Transform logarithmic function graphs

Ontario Grade 12 Mathematics

Exponential and Logarithmic Functions

Reading and applying shifts, stretches, and reflections

A logarithmic graph can be shifted, stretched, or reflected by changing its equation. To sketch the new graph, start with the basic logarithmic shape, track how its points move, and check its vertical asymptote and domain. This lesson focuses on those graph transformations.

What you will learn

Identify the transformations in a logarithmic function.
Use key points and the vertical asymptote to sketch a transformed graph.
Determine a logarithmic graph's domain and describe how its shape changes.

1. Prerequisite bridge: the basic logarithmic graph

A logarithm answers an exponent question. For example,

\log_3 9=2

because

3^2=9

. The basic logarithmic function is

y=\log_b x

, where the base satisfies

b>0

and

b\ne1

.

The input

x

must be positive. So the graph has no points at or to the left of the vertical line

x=0

. This line is a vertical asymptote: the graph approaches it but does not meet it.

For a base greater than

1

, the basic graph rises from left to right. Useful points include

(1,0)

and

(b,1)

, since

\log_b 1=0

and

\log_b b=1

. These points help anchor a sketch.

y=\log_b x

The parent graph is $y=\log_b x$ .
Its domain is $x>0$ , and its vertical asymptote is $x=0$ .
When $b>1$ , the graph increases from left to right.

2. Plain language: what changes in the equation?

A transformed logarithmic function can be written as

y=a\log_b(k(x-d))+c

. The constants change the parent graph in predictable ways. The number

d

shifts the graph horizontally, and

c

shifts it vertically.

The factor

a

acts outside the logarithm. If its absolute value is greater than

1

, it stretches the graph vertically; if it is between

0

and

1

, it compresses the graph vertically. A negative

a

reflects the graph across the horizontal axis before the vertical shift.

The factor

k

acts on the input inside the logarithm. Its horizontal scale factor is

1/|k|

: values of |k| greater than

1

compress the graph horizontally, while values between

0

and

1

stretch it horizontally. A negative

k

reflects the graph horizontally as well. Track the expression inside the logarithm carefully because it determines which side of the asymptote is in the domain.

y=a\log_b(k(x-d))+c

$d$ gives the horizontal shift and $c$ gives the vertical shift.
$a$ controls vertical stretch or compression and may reflect the graph.
$k$ controls horizontal stretch or compression and may reflect the graph.

3. Multiple representations: points, asymptote, and domain

A reliable way to transform a graph is to map points from the parent graph. For a parent point

(x,y)

, the corresponding point on the transformed graph is

(d+x/k,\;c+ay)

. The input is adjusted by the inside factor and horizontal shift; the output is adjusted by the outside factor and vertical shift.

The vertical asymptote of the parent graph occurs where its input is zero. For the transformed function, set

k(x-d)=0

. Since

k

is nonzero, the asymptote is

x=d

. The domain comes from requiring the logarithm's input to be positive:

k(x-d)>0

. If

k

is positive, the domain is

x>d

. If

k

is negative, it is

x<d

.

A sketch should show the asymptote as a dashed vertical line, plot transformed points, and draw the logarithmic curve on the allowed side. The curve approaches the asymptote but never crosses it. The sign of

a

and the base also help determine whether the graph rises or falls.

(x,y)\mapsto\left(d+\frac{x}{k},\;c+ay\right)

Map each parent point with $(x,y)\mapsto(d+x/k,\;c+ay)$ .
The vertical asymptote is $x=d$ .
The domain is determined by $k(x-d)>0$ .

4. From equation to sketch

Use a consistent order when sketching. First identify the parent function and its familiar points. Next read the horizontal and vertical changes from the equation. Then find the asymptote and domain, map the points, and draw the curve on the correct side.

When the input is written in a form such as

k(x-d)

, use that form directly to locate the asymptote. Avoid treating every number inside the logarithm as a simple shift. For example, the horizontal factor changes point locations as well as the graph's width.

Check that every plotted point has a positive logarithm input. A point outside the domain cannot belong to the graph, even if it appears to fit a careless transformation.

k(x-d)>0

Use the input of the logarithm to find the domain and asymptote.
Use mapped points to preserve the graph's shape and placement.
Keep the sketch on the side allowed by the domain.

Point mapping for the worked graph

Parent point	Transformed point
$(1,0)$	$(-1/2,1)$
$(3,1)$	$(1/2,-1)$
$(1/3,-1)$	$(-5/6,3)$

Worked example

Sketch a transformed logarithmic graph

Describe the transformations, domain, vertical asymptote, and three useful points for

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

.

Identify the parent graph
The parent function is $y=\log_3 x$ . Its convenient points are $(1,0)$ , $(3,1)$ , and $(1/3,-1)$ . The base is greater than $1$ , so the parent graph rises from left to right.
$y=\log_3 x$
Read the transformations
Rewrite the inside as $2(x-(-1))$ . Thus $d=-1$ , $k=2$ , $a=-2$ , and $c=1$ . The graph shifts left $1$ , compresses horizontally by a factor of $1/2$ , reflects across the horizontal axis, stretches vertically by a factor of $2$ , and shifts up $1$ .
$d=-1,\quad k=2,\quad a=-2,\quad c=1$
Find the asymptote and domain
The asymptote is at $x=d=-1$ . The logarithm's input must be positive. Since the factor $2$ is positive, the graph exists to the right of the asymptote.
$2(x+1)>0\quad\Longrightarrow\quad x>-1$
Map the parent points
For each parent point, use the input mapping $x_{\text{new}}=d+x/k$ and output mapping $y_{\text{new}}=c+ay$ . Applying these rules gives three points on the transformed graph.
$(1,0)\mapsto\left(-\frac12,1\right),\quad(3,1)\mapsto\left(\frac12,-1\right),\quad\left(\frac13,-1\right)\mapsto\left(-\frac56,3\right)$
Sketch the curve
Draw the vertical asymptote at $x=-1$ , then plot the three mapped points to its right. The reflected graph falls as $x$ increases. Draw a smooth logarithmic curve through the points that approaches the asymptote without touching it.
$x=-1$

Answer: The graph has domain

x>-1

and vertical asymptote

x=-1

. It passes through

(-1/2,1)

,

(1/2,-1)

, and

(-5/6,3)

, and falls from left to right.

Check: Substituting each mapped input into

2(x+1)

gives, respectively,

1

,

3

, and

1/3

, matching the parent inputs. All three inputs are positive.

Common mistakes and how to avoid them

Saying the horizontal scale factor is |k|.

Correction: For an input of

k(x-d)

, the horizontal scale factor is

1/|k|

. A larger |k| makes the graph narrower.

Using

x=d

as the domain boundary without checking which side is allowed.

Correction: Solve

k(x-d)>0

. The sign of

k

determines whether the domain is to the right or left of the asymptote.

Reflecting the graph across the horizontal axis when

k

is negative.

Correction: A negative outside factor

a

reflects across the horizontal axis. A negative inside factor

k

changes the horizontal orientation and the allowed side.

Drawing the graph through its vertical asymptote.

Correction: The asymptote marks a boundary the logarithmic graph approaches but does not reach.

Lesson summary

Start with the parent graph $y=\log_b x$ and its key points.
For $y=a\log_b(k(x-d))+c$ , $a$ and $c$ change vertical position or shape; $k$ and $d$ change horizontal position or shape.
The asymptote is $x=d$ , and the domain is found from $k(x-d)>0$ .
Map parent points, check the domain, and sketch the curve on the permitted side.

Check your understanding

Question 1

For

y=\log_2(3(x-4))-2

, what is the vertical asymptote?

$x=4$
$x=-4$
$x=3$
$y=-2$

Show answer and explanation

x=4

The inside is

3(x-4)

, so it is zero at

x=4

. That gives the vertical asymptote.

Question 2

What is the domain of

y=\log_5(-(x+2))+1

?

$x>-2$
$x<-2$
$x>2$
All real $x$

Show answer and explanation

x<-2

The logarithm input must be positive:

-(x+2)>0

. Solving gives

x<-2

.

Question 3

For

y=\frac12\log_3 x

, what happens to the parent graph?

It is stretched vertically by a factor of $2$ .
It is compressed vertically by a factor of $1/2$ .
It is reflected across the horizontal axis.
It shifts right by $1/2$ .

Show answer and explanation

It is compressed vertically by a factor of

1/2

.

The outside multiplier is positive and less than

1

, so it compresses the graph vertically by a factor of

1/2

.

Key terms

Logarithm: An exponent that tells what power of a base produces a given value.
Vertical asymptote: A vertical line that a graph approaches but does not reach.
Domain: The set of input values for which a function is defined.
Reflection: A flip of a graph across a line, such as a horizontal or vertical axis.

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation A2.3. 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.

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