C2.2 Analyze linear-over-linear rational functions

Ontario Grade 12 Mathematics

Polynomial and Rational Functions

Connect equations, key features, tables, and graphs

A rational function is a function written as one polynomial divided by another. In this lesson, we focus on functions whose numerator and denominator are both linear. Their graphs have a recognizable shape, but the denominator creates an input that is not allowed. We will connect that restriction to the graph, then use intercepts, asymptotes, and a table to describe the function. We will use average rates of change from pairs of values; we will not use derivative rules.

What you will learn

Identify the domain restrictions and intercepts of a linear-over-linear rational function.
Find and interpret vertical and horizontal asymptotes.
Rewrite a function to reveal its graph’s shape and position.
Use a table and numerical rates of change to describe how the function behaves.

1. Prerequisite bridge: restrictions and intercepts

A linear expression has the form

mx+b

, where

m

and

b

are numbers. For example,

2x+1

is linear. A rational expression is a quotient, or division, of expressions. A linear-over-linear rational function has a linear numerator and a linear denominator, such as

f(x)=\frac{2x+1}{x-3}

.

Division by zero is not defined. So the denominator cannot equal zero. This creates a domain restriction: a value that cannot be used as an input. For this function,

x=3

is not in the domain because it makes the denominator zero.

An

x

-intercept is a point where the graph crosses the

x

-axis. Its output is zero, so set the numerator equal to zero, as long as that input is allowed. A

y

-intercept is where the graph crosses the

y

-axis. Its input is zero, so substitute

x=0

if zero is in the domain.

f(x)=\frac{ax+b}{cx+d}

The denominator identifies inputs that are not allowed.
Find an $x$ -intercept by setting the numerator to zero and checking the denominator.
Find a $y$ -intercept by evaluating the function at $x=0$ , when allowed.

2. Plain language: the graph’s key features

For a function in the form

f(x)=\frac{ax+b}{cx+d}

, first find where the denominator is zero. If that value is not cancelled by a matching factor in the numerator, the graph has a vertical asymptote there. An asymptote is a line that the graph gets closer to as it continues; the graph does not cross a vertical asymptote because the function is undefined at that input.

When the numerator and denominator have the same degree, the horizontal asymptote is the ratio of their leading coefficients. The degree of a linear expression is one, and its leading coefficient is the number multiplying

x

. The horizontal asymptote describes the output the function approaches far to the left or right.

A useful rewrite makes both asymptotes visible: divide to express the function as a constant plus a fraction with a constant numerator. In the form

f(x)=q+\frac{r}{x-h}

, the vertical asymptote is

x=h

, and the horizontal asymptote is

y=q

. The graph has two branches, one on each side of the vertical asymptote.

\frac{ax+b}{cx+d}=q+\frac{r}{x-h}

A vertical asymptote comes from a denominator zero that remains after simplifying.
For linear-over-linear functions, the horizontal asymptote is the ratio of the coefficients of $x$ .
The rewritten form shows the asymptotes directly.

3. Multiple representations: equation, table, and graph behaviour

An equation gives exact features. A table gives selected input-output pairs. A graph shows how those pairs fit together and how the branches approach the asymptotes. These representations support one another, but a short table alone cannot establish every feature of a graph.

The average rate of change between two inputs is the change in output divided by the change in input. It is the slope of the line through the two corresponding points on the graph. For a straight-line function, this rate is constant. For a linear-over-linear function, it generally changes from one interval to another.

When making a table, do not include a restricted input. It is often helpful to choose inputs on both sides of the vertical asymptote. Values close to the asymptote can show that outputs become large in size, while values far from it can show the function approaching its horizontal asymptote.

average rate of change=

\frac{f(x_2)-f(x_1)}{x_2-x_1}

Use the equation to locate exact features and the table to compare selected values.
Average rate of change compares two points, not one point.
A restricted input must not appear in a table of function values.

4. Guided example and application

Consider

f(x)=\frac{2x+1}{x-3}

. We will identify its restriction, rewrite it, and find its intercepts before using a table. Together, these features give a reliable description of the graph without relying on a sketch alone.

The denominator is zero at

x=3

, so the function is undefined there. The numerator is not zero at that input, so there is no cancellation. Rewriting the function gives

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

. This makes the vertical and horizontal asymptotes clear.

For the

x

-intercept, solve

2x+1=0

. The resulting input is allowed, so it gives an intercept. For the

y

-intercept, substitute zero. The table includes values on both sides of the vertical asymptote. The average rate from

x=4

to

x=5

differs from the rate from

x=5

to

x=6

, showing that the rate is not constant.

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

The domain is all real inputs except $3$ .
The asymptotes are $x=3$ and $y=2$ .
The intercepts are $\left(-\frac{1}{2},0\right)$ and $\left(0,-\frac{1}{3}\right)$ .

Selected values for the worked example

Input $x$	Output $f(x)$
0	$-\frac{1}{3}$
1	$-\frac{3}{2}$
2	$-5$
3	undefined
4	$9$
5	$5$
6	$\frac{13}{3}$

Worked example

Analyze a linear-over-linear function

Analyze

f(x)=\frac{2x+1}{x-3}

. Find its domain restriction, asymptotes, intercepts, selected values, and two average rates of change.

Find the restriction
Set the denominator equal to zero. The resulting input makes the original quotient undefined. Since the numerator is not zero there, the factor does not cancel.
$x-3=0\Rightarrow x=3$
Rewrite to reveal the asymptotes
Rewrite the numerator as twice the denominator plus seven. This expresses the function as a constant plus a fraction. The denominator of that fraction reveals the vertical asymptote, and the constant reveals the horizontal asymptote.
$\frac{2x+1}{x-3}=2+\frac{7}{x-3}$
Find the intercepts
Set the numerator to zero for the $x$ -intercept. Then evaluate the function at zero for the $y$ -intercept. Both inputs are allowed.
$2x+1=0\Rightarrow x=-\frac{1}{2},\qquad f(0)=-\frac{1}{3}$
Compare values and rates
Use the rewritten form to evaluate the function at inputs on both sides of the restriction. Then apply the average-rate formula to each interval. The outputs and rates show that the graph does not behave like a straight line.
$\begin{array}{c|rrrrrr}x&0&1&2&4&5&6\\\hline f(x)&-\frac13&-\frac32&-5&9&5&\frac{13}{3}\end{array},\qquad \frac{5-9}{5-4}=-4,\quad \frac{\frac{13}{3}-5}{6-5}=-\frac23$

Answer: The domain is all real numbers except

3

. The vertical asymptote is

x=3

, and the horizontal asymptote is

y=2

. The intercepts are

\left(-\frac{1}{2},0\right)

and

\left(0,-\frac{1}{3}\right)

. The average rates of change on the intervals from

4

to

5

and from

5

to

6

are

-4

and

-\frac{2}{3}

, respectively.

Check: At

x=3

, the original denominator is zero, so no function value exists. For large positive or negative inputs, the fraction

\frac{7}{x-3}

approaches zero, which agrees with the horizontal asymptote

y=2

.

Common mistakes and how to avoid them

Including the value that makes the denominator zero in the domain.

Correction: Solve denominator equals zero and exclude that input from the domain.

Finding an

x

-intercept from the denominator.

Correction: Set the numerator equal to zero, then confirm the denominator is not zero at that input.

Calling the ratio of leading coefficients the vertical asymptote.

Correction: That ratio gives the horizontal asymptote for a linear-over-linear function. The denominator zero gives the vertical asymptote when no factor cancels.

Assuming a changing function has one constant rate of change.

Correction: Calculate average rates on the stated intervals. Different intervals can give different rates.

Lesson summary

For $f(x)=\frac{ax+b}{cx+d}$ , exclude inputs that make the denominator zero.
The numerator zero gives an $x$ -intercept when that input is allowed; evaluating at zero gives the $y$ -intercept when defined.
For a linear-over-linear function without cancellation, the denominator zero gives the vertical asymptote, and the ratio of leading coefficients gives the horizontal asymptote.
A rewrite as a constant plus a fraction can show both asymptotes directly.
Tables and average rates of change help describe how the function behaves over selected intervals.

Check your understanding

Question 1

For

g(x)=\frac{3x-2}{x+4}

, which input is excluded from the domain?

$x=-4$
$x=\frac{2}{3}$
$x=4$
correctIndexأن

Show answer and explanation

x=-4

The denominator is zero when

x+4=0

, so

x=-4

is excluded.

Question 2

What is the horizontal asymptote of

g(x)=\frac{3x-2}{x+4}

?

$y=-4$
$y=3$
$x=3$
correctIndex

Show answer and explanation

y=3

The leading coefficients are

3

and

1

. Their ratio is

3

, so the horizontal asymptote is

y=3

.

Question 3

For

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

, what is the

x

-intercept?

$(5,0)$
$(-5,0)$
$(0,-\frac{5}{2})$
correctIndex

Show answer and explanation

(-5,0)

Set the numerator equal to zero:

x+5=0

, so

x=-5

. The denominator is nonzero at this input.

Key terms

Rational function: A function written as a quotient of polynomial expressions.
Domain restriction: An input that must be excluded because it makes the function undefined.
Intercept: A point where a graph crosses an axis.
Asymptote: A line that a graph approaches as part of its behaviour.
Average rate of change: The change in output divided by the change in input between two points.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation C2.2. It is a study resource, not an official curriculum publication.

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C2.2 · Analyze linear-over-linear rational functions

Ontario Grade 12 Mathematics

Connect equations, key features, tables, and graphs

What you will learn

1. Prerequisite bridge: restrictions and intercepts

2. Plain language: the graph’s key features

3. Multiple representations: equation, table, and graph behaviour

4. Guided example and application

Selected values for the worked example

Analyze a linear-over-linear function

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MHF4U

About this lesson and its review