Transformations of graphs

IB Mathematics: Applications and Interpretation HL· Unit 2: Functions, Topic 3: Transformations of graphs· 20 min read

1. Translations of Graphs★☆☆☆☆⏱ 10 min

📘 Definition

Translation

Vertical: $y = f (x) + k$ , Horizontal: $y = f (x - h)$

A rigid transformation that shifts all points on the graph by the same distance in a fixed direction. It does not change the shape, size or orientation of the graph.

Example:

Shifting a parabola 2 units left and 3 units up

📐 Worked Example

Given $f (x) = x^{2}$ , write the equation of $f (x)$ translated 3 units up and 2 units to the left. State the coordinates of the new vertex.

1
Recall the horizontal translation rule: shifting 2 units left means replacing $x$ with $x + 2$ :
2
$f(x+2) = (x+2)^2$
3
A vertical translation 3 units up adds +3 outside the function:
4
Final equation is $y = f (x + 2) + 3$ :
5
$y = (x+2)^2 + 3$
6
Original vertex is $(0, 0)$ . Shifting left 2 and up 3 gives new vertex:
7
$(-2, 3)$

Exam tip:

Always confirm the direction of horizontal translations, they are the most common mistake on exam questions.

2. Reflections of Graphs★★☆☆☆⏱ 10 min

📘 Definition

Reflection

Over x-axis: $y = - f (x)$ , Over y-axis: $y = f (- x)$

A transformation that flips a graph across a given line of reflection, producing a mirror image of the original graph.

📐 Worked Example

The original function $g (x) = e^{x}$ has a y-intercept at $(0, 1)$ and horizontal asymptote $y = 0$ . Find the new intercept and asymptote after reflection over the x-axis.

1
Reflection over the x-axis multiplies all function values by -1, so the new function is:
2
$y = -g(x) = -e^x$
3
Substitute $x = 0$ to find the new y-intercept:
4
$y = -e^0 = -1$
5
The asymptote $y = 0$ remains unchanged because $- 0 = 0$ .
6
Final answer: intercept at $(0, - 1)$ , asymptote $y = 0$

Exam tip:

Label which axis you are reflecting over to avoid mixing up the transformation rule.

3. Stretches and Compressions★★★☆☆⏱ 15 min

📘 Definition

Stretch / Compression

Vertical stretch by $a$ : $y = a f (x)$ , Horizontal stretch by $b$ : $y = f (\frac{x}{b})$

A non-rigid transformation that resizes a graph proportionally along the x or y axis, changing the distance between points but not the overall shape.

📐 Worked Example

Given $h (x) = sin x$ , which has amplitude 1 and period $2 π$ , find the equation after a vertical stretch by factor 3 and a horizontal compression by factor 2. State the new amplitude and period.

1
Vertical stretch by factor 3 multiplies the entire function by 3:
2
$y = 3 h(x)$
3
A horizontal compression by factor 2 is equivalent to a horizontal stretch by $\frac{1}{2}$ , so replace $x$ with $\frac{x}{1/2} = 2 x$ :
4
$y = 3 \sin(2x)$
5
New amplitude = 3 × original amplitude = 3, new period = original period × 1/2 = $π$

4. Combined Transformations★★★★☆⏱ 15 min

When multiple transformations are applied, order matters. We always apply all transformations inside $f (x)$ (horizontal transformations) first, then stretches/compressions, then transformations outside $f (x)$ (vertical transformations), following the standard order of operations.

📐 Worked Example

Starting from $f (x) = x^{2}$ , write the equation after shifting 1 unit right, stretching vertically by factor 2, then shifting 4 units down.

1
Step 1: Shift 1 unit right (inside $f (x)$ first): replace $x$ with $x - 1$ :
2
$f(x-1) = (x-1)^2$
3
Step 2: Stretch vertically by factor 2 (multiply by 2):
4
$2(x-1)^2$
5
Step 3: Shift 4 units down (vertical transformation last): subtract 4:
6
$y = 2(x-1)^2 - 4$
7
If we had done the shift before the stretch, we would get the incorrect result $y = 2 (x - 1)^{2} - 8$ .

✓ Quick check

Test your understanding of order for combined transformations

What is the correct order of transformations to get $y = 3 f (2 x + 4) + 1$ from $y = f (x)$ ?
- A. Shift left 4, horizontal stretch by $\frac{1}{2}$ , vertical stretch by 3, shift up 1
- B. Shift right 4, horizontal stretch by 2, vertical stretch by 3, shift up 1
- C. Horizontal stretch by $\frac{1}{2}$ , shift left 2, vertical stretch by 3, shift up 1
Reveal answer
C —
Correct! First factor out the coefficient of $x$ inside $f$ : $y = 3 f (2 (x + 2)) + 1$ , so we handle the stretch first, then the shift.

Exam tip:

Always factor out the coefficient of $x$ inside $f (x)$ before identifying the horizontal shift, this will save you from common mistakes.

5. Common Pitfalls

Wrong move:

Interpret $f (x + 3)$ as a shift 3 units right

Why:

Changes inside $f (x)$ are opposite the sign

Correct move:

$f (x + 3)$ shifts the graph 3 units left

Wrong move:

Call $y = f (2 x)$ a horizontal stretch by factor 2

Why:

Stretch rules are swapped for horizontal transformations

Correct move:

$y = f (2 x)$ is a horizontal compression by factor 2 (stretch by $\frac{1}{2}$ )

Wrong move:

Apply vertical transformations before horizontal transformations

Why:

Order of operations requires handling inside the function first

Correct move:

Always apply all transformations inside $f (x)$ before transformations outside $f (x)$

Wrong move:

Forget to scale all y-values for a vertical stretch, only scaling the leading term

Why:

Every point on the graph is stretched vertically, not just the curve shape

Correct move:

Multiply all y-coordinates, including intercepts, by the vertical stretch factor

Wrong move:

Mix up reflection rules: $y = - f (x)$ is reflection over y-axis

Why:

Confusion between which coordinate is changed

Correct move:

$y = - f (x)$ flips the sign of y, so it is reflection over the x-axis

6. Quick Reference Cheatsheet

Transformation	Notation	Effect on $(x, y)$
$k$ units up vertical translation	$y = f(x) + k	$(x, y) \to (x, y + k)$
$h$ units right horizontal translation	$y = f (x - h)$	$(x, y) \to (x + h, y)$
$h$ units left horizontal translation	$y = f (x + h)$	$(x, y) \to (x - h, y)$
Reflection over x-axis	$y = - f (x)$	$(x, y) \to (x, - y)$
Reflection over y-axis	$y = f (- x)$	$(x, y) \to (- x, y)$
Vertical stretch by factor $a$	$y = a f (x)$	$(x, y) \to (x, a y)$
Horizontal stretch by factor $b$	$y = f (x / b)$	$(x, y) \to (b x, y)$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

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What's Next

Transformations of graphs are a foundational tool for all function work in IB AI HL. You will use them constantly when working with trigonometric, exponential and logarithmic models for real-world data, when sketching graphs for optimisation, and when solving inverse function problems. Mastery of transformations will save you time on both paper 1 and paper 2 exam questions, and is essential for internal assessment modelling tasks.