Graph and analyze rational functions with numerator and denominator of degree at most two

Alberta Grade 12 Mathematics

Relations and Functions

Domain, intercepts, asymptotes, and holes

A rational function is a fraction in which the numerator and denominator are polynomials. For example,

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

is rational. A denominator cannot equal zero, so

x=3

is not an allowed input. This restriction remains even if a factor later cancels during simplification. In this lesson, the numerator and denominator have degree at most two. You will use restrictions, intercepts, asymptotes, and holes to analyze their graphs.

What you will learn

State the domain of a rational function by identifying denominator zeros.
Find valid intercepts and identify vertical and horizontal asymptotes.
Recognize a hole caused by a shared factor and find its coordinates.
Use key features and sign checks to describe a rational-function graph.

1. Review: restrictions, factoring, and intercepts

A polynomial is an expression made from constants and whole-number powers of a variable. Its degree is its greatest exponent with a nonzero coefficient. For example,

x^2-4

has degree two.

A rational function can be written as

f(x)=\frac{N(x)}{D(x)}

, where

N(x)

is the numerator and

D(x)

is the denominator. The domain is the set of inputs for which the function is defined. Find restrictions by solving

D(x)=0

in the original function. Those inputs are excluded.

Factoring rewrites a polynomial as a product. For example,

x^2-4=(x-2)(x+2)

. Factoring makes it easier to identify denominator zeros and factors shared by numerator and denominator.

An

x

-intercept is a point where the graph meets the

x

-axis. Its function value is zero, so a numerator zero can give an intercept if that input is allowed. A

y

-intercept is where the graph meets the

y

-axis. Find it by evaluating the function at

x=0

, if zero is in the domain.

f(x)=\frac{N(x)}{D(x)}

Find excluded inputs from the original denominator.
Check that a proposed intercept is allowed by the domain.
The $y$ -intercept is found by evaluating at zero, not by setting the numerator to zero.

2. Asymptotes and holes

An asymptote is a line that a graph approaches. A vertical asymptote is often found at a denominator zero whose factor remains after simplifying. The function is undefined at that input.

A horizontal asymptote describes the graph’s height far to the left and far to the right. Compare the degrees of the numerator and denominator. If the numerator degree is smaller, the horizontal asymptote is

y=0

. If the degrees are equal, divide the numerator’s leading coefficient by the denominator’s leading coefficient. That ratio is the horizontal asymptote’s height.

When the numerator degree is greater than the denominator degree, these rules do not give a horizontal asymptote. This lesson focuses on horizontal asymptotes when the numerator degree is no greater than the denominator degree.

A shared factor can cancel when simplifying. However, its zero remains excluded if it made the original denominator zero. If the simplified expression has a finite value at that input, the graph has a hole there instead of a vertical asymptote. Substitute the excluded input into the simplified expression to find the hole’s height.

For equal degrees, the horizontal asymptote is the ratio of the leading coefficients.

An uncancelled denominator zero gives a vertical asymptote.
A cancelled denominator factor can leave a hole, and its input is still excluded.
For equal degrees, the horizontal asymptote is the ratio of leading coefficients.

3. Organize features and sketch

Record the domain restrictions first. Factor the numerator and denominator when possible. If factors cancel, keep the original restrictions beside the simplified expression.

Find the valid intercepts, then identify vertical and horizontal asymptotes and any hole. These features give a framework for a sketch. A graph cannot include an excluded point, and a vertical asymptote is not part of the graph.

A sign check helps locate a branch above or below the

x

-axis. Choose a test input in an interval between important

x

-values and evaluate the function. A positive value places the graph above the axis at that input; a negative value places it below.

Sketch the branches using the features and sign checks. A graphing window displays only part of the full graph. Use the algebraic features to describe the graph beyond the window as well.

D_f=\{x∈

\mathbb{R}

:D(x)\ne0\}

Keep restrictions from the original function after simplifying.
Use a test input to check the sign on an interval.
Label intercepts, asymptotes, holes, and excluded inputs clearly.

Worked example

A linear numerator and denominator

Analyze and sketch

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

. State the domain, intercepts, and asymptotes.

A rational function with two asymptotes

The graph shows only a window of the full domain. The vertical asymptote is at $x=3$.

Find the restriction
The denominator is zero when $x=3$ . Since this factor does not cancel, $x=3$ is excluded and gives a vertical asymptote.
$x-3=0\implies x=3$
Find the intercepts
Set the numerator equal to zero for the $x$ -intercept. For the $y$ -intercept, evaluate at zero; zero is allowed.
$2x+1=0\implies x=-\frac12;\quad f(0)=-\frac13$
Find the horizontal asymptote
Both polynomials have degree one. Divide the leading coefficient of the numerator by that of the denominator to get the horizontal asymptote.
$y=\frac21=2$

Answer: The domain is

x\ne3

. The intercepts are

(-\frac12,0)

and

(0,-\frac13)

. The vertical asymptote is

x=3

, and the horizontal asymptote is

y=2

.

Check: The graph approaches the line

x=3

near the excluded input and approaches height

2

far to either side.

Worked example

A cancelled factor and a hole

Analyze

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

. Find its domain, intercepts, asymptotes, and hole.

A rational function with a hole

The graph shows only a window of the full domain. The cancelled input $x=2$ is a hole at height $\frac43$.

Factor and record restrictions
The original denominator is zero at $x=2$ and $x=-1$ . Both inputs are excluded, even though a factor will cancel.
$g(x)=\frac{(x-2)(x+2)}{(x-2)(x+1)}$
Simplify and find the hole
For allowed inputs, cancel the shared factor. The simplified expression gives the height at the excluded input $x=2$ , which locates the hole.
$\frac{x+2}{x+1};\quad \frac{2+2}{2+1}=\frac43$
Find intercepts and asymptotes
The remaining numerator is zero at $x=-2$ , which is allowed. Evaluating at zero gives the $y$ -intercept. The simplified numerator and denominator have equal degree and leading coefficient one.
$(-2,0);\quad g(0)=2;\quad y=1$

Answer: The domain is

x\ne2,-1

. The intercepts are

(-2,0)

and

(0,2)

. There is a vertical asymptote at

x=-1

, a horizontal asymptote at

y=1

, and a hole at

(2,\frac43)

.

Check: The point at

x=2

is missing even though the simplified expression has value

\frac43

there.

Worked example

A sign check between two vertical asymptotes

Analyze

h(x)=\frac{x+1}{x^2-4}

. State its domain, intercepts, and asymptotes. Use sign checks to describe the graph between its vertical asymptotes.

A rational function with two vertical asymptotes

The graph shows only a window of the full domain. The vertical asymptotes are at $x=-2$ and $x=2$.

Find the domain and vertical asymptotes
Factor the denominator. Neither factor cancels with the numerator, so both zeros are excluded and give vertical asymptotes.
$x^2-4=(x-2)(x+2);\quad x\ne-2,2$
Find intercepts and the horizontal asymptote
The numerator is zero at $x=-1$ , giving an allowed $x$ -intercept. At zero, the function value is $-\frac14$ . The numerator degree is less than the denominator degree, so the horizontal asymptote is the $x$ -axis.
$(-1,0);\quad h(0)=-\frac14;\quad y=0$
Check the two parts of the middle interval
Test $x=-\frac32$ in the interval $(-2,-1)$ . Both the numerator and denominator are negative, so the function is positive. Test $x=0$ in $(-1,2)$ ; the function is negative. The graph crosses the axis at $x=-1$ .
$h\left(-\frac32\right)=\frac{-\frac12}{-\frac74}=\frac27;\quad h(0)=-\frac14$

Answer: The domain is

x\ne-2,2

. The intercepts are

(-1,0)

and

(0,-\frac14)

. The vertical asymptotes are

x=-2

and

x=2

, and the horizontal asymptote is

y=0

. Between the asymptotes, the graph is above the

x

-axis on

(-2,-1)

and below it on

(-1,2)

.

Check: The sign changes at the numerator zero

x=-1

. The function is undefined at both denominator zeros.

Common mistakes and how to avoid them

Cancelling a shared factor and then including its zero in the domain.

Correction: Keep every restriction found from the original denominator. A cancelled factor can leave a hole, not a restored input.

Calling every denominator zero a vertical asymptote.

Correction: Check whether the factor cancels. A cancelled factor can create a hole instead.

Using an excluded input as an

x

-intercept.

Correction: Check the original domain before including an intercept.

Reversing the coefficient ratio for a horizontal asymptote when degrees are equal.

Correction: Divide the numerator’s leading coefficient by the denominator’s leading coefficient.

Lesson summary

Exclude every zero of the original denominator from the domain.
Factor to find valid numerator zeros, shared factors, and uncancelled denominator zeros.
Use valid numerator zeros for $x$ -intercepts and evaluate at zero for the $y$ -intercept.
An uncancelled denominator zero gives a vertical asymptote; a cancelled factor can leave a hole.
Compare degrees to find a horizontal asymptote when the numerator degree is no greater than the denominator degree.

Check your understanding

Question 1

For

p(x)=\frac{x-4}{x^2-9}

, which values are excluded from the domain?

$x=4$ only
$x=-3$ and $x=3$
$x=-3$ , $x=3$ , and $x=4$
No values

Show answer and explanation

x=-3

and

x=3

The denominator is zero at

x=-3

and

x=3

. The numerator zero at

x=4

gives an intercept, not a restriction.

Question 2

For

q(x)=\frac{x^2-1}{x^2+x}

, what happens at

x=-1

?

There is a vertical asymptote.
There is a hole at $(-1,2)$ .
The graph has an $x$ -intercept there.
The function is defined there.

Show answer and explanation

There is a hole at

(-1,2)

.

Factoring gives

\frac{(x-1)(x+1)}{x(x+1)}

. The factor cancels, but

x=-1

remains excluded. The simplified expression has value

2

there, so the hole is at

(-1,2)

.

Question 3

What is the horizontal asymptote of

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

?

$y=0$
$y=\frac23$
$y=\frac32$
There is no horizontal asymptote.

Show answer and explanation

y=\frac32

The numerator and denominator have equal degree. Their leading coefficients are

3

and

2

, so the horizontal asymptote is

y=\frac32

.

Key terms

Rational function: A function written as one polynomial divided by another polynomial.
Domain: The set of input values for which a function is defined.
Intercept: A point where a graph meets an axis.
Asymptote: A line that a graph approaches.
Hole: A missing graph point caused by a cancelled factor, while the original denominator still excludes that input.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Mathematics (Math 30-1), expectation RF14. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

Official curriculum reference

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RF14 · Graph and analyze rational functions with numerator and denominator of degree at most two

Alberta Grade 12 Mathematics

Domain, intercepts, asymptotes, and holes

What you will learn

1. Review: restrictions, factoring, and intercepts

2. Asymptotes and holes

3. Organize features and sketch

A linear numerator and denominator

A cancelled factor and a hole

A sign check between two vertical asymptotes

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through Math 30-1

About this lesson and its review