Function Transformations: a, k, d and c Explained

Ontario Grade 11 Mathematics

Characteristics of Functions

How a, k, d, and c reshape any function

Every function you study in Grade 11 — quadratic, square root, trigonometric — can be stretched, flipped, and shifted using one master template:

y = af(k(x - d)) + c

. The four constants

a

,

k

,

d

, and

c

are called transformation parameters. Changing even one of them reshapes or repositions the entire graph in a predictable way. This lesson builds each parameter from scratch, connects it to a picture, and then combines them so you can handle any transformation question with confidence.

What you will learn

Identify what each parameter (a, k, d, c) does to the graph of a base function.
Distinguish between transformations that change the shape of a graph (stretches and reflections) and those that move it (translations).
Apply y = af(k(x − d)) + c to describe or sketch the image of a given base function.
Determine parameter values from a written description or a transformed graph.
Recognize how two or more parameters combine to produce a single transformed graph.

Prerequisite Bridge: What Is a Base Function?

Before looking at transformations, you need a starting point. A base function is the simplest version of a family of functions — no stretching, no shifting. Examples you already know from Grade 10 include

f(x) = x^2

(a parabola opening upward with vertex at the origin),

f(x) = \sqrt{x}

(a curve starting at the origin going right and up), and

f(x) = \sin(x)

(a wave that repeats every 360°).

When you apply the template

y = af(k(x - d)) + c

, the letter

f

represents whichever base function you are working with. The parameters

a

,

k

,

d

, and

c

are constants — fixed numbers you substitute in. Your job in any problem is to read off those numbers and translate them into a description of what the graph looks like.

A base function is the simplest form: no parameters other than $x$ .
The template $y = af(k(x - d)) + c$ wraps around any base function $f$ .
The four parameters $a$ , $k$ , $d$ , $c$ each control exactly one type of change.

The Vertical Parameters: a and c

The parameter

c

is the easiest to understand. Adding

c

outside the function shifts the entire graph up or down without changing its shape. If

c > 0

, every point moves up by

c

units. If

c < 0

, every point moves down by |c| units. This is called a vertical translation.

The parameter

a

is multiplied by the function's output. Because it acts on the

y

-value, it causes a vertical stretch or compression. If

|a| > 1

, the graph is stretched away from the

x

-axis — points that were close to the axis move farther away. If

0 < |a| < 1

, the graph is compressed toward the

x

-axis. When

a

is negative, the graph is also reflected across the

x

-axis, meaning it flips upside-down. The value |a| is often called the vertical stretch factor.

A key detail: vertical stretches and reflections happen before the vertical translation. Think of it as 'reshape first, then slide.' For example,

y = -2f(x) + 3

first stretches the graph by a factor of

2

, then reflects it over the

x

-axis, and finally shifts it up

3

units.

y = af(x) + c

$c > 0$ : graph moves up; $c < 0$ : graph moves down.
$|a| > 1$ : vertical stretch; $0 < |a| < 1$ : vertical compression.
$a < 0$ : reflection in the $x$ -axis (graph flips vertically).
Apply $a$ before applying $c$ when sketching.

The Horizontal Parameters: k and d

The parameter

d

appears inside the function as

(x - d)

. It causes a horizontal translation — a left or right slide. The direction is the opposite of what the sign suggests:

(x - d)

with

d > 0

shifts the graph right by

d

units, and

(x - d)

with

d < 0

(which looks like

(x + |d|)

) shifts the graph left by |d| units. A useful memory trick: ask 'what value of

x

makes the bracket equal zero?' That value is

d

, and it tells you where a key reference point (like the vertex or starting point) moves to.

The parameter

k

is multiplied by

x

inside the function. Because it acts on the input, it causes a horizontal stretch or compression, but in the opposite direction from what you might expect. A factor of

|k| > 1

compresses the graph horizontally toward the

y

-axis — the graph gets narrower. A factor of

0 < |k| < 1

stretches the graph horizontally away from the

y

-axis — the graph gets wider. The horizontal stretch factor applied to each

x

-coordinate is

\frac{1}{|k|}

. When

k < 0

, the graph is also reflected across the

y

-axis.

Both

k

and

d

affect the

x

-coordinates of every point. To find the new

x

-coordinate of any point

(x_0, y_0)

on the base graph, calculate

x_{\text{new}} = \frac{x_0}{k} + d

. This single formula captures both the horizontal stretch and the horizontal translation together.

x_{\text{new}} = \frac{x_0}{k} + d

$d > 0$ : shift right; $d < 0$ : shift left — always opposite to the sign inside the bracket.
$|k| > 1$ : horizontal compression; $0 < |k| < 1$ : horizontal stretch.
$k < 0$ : reflection in the $y$ -axis.
New $x$ -coordinate of any point: $x_{\text{new}} = \frac{x_0}{k} + d$ .

Combining All Four Parameters

When all four parameters appear together, use a consistent mapping rule for every key point

(x_0, y_0)

on the base graph. The new coordinates are

x_{\text{new}} = \frac{x_0}{k} + d

and

y_{\text{new}} = a \cdot y_0 + c

. This means horizontal and vertical transformations are independent — you can work out the new

x

and new

y

separately and then combine them.

A reliable sketching order is: (1) identify the base function and its key points; (2) apply the horizontal stretch or compression using

k

; (3) apply the horizontal translation using

d

; (4) apply the vertical stretch, compression, or reflection using

a

; (5) apply the vertical translation using

c

. You will get the same final result regardless of whether you process horizontal or vertical transformations first, as long as you keep the two directions separate.

For trigonometric functions like

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

, the parameter |a| gives the amplitude (the distance from the midline to a peak), the period becomes

\frac{360°}{|k|}

,

d

is the phase shift (horizontal slide), and

c

is the equation of the midline. These are just the same four parameters applied to

f(x) = \sin(x)

, so no new rules are needed.

(x_0,\, y_0) \rightarrow \left(\frac{x_0}{k}+d,\; ay_0+c\right)

Mapping rule: $(x_0, y_0) \rightarrow \left(\frac{x_0}{k} + d,\; ay_0 + c\right)$ .
Work horizontal and vertical transformations independently.
For $y = a\sin(k(x-d))+c$ : amplitude $= |a|$ , period $= \frac{360°}{|k|}$ , midline $y = c$ .
Sketching order: base → horizontal stretch → horizontal shift → vertical stretch/reflect → vertical shift.

Reading Parameters from a Graph or Description

Sometimes you are given the transformed graph and asked to find the equation. Start by identifying the base function family (parabola, square root, sine, etc.). Then locate a key reference point on the base graph — the vertex of a parabola, the endpoint of a square root curve, or a midline crossing of a sine curve — and see where it has moved on the transformed graph. The horizontal shift of that reference point gives

d

, and the vertical shift gives

c

.

Next, measure the vertical scale. Compare the

y

-distance between two key points on the transformed graph to the corresponding distance on the base graph. Their ratio is |a|. If the graph has flipped relative to the base, then

a

is negative. Similarly, compare the horizontal distance between two key features (such as two consecutive peaks of a sine curve) to find the period, then use the relationship between period and

k

to solve for |k|. That relationship is shown in the formula field below.

Always write your final equation in the form

y = af(k(x - d)) + c

with the bracket factored correctly. For example,

y = 3\sin(2x - 90°) + 1

should be rewritten as

y = 3\sin(2(x - 45°)) + 1

so that

d = 45°

is clearly visible. Forgetting to factor out

k

before reading off

d

is the single most common error in this topic.

|k| = \frac{360°}{\text{period}}

Identify the base function family first.
Shift of a key reference point gives $d$ (horizontal) and $c$ (vertical).
Ratio of vertical distances gives |a|; sign of $a$ depends on whether the graph flipped.
Use the period and the formula for |k| for periodic functions.
Always factor $k$ out of the bracket before reading $d$ .

Summary of the Four Transformation Parameters

Parameter	Location in equation	Effect on graph	Key detail
$a$	Multiplies $f(\ldots)$	Vertical stretch ( $\|a\|>1$ ) or compression ( $\|a\|<1$ )	Negative $a$ reflects over the $x$ -axis
$k$	Multiplies $x$ inside $f$	Horizontal compression ( $\|k\|>1$ ) or stretch ( $\|k\|<1$ )	Negative $k$ reflects over the $y$ -axis
$d$	Subtracted from $x$ inside $f$	Horizontal translation: right if $d>0$ , left if $d<0$	Factor $k$ out first to read $d$ correctly
$c$	Added outside $f$	Vertical translation: up if $c>0$ , down if $c<0$	Shifts the midline or vertex

Worked example

Sketching a Transformed Square-Root Function

The base function is

f(x) = \sqrt{x}

. Describe all transformations and find the image of the point

(9, 3)

under

y = -\frac{1}{2}f(2(x - 4)) + 1

.

Read off each parameter
Compare $y = -\frac{1}{2}f(2(x - 4)) + 1$ to the template $y = af(k(x-d)) + c$ . The bracket is already factored, so reading the values directly gives $a = -\frac{1}{2}$ , $k = 2$ , $d = 4$ , and $c = 1$ .
$a = -\frac{1}{2},\quad k = 2,\quad d = 4,\quad c = 1$
Describe each transformation in words
Because $k = 2$ and $|k| > 1$ , the graph is compressed horizontally by a factor of $\frac{1}{2}$ (it gets narrower). Because $d = 4 > 0$ , the graph shifts right 4 units. Because $|a| = \frac{1}{2} < 1$ , the graph is compressed vertically by a factor of $\frac{1}{2}$ . Because $a < 0$ , the graph is also reflected over the $x$ -axis. Because $c = 1 > 0$ , the graph shifts up 1 unit.
Apply the horizontal mapping to the x-coordinate
Use the formula $x_{\text{new}} = \frac{x_0}{k} + d$ with $x_0 = 9$ , $k = 2$ , and $d = 4$ . Dividing 9 by 2 gives 4.5, and adding 4 gives 8.5.
$x_{\text{new}} = \frac{9}{2} + 4 = 4.5 + 4 = 8.5$
Apply the vertical mapping to the y-coordinate
Use the formula $y_{\text{new}} = a \cdot y_0 + c$ with $a = -\frac{1}{2}$ , $y_0 = 3$ , and $c = 1$ . Multiplying $-\frac{1}{2}$ by 3 gives $-1.5$ , and adding 1 gives $-0.5$ .
$y_{\text{new}} = -\frac{1}{2}(3) + 1 = -1.5 + 1 = -0.5$
State the image point and verify
Combining the two new coordinates, the image of $(9, 3)$ under this transformation is $(8.5, -0.5)$ . To verify, substitute $x = 8.5$ into the equation: first compute $2(8.5 - 4) = 2(4.5) = 9$ , then $\sqrt{9} = 3$ , then $-\frac{1}{2}(3) + 1 = -1.5 + 1 = -0.5$ . This matches, confirming the image point is correct.
$(9,\, 3) \rightarrow (8.5,\; -0.5)$

Answer: The image of

(9, 3)

is

(8.5, -0.5)

. Transformations applied: horizontal compression by factor

\frac{1}{2}

, right shift 4 units, vertical compression by factor

\frac{1}{2}

, reflection in the

x

-axis, and upward shift of 1 unit.

Check: Substituting

x = 8.5

into

y = -\frac{1}{2}\sqrt{2(8.5-4)}+1

gives

y = -\frac{1}{2}\sqrt{9}+1 = -\frac{1}{2}(3)+1 = -0.5

. Confirmed.

Worked example

Finding the Equation of a Transformed Sine Function

A sinusoidal graph has a maximum point at

(75°, 5)

and a minimum point at

(165°, -1)

. The graph has not been reflected. Write its equation in the form

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

.

Find the amplitude a
The amplitude is half the total vertical distance between the maximum and the minimum. Subtract the minimum $y$ -value from the maximum $y$ -value and divide by 2. Since the graph has not been reflected, $a$ is positive.
$a = \frac{5 - (-1)}{2} = \frac{6}{2} = 3$
Find the midline value c
The midline is the horizontal line exactly halfway between the maximum and minimum. Average the two $y$ -values to find it. The midline equation is $y = 2$ , so $c = 2$ .
$c = \frac{5 + (-1)}{2} = \frac{4}{2} = 2$
Find the period and then k
The horizontal distance from a maximum to the very next minimum is exactly half a period. The maximum is at $x = 75°$ and the minimum is at $x = 165°$ , so half a period equals $165° - 75° = 90°$ . Therefore the full period is CAD 180°. Use the period formula to find $k$ .
$k = \frac{360°}{180°} = 2$
Find the phase shift d
For the base function $y = \sin(x)$ , the first maximum after the origin occurs at $x = 90°$ . After a horizontal compression by factor $k = 2$ , the maximum of $y = \sin(2x)$ moves to $x = \frac{90°}{2} = 45°$ . The actual maximum on the transformed graph is at $x = 75°$ , which is 30° to the right of 45°. Therefore the graph has been shifted right by $d = 30°$ .
$d = 75° - 45° = 30°$
Write the final equation and verify both key points
Substitute all four parameters into the template to get the equation. Then check both given points. At $x = 75°$ : compute $2(75° - 30°) = 2(45°) = 90°$ , so $y = 3\sin(90°) + 2 = 3(1) + 2 = 5$ . This matches the maximum. At $x = 165°$ : compute $2(165° - 30°) = 2(135°) = 270°$ , so $y = 3\sin(270°) + 2 = 3(-1) + 2 = -1$ . This matches the minimum.
$y = 3\sin(2(x - 30°)) + 2$

Answer:

y = 3\sin(2(x - 30°)) + 2

Check: At

x = 75°

:

y = 3\sin(2(45°)) + 2 = 3\sin(90°) + 2 = 3 + 2 = 5

✓. At

x = 165°

:

y = 3\sin(2(135°)) + 2 = 3\sin(270°) + 2 = -3 + 2 = -1

✓.

Common mistakes and how to avoid them

Reading

d

directly from an un-factored bracket. For example, writing

d = 90°

from

y = \sin(2x - 90°)

instead of first factoring to get

y = \sin(2(x - 45°))

, which gives

d = 45°

.

Correction: Always factor

k

out of the bracket completely before identifying

d

. The expression inside the function must look like

k(x - d)

.

Confusing the direction of horizontal translations. Seeing

(x - 4)

and shifting left instead of right.

Correction: The graph shifts in the direction that makes the bracket equal zero. If

(x - 4) = 0

, then

x = 4

, so the shift is right 4 units.

Applying the horizontal stretch factor in the wrong direction — multiplying

x

-coordinates by

k

instead of dividing.

Correction: The

x

-coordinates of key points are divided by

k

(equivalently, multiplied by

\frac{1}{k}

), not multiplied by

k

. Use

x_{\text{new}} = \frac{x_0}{k} + d

.

Forgetting that

a < 0

causes a reflection and treating

a = -3

as only a vertical stretch by factor 3.

Correction: Always check the sign of

a

separately. |a| gives the stretch factor, and the negative sign means the graph flips over the

x

-axis.

Mixing up which parameters affect the

x

-coordinates and which affect the

y

-coordinates — for example, thinking

c

shifts the graph horizontally.

Correction:

a

and

c

are outside the function, so they only change

y

-values.

k

and

d

are inside the function, so they only change

x

-values.

Lesson summary

The template $y = af(k(x - d)) + c$ applies to any base function $f$ and uses four parameters to transform its graph.
Parameters $a$ and $c$ act on the output ( $y$ -values): $a$ stretches or compresses vertically and reflects if negative, while $c$ translates vertically.
Parameters $k$ and $d$ act on the input ( $x$ -values): $k$ stretches or compresses horizontally and reflects if negative, while $d$ translates horizontally.
The mapping rule $(x_0, y_0) \rightarrow \left(\frac{x_0}{k} + d,\; ay_0 + c\right)$ gives the exact image of any point on the base graph.
For sinusoidal functions, |a| is the amplitude, $\frac{360°}{|k|}$ is the period, $d$ is the phase shift, and $c$ is the midline value.
Always factor $k$ out of the bracket before reading the phase shift $d$ ; failing to do so is the most common source of errors in this topic.

Check your understanding

Question 1

The function

y = f(x)

has a point at

(4, 2)

. What are the coordinates of the corresponding point on

y = 3f(2(x - 1)) + 4

?

$(3, 10)$
$(9, 10)$
$(3, 2)$
$(9, 4)$

Show answer and explanation

(3, 10)

Use the mapping rule. New

x

:

\frac{4}{2} + 1 = 2 + 1 = 3

. New

y

:

3(2) + 4 = 6 + 4 = 10

. The image point is

(3, 10)

.

Question 2

Which transformation does

k = \frac{1}{3}

(with

k > 0

) produce?

A horizontal compression by a factor of $\frac{1}{3}$
A vertical stretch by a factor of $3$
A horizontal stretch by a factor of $3$
A vertical compression by a factor of $\frac{1}{3}$

Show answer and explanation

A horizontal stretch by a factor of

3

Because

k

is inside the function, it affects

x

-coordinates. The horizontal scale factor is

\frac{1}{k} = \frac{1}{\frac{1}{3}} = 3

, so every

x

-coordinate is multiplied by 3. This stretches the graph horizontally by a factor of 3.

Question 3

A sine curve has a maximum

y

-value of

7

and a minimum

y

-value of

-3

. What is the value of

c

(the midline)?

$c = 5$
$c = 7$
$c = 2$
$c = 10$

Show answer and explanation

c = 2

The midline is the average of the maximum and minimum

y

-values:

c = \frac{7 + (-3)}{2} = \frac{4}{2} = 2

.

Question 4

A student rewrites

y = \sin(3x - 60°)

as

y = \sin(3(x - 60°))

. What error did the student make?

The student reflected the graph incorrectly.
The student changed the amplitude by mistake.
The student did not factor $k$ out of the bracket correctly, giving the wrong value of $d$ .
The student applied the vertical shift to the wrong parameter.

Show answer and explanation

The student did not factor

k

out of the bracket correctly, giving the wrong value of

d

.

Factoring 3 out of

(3x - 60°)

gives

3(x - 20°)

, so

d = 20°

. The student wrote

d = 60°

by forgetting to divide 60° by

k = 3

. The correct rewritten form is

y = \sin(3(x - 20°))

.

Key terms

Base function: The simplest form of a function family, with no transformation parameters applied — for example, $f(x) = x^2$ or $f(x) = \sin(x)$ .
Transformation parameter: One of the constants $a$ , $k$ , $d$ , or $c$ in $y = af(k(x-d)) + c$ that changes how the base function's graph looks or where it sits.
Vertical stretch / compression: A change controlled by |a| that pulls the graph away from (stretch, $|a|>1$ ) or pushes it toward (compression, $|a|<1$ ) the $x$ -axis.
Horizontal stretch / compression: A change controlled by |k| that pushes the graph toward (compression, $|k|>1$ ) or pulls it away from (stretch, $|k|<1$ ) the $y$ -axis. The $x$ -coordinates are multiplied by $\frac{1}{|k|}$ .
Reflection: A flip of the graph. A negative value of $a$ reflects over the $x$ -axis; a negative value of $k$ reflects over the $y$ -axis.
Translation: A slide of the entire graph without changing its shape. $d$ controls horizontal translation and $c$ controls vertical translation.
Amplitude: For a sinusoidal function in the form $y = a\sin(k(x-d))+c$ , the amplitude is |a| — the distance from the midline to a maximum or minimum point.
Phase shift: The horizontal translation of a periodic function, given by $d$ in $y = a\sin(k(x-d))+c$ . It shows how far the standard cycle has slid left or right.

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A1.8 · Function Transformations: a, k, d and c Explained

Ontario Grade 11 Mathematics

How a, k, d, and c reshape any function

What you will learn

Prerequisite Bridge: What Is a Base Function?

The Vertical Parameters: a and c

The Horizontal Parameters: k and d

Combining All Four Parameters

Reading Parameters from a Graph or Description

Summary of the Four Transformation Parameters

Sketching a Transformed Square-Root Function

Finding the Equation of a Transformed Sine Function

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MCR3U

Tutor support for A1.8 · Investigate transformation parameters in y = af(k(x − d)) + c

About this lesson and its review