SL 2.11 Transform graphs by translations, reflections, and stretches | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Functions

Translations, reflections, and stretches for IB Mathematics: Analysis and Approaches SL

A graph transformation changes a graph’s position, orientation, or scale. The starting graph is often written as

y=f(x)

, where

f

is a rule and

x

is an input. For instance, if

f(x)=x^2

, then

y=f(x)

is the graph of

y=x^2

. Before transforming a graph, it helps to recall that replacing the input changes horizontal features, while multiplying the output changes vertical features. We will use that distinction to predict how points, intercepts, and turning points move.

What you will learn

Describe how translations, reflections, and stretches change a graph.
Use function notation to write transformed graphs and identify their key features.
Relate points on an original graph to corresponding points on a transformed graph.
Use a graphing tool to check a transformation while explaining it mathematically.

1. Prior knowledge: inputs, outputs, and points

A point

(u,v)

on

y=f(x)

means that the input

u

produces the output

v

, so

v=f(u)

. This point-based view is useful because a transformation can be described by tracking where each original point goes.

For example, adding a constant to the output moves every point vertically. Replacing

x

by

x-h

changes which input produces a given output, so it moves horizontal features. The signs can feel counterintuitive:

f(x-h)

shifts the graph right by

h

, not left.

y=f(x)

On $y=f(x)$ , $x$ is the input and $y$ is the output.
The point $(u,f(u))$ lies on the original graph.
Check horizontal changes by tracking inputs; check vertical changes by tracking outputs.

2. Translations and reflections

A translation moves every point the same distance without changing the graph’s shape. The graph

y=f(x-h)+k

is the graph of

y=f(x)

shifted right by

h

and up by

k

. A negative value of

h

shifts it left; a negative value of

k

shifts it down.

Reflections reverse a graph across a line. The graph

y=-f(x)

reflects across the horizontal axis, since each output changes sign. The graph

y=f(-x)

reflects across the vertical axis, since each input changes sign. If the original graph contains

(u,v)

, these reflections send it to

(u,-v)

and

(-u,v)

respectively.

A horizontal reflection and a horizontal translation can be described together using

f(-(x-h))

: the reflection is across the vertical line

x=h

. It is safest to identify the expression inside

f

and test a point rather than relying on a verbal description alone.

y=f(x-h)+k

$f(x-h)+k$ shifts right by $h$ and up by $k$ .
$-f(x)$ reflects across the horizontal axis.
$f(-x)$ reflects across the vertical axis.

3. Stretches and a point-mapping method

A vertical stretch by factor

a

multiplies every output by

a

, giving

y=af(x)

. When

a>1

, distances from the horizontal axis increase; when

0<a<1

, they decrease. A negative multiplier also reflects the graph across that axis.

Horizontal stretches are controlled by the input. For

y=f(x/s)

, where

s>0

, each original horizontal coordinate is multiplied by

s

. This is a horizontal stretch by factor

s

when

s>1

, and a horizontal compression when

0<s<1

. In contrast,

f(sx)

scales horizontal distances by

1/s

for

s>0

.

A combined form is

y=A f((x-h)/s)+k

, with

s>0

. If

(u,v)

is on the original graph, it becomes

(h+su, hinspace k+Av)

. This mapping makes the effects explicit: horizontal scale and shift act on the input coordinate, and vertical scale and shift act on the output coordinate. The domain also changes: original inputs

u

become

x=h+su

.

To include a horizontal reflection, change the input to

-(x-h)/s

. A graphing tool can help verify the direction and scale, but the point mapping explains why the result has that shape.

y=A f\left(\frac{x-h}{s}\right)+k

$af(x)$ scales outputs; $f(x/s)$ scales horizontal distances by $s$ .
Use an original point $(u,v)$ to track the transformed point.
Consider the original domain when deciding which transformed inputs are allowed.

4. Graphing technology and exam-style reasoning

For a technology check, enter the original and transformed functions as separate graphs and use the same viewing window. Choose a clear scale on both axes. Compare recognisable features such as a vertex, intercept, or endpoint with the point mapping. If the graph does not match your prediction, check the signs inside the function and whether a horizontal factor has been inverted.

A calculator display is a check, not a justification. In a written solution, state the transformation and show how at least one feature or point moves. For a restricted-domain function, enter or observe the domain carefully: a window may hide part of the graph, and a plotted curve should not be assumed to continue beyond its defined inputs.

In an exam-style question, it is efficient to identify the original function, write the transformed expression, state the shifts or scales, and then give the requested feature. When several changes occur, use the point mapping to avoid mixing their effects.

Use technology to compare predicted and displayed features.
State the mathematical reason for a graph feature; do not rely on the screen alone.
Check domains and choose a viewing window that shows the relevant parts.

Worked example

Translate a parabola

The original graph is

y=f(x)=x^2

. Describe and sketch

y=(x-2)^2-3

by tracking its vertex and two further points.

Identify the translation
The expression is $f(x-2)-3$ . Compared with $f(x)$ , it is shifted right by $2$ and down by $3$ .
$y=f(x-2)-3$
Track useful points
The original vertex $(0,0)$ moves to $(2,-3)$ . The original points $(1,1)$ and $(-1,1)$ move to $(3,-2)$ and $(1,-2)$ , respectively.
$(u,v)\mapsto(u+2,v-3)$
Sketch and check
Plot the three transformed points and draw the upward-opening parabola through them. A graphing tool should show its vertex at $(2,-3)$ and symmetry about the vertical line $x=2$ .

Answer: The graph is the original parabola translated right by

2

and down by

3

. Its vertex is

(2,-3)

.

Check: At the vertex input

x=2

, the transformed output is

(2-2)^2-3=-3

, as predicted.

Worked example

Reflect and stretch an absolute-value graph

Let

f(x)=|x|

. Describe

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

and find its vertex and the images of

(1,1)

and

(-1,1)

.

Read the horizontal change
The input is $x/3$ , so the graph is stretched horizontally by factor $3$ . The points’ horizontal coordinates are multiplied by $3$ .
$(u,v)\mapsto(3u,v)$
Read the vertical changes
Multiplication by $-2$ stretches vertical distances by factor $2$ and reflects across the horizontal axis. Adding $1$ then shifts the result up by $1$ .
$(x,y)\mapsto(x,-2y+1)$
Combine the point changes
The vertex $(0,0)$ stays at horizontal coordinate $0$ and moves to output $1$ . The point $(1,1)$ goes to $(3,-1)$ , while $(-1,1)$ goes to $(-3,-1)$ .
$(u,v)\mapsto(3u,-2v+1)$

Answer: The graph is a horizontally stretched absolute-value graph, reflected across the horizontal axis, vertically stretched by factor

2

, and shifted up by

1

. Its vertex is

(0,1)

.

Check: Substitution gives

y=-2|0/3|+1=1

at the vertex, and

y=-2|3/3|+1=-1

at

x=3

.

Worked example

Combine a reflection, stretch, and translation

The original graph

y=f(x)

contains

(-2,3)

and

(1,-1)

. Find the corresponding points on

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

.

Match the input to an original input
For an original input $u$ , set $-(x-1)/2=u$ . Solving gives $x=1-2u$ , so the horizontal change reflects the inputs and stretches their distances from $x=1$ by factor $2$ .
$x=1-2u$
Transform the outputs
An original output $v$ becomes $2v+4$ . This is a vertical stretch by factor $2$ followed by a shift up by $4$ .
$y=2v+4$
Apply the mapping
For $(-2,3)$ , the new coordinates are $x=1-2(-2)=5$ and $y=2(3)+4=10$ . For $(1,-1)$ , they are $x=1-2(1)=-1$ and $y=2(-1)+4=2$ .
$(u,v)\mapsto(1-2u,2v+4)$

Answer: The corresponding points are

(5,10)

and

(-1,2)

. The input reflection is about the vertical line

x=1

, with horizontal stretch factor

2

.

Check: Substituting

x=5

makes the function input

-(5-1)/2=-2

; substituting

x=-1

makes it

-(-1-1)/2=1

. These are the stated original inputs.

Common mistakes and how to avoid them

Treating

f(x-4)

as a shift left by

4

.

Correction: It shifts right by

4

. The input

x-4

equals the original input

u

when the new coordinate is

x=u+4

.

Calling

f(3x)

a horizontal stretch by factor

3

.

Correction: For positive

3

,

f(3x)

scales horizontal distances by

1/3

, so it is a horizontal compression. Use

f(x/3)

for a stretch by factor

3

.

Applying a vertical reflection to the input rather than the output.

Correction:

-f(x)

reflects outputs across the horizontal axis;

f(-x)

reflects inputs across the vertical axis.

Using a transformation rule without checking which points are in the original domain.

Correction: A transformed point comes from an original input that is allowed for

f

. Map the domain along with the graph.

Lesson summary

Translations change position: $f(x-h)+k$ shifts right by $h$ and up by $k$ .
Reflections change orientation: $-f(x)$ reflects across the horizontal axis and $f(-x)$ across the vertical axis.
Vertical factors multiply outputs; horizontal factors inside the function scale input distances in the reciprocal way.
A point mapping is a reliable way to combine transformations and check graph features.

Check your understanding

Question 1

The point

(2,5)

lies on

y=f(x)

. What point does it become on

y=f(x+3)-2

?

$(5,3)$
$(-1,3)$
$(-1,7)$
$(5,7)$

Show answer and explanation

(-1,3)

The graph shifts left by

3

and down by

2

, so

(2,5)

becomes

(-1,3)

.

Question 2

Which transformation does

y=f(x/4)

apply to

y=f(x)

?

Horizontal stretch by factor $4$
Horizontal compression by factor $4$
Vertical stretch by factor $4$
Reflection across the horizontal axis

Show answer and explanation

Horizontal stretch by factor

4

An original input

u

appears at the new coordinate

x=4u

, so horizontal distances are multiplied by

4

.

Question 3

The original point

(3,-2)

is on

y=f(x)

. Find its image on

y=-f(x)+1

.

$(3,-1)$
$(-3,3)$
$(3,3)$
$(4,-2)$

Show answer and explanation

(3,3)

The input stays the same, while the output becomes

-(-2)+1=3

. The image is

(3,3)

.

Key terms

Translation: A movement that shifts every point of a graph the same distance and direction.
Reflection: A reversal of a graph across a line, such as the horizontal or vertical axis.
Stretch: A change that multiplies distances from an axis or line by a fixed factor.
Domain: The set of input values for which a function is defined.

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Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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SL 2.11 · Transform graphs by translations, reflections, and stretches

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Translations, reflections, and stretches for IB Mathematics: Analysis and Approaches SL

What you will learn

1. Prior knowledge: inputs, outputs, and points

2. Translations and reflections

3. Stretches and a point-mapping method

4. Graphing technology and exam-style reasoning

Translate a parabola

Reflect and stretch an absolute-value graph

Combine a reflection, stretch, and translation

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review