features

Rewrite a linear-over-linear function

A cancellation and a missing point

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review