A2.1 Graph logarithmic functions and identify key features

Ontario Grade 12 Mathematics

Exponential and Logarithmic Functions

Ontario Grade 12 Mathematics — A2.1

A logarithm answers a question about an exponent. For example,

\log_2 8=3

because

2^3=8

. This connection helps us graph logarithmic functions. Points on an exponential graph become points on its inverse logarithmic graph, with the coordinates switched. In this lesson, you will use that relationship, a value table, and key graph features to sketch and describe logarithmic functions.

What you will learn

Connect logarithms to powers and use that connection to graph logarithmic functions.
Identify a logarithmic graph’s domain, range, vertical asymptote, intercepts, and direction of change.
Use a value table and transformations to sketch and describe logarithmic graphs.

1. Prerequisite bridge: logarithms and powers

A power has a base and an exponent. In

2^3=8

, the base is

2

and the exponent is

3

. A logarithm asks which exponent produces a given number. Thus,

\log_2 8=3

.

In

\log_b x

,

b

is the base and

x

is the input. The logarithm is defined when

x

is positive, and the base must be positive and not equal to

1

. In this lesson, the bases are greater than

1

. The defining relationship is that

\log_b x=y

exactly when

b^y=x

.

The domain is the set of allowed input values. The range is the set of possible output values. An intercept is a point where a graph meets an axis. For the logarithmic graphs in this lesson, a vertical asymptote is a vertical line that the graph approaches but does not meet. It marks the boundary of the domain.

\log_b x=y\iff b^y=x

A logarithm gives an exponent.
The input of a logarithm must be positive.
Use the relationship between logarithms and powers to find graph points.

2. From a value table to a graph

Start with

y=\log_2 x

. To make a table, choose simple exponents and use powers of

2

to find the matching inputs. If the exponent is

-2

, then the input is

2^{-2}=\frac14

. This gives the point

(\frac14,-2)

.

The graph also passes through

(1,0)

because

2^0=1

. It passes through

(2,1)

because

2^1=2

, and through

(4,2)

because

2^2=4

. Plotting points such as these shows the graph’s shape. The graph increases from left to right: larger positive inputs give larger outputs.

The inputs must be positive, so the domain is

x>0

. The output can be any real number, so the range is all real numbers. As the input approaches

0

from the positive side, the output decreases without bound. The graph approaches the vertical line

x=0

but does not meet it. This line is the vertical asymptote. The graph crosses the

x

-axis at

(1,0)

and has no

y

-intercept because

x=0

is not in its domain.

b^y=x

For $y=\log_b x$ with $b>1$ , the domain is $x>0$ and the range is all real numbers.
The vertical asymptote is $x=0$ .
The graph has an $x$ -intercept at $(1,0)$ and no $y$ -intercept.

3. Transformations and key features

A transformation moves or changes a graph. In

y=\log_b(x-h)+k

, the horizontal shift is controlled by

h

, and the vertical shift is controlled by

k

. The input of the logarithm must remain positive, so require

x-h>0

. This gives a domain of

x>h

and a vertical asymptote at

x=h

.

The range remains all real numbers. The basic point

(1,0)

moves to

(h+1,k)

. To find an

x

-intercept, set

y=0

and solve for

x

. To check for a

y

-intercept, set

x=0

and make sure that this input is in the domain before evaluating the function.

For

y=a\log_b(x-h)+k

, the factor

a

changes the vertical scale and may reflect the graph. When

b>1

, the basic logarithmic graph increases. If

a>0

, it still increases; if

a<0

, it decreases. For

a\ne0

, the range remains all real numbers, and the vertical asymptote and domain are unchanged by this outside factor.

y=a\log_b(x-h)+k

The logarithm’s input gives the domain and the vertical asymptote.
Horizontal and vertical shifts move graph points and the asymptote.
For a base greater than $1$ , a negative outside factor makes the graph decrease.

4. A reliable graphing process

First, find the domain by requiring the logarithm’s input to be positive. The boundary value gives the vertical asymptote for the logarithmic graphs in this lesson. Next, use convenient powers of the base to make points. Include the point that comes from an exponent of

0

, since it often makes a useful anchor.

Then identify the direction of change and any shifts or reflections. Plot the points on the allowed side of the vertical asymptote, and draw a smooth curve through them that approaches the asymptote without meeting it. Finally, find intercepts by testing

y=0

and

x=0

, checking the domain each time.

Find the domain before plotting.
Use powers of the base to create accurate points.
Check intercepts against the domain and describe the graph’s direction.

Points for the worked example

Exponent used	Input $x$	Output $y$	Point
$-1$	$\frac73$	$2$	$\left(\frac73,2\right)$
$0$	$3$	$1$	$(3,1)$
$1$	$5$	$0$	$(5,0)$

Worked example

Graph and describe a transformed logarithmic function

Sketch

y=-\log_3(x-2)+1

. Identify its domain, range, vertical asymptote, intercepts, and direction of change.

Find the domain and asymptote
The logarithm’s input must be positive. Solving $x-2>0$ gives the allowed inputs. The boundary value is the vertical asymptote for this logarithmic graph.
$x>2,\quad x=2$
Build points from powers
Choose exponents $-1$ , $0$ , and $1$ . Their corresponding logarithm inputs are $\frac13$ , $1$ , and $3$ . Since the input is $x-2$ , add $2$ to each value to find $x$ . Then apply the negative sign and add $1$ to find $y$ .
$\left(\frac73,2\right),\quad (3,1),\quad (5,0)$
Identify the range and direction
A logarithm with base $3$ can produce any real output. Multiplying by $-1$ and adding $1$ still allows every real output. The negative factor reflects the increasing basic graph, so this graph decreases from left to right.
$y∈\mathbb{R}$
Find the intercepts
The point table gives the $x$ -intercept at $(5,0)$ . A $y$ -intercept would require $x=0$ , but $0$ is outside the domain $x>2$ . Therefore, the graph has no $y$ -intercept.
$(5,0)$

Answer: The graph decreases, has domain

x>2

, range all real numbers, and vertical asymptote

x=2

. Its

x

-intercept is

(5,0)

, and it has no

y

-intercept.

Check: At

x=3

, the logarithm input is

1

, so

y=-\log_3 1+1=1

. This confirms the point

(3,1)

. The other points follow from

3^{-1}=\frac13

and

3^1=3

.

Common mistakes and how to avoid them

Allowing zero or negative values inside a logarithm.

Correction: Require the logarithm’s input to be greater than zero. For example,

x-2>0

gives

x>2

.

Giving the vertical asymptote the wrong sign after a horizontal shift.

Correction: Set the logarithm’s input equal to zero to locate the boundary. For an input of

x-h

, the asymptote is

x=h

.

Assuming every logarithmic graph increases.

Correction: Check the outside factor. For a base greater than

1

, a negative factor reflects the basic increasing graph, making it decrease.

Claiming a

y

-intercept without checking whether

x=0

is allowed.

Correction: Test

x=0

against the domain before finding the output. If

0

is outside the domain, the graph has no

y

-intercept.

Lesson summary

Use $\log_b x=y$ exactly when $b^y=x$ to make points.
For $y=\log_b x$ with $b>1$ , the domain is $x>0$ , the range is all real numbers, and the vertical asymptote is $x=0$ .
For $y=a\log_b(x-h)+k$ , use the input condition to find the domain and asymptote. The shifts move the graph, and the sign of $a$ sets its direction when $b>1$ .
Find intercepts by testing $y=0$ or $x=0$ , and check that the input is in the domain.

Check your understanding

Question 1

For

y=\log_5(x+4)-2

, which statement gives the domain and vertical asymptote?

Domain $x>-4$ ; asymptote $x=-4$
Domain $x>-4$ ; asymptote $x=4$
Domain $x\geq-4$ ; asymptote $x=-4$
Domain all real numbers; asymptote $x=0$

Show answer and explanation

Domain

x>-4

; asymptote

x=-4

The logarithm’s input must satisfy

x+4>0

, so

x>-4

. The boundary is the vertical asymptote

x=-4

.

Question 2

For

y=-\log_2 x

, which description is correct?

It increases and has range $y>0$ .
It decreases and has range all real numbers.
It increases and has vertical asymptote $y=0$ .
It decreases and has domain all real numbers.

Show answer and explanation

It decreases and has range all real numbers.

The basic graph for base

2

increases, and the negative factor reflects it so it decreases. The domain is

x>0

, the range is all real numbers, and the vertical asymptote is

x=0

.

Key terms

Logarithm: The exponent that a base must have to produce a given positive number.
Domain: The set of input values for which a function is defined.
Range: The set of output values a function can produce.
Intercept: A point where a graph meets the $x$ -axis or the $y$ -axis.
Vertical asymptote: For the logarithmic graphs in this lesson, a vertical line that the graph approaches but does not meet.
Transformation: A shift, stretch, or reflection that changes a graph’s position or shape.

Continue through MHF4U

View the complete MHF4U Ontario Grade 12 Mathematics curriculum and lessons

About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation A2.1. It is a study resource, not an official curriculum publication.

Official curriculum reference

Report a correction or ask a question

A2.1 · Graph logarithmic functions and identify key features

Ontario Grade 12 Mathematics

Ontario Grade 12 Mathematics — A2.1

What you will learn

1. Prerequisite bridge: logarithms and powers

2. From a value table to a graph

3. Transformations and key features

4. A reliable graphing process

Points for the worked example

Graph and describe a transformed logarithmic function

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through MHF4U

About this lesson