Transformations of graphs of functions

IB Mathematics: Analysis and Approaches SL· 35 min read

1. Vertical Transformations★★☆☆☆⏱ 10 min

📘 Definition

Vertical Transformation

Original function $y = f (x)$

Algebraic changes outside the function $f (x)$ modify the output $y$ -value, resulting in transformation parallel to the $y$ -axis.

Example:

$y = 2 f (x)$ , $y = f (x) + 3$ , $y = - f (x)$

All vertical transformations behave intuitively, matching the sign of the change. Adding a constant shifts the graph up, subtracting shifts it down. Multiplying by a constant greater than 1 stretches vertically, between 0 and 1 compresses vertically. Multiplying by $- 1$ reflects over the $x$ -axis.

📐 Worked Example

Original function is $f (x) = x^{2}$ . Write the equation and describe the transformation for $y = 3 f (x) - 2$ .

1
1. Identify transformations: the coefficient 3 outside $f (x)$ is a vertical stretch, the $- 2$ is a vertical shift.
2
1. Apply the vertical stretch by scale factor 3 parallel to the $y$ -axis:
3
$y = 3x^2$
4
1. Subtract 2 to apply the shift down 2 units:
5
$y = 3x^2 - 2$
6
Final result: original parabola stretched vertically by factor 3, then shifted down 2 units.

Exam tip:

Changes outside the $f (\cdot)$ bracket always affect the $y$ -coordinate (vertical direction).

2. Horizontal Transformations★★★☆☆⏱ 15 min

📘 Definition

Horizontal Transformation

Original function $y = f (x)$

Algebraic changes inside the function argument modify the input $x$ -value, resulting in transformation parallel to the $x$ -axis, opposite in direction to the sign.

Example:

$y = f (2 x)$ , $y = f (x + 3)$ , $y = f (- x)$

Horizontal transformations are counter-intuitive: adding a positive constant to $x$ shifts the graph left, subtracting shifts it right. Multiplying $x$ by a constant greater than 1 compresses horizontally by the reciprocal of the constant, between 0 and 1 stretches horizontally. Multiplying by $- 1$ reflects over the $y$ -axis.

📐 Worked Example

Original function is $f (x) = sin x$ . Describe the transformation from $f (x)$ to $g (x) = sin (2 x - \frac{π}{3})$ .

1
1. Factorize the expression inside the sine function to isolate transformations of $x$ :
2
$2x - \frac{\pi}{3} = 2\left(x - \frac{\pi}{6}\right)$
3
1. Identify the two transformations: factor 2 inside gives horizontal compression, $- \frac{π}{6}$ gives a horizontal shift.
4
1. First transformation: horizontal compression by scale factor $\frac{1}{2}$ parallel to the $x$ -axis.
5
1. Second transformation: horizontal shift to the right by $\frac{π}{6}$ units.

3. Reflections★★☆☆☆⏱ 8 min

Reflections flip a graph over an axis, and are one of the simplest transformation types. There are two core reflections you need to know for IB exams:

Reflection over the $x$ -axis: $y = - f (x)$ , flips the sign of all $y$ -values, every point $(a, b)$ becomes $(a, - b)$
Reflection over the $y$ -axis: $y = f (- x)$ , flips the sign of all $x$ -values, every point $(a, b)$ becomes $(- a, b)$

📐 Worked Example

The point $(2, - 3)$ lies on the graph of $y = f (x)$ . Find the coordinates of the transformed point after reflection over the $y$ -axis.

1
Reflection over the $y$ -axis means the new function is $y = f (- x)$ .
2
To get the original output $y$ , set $- x = 2$ , so $x = - 2$ .
3
The $y$ -value stays unchanged, so $y = - 3$ .
4
Final transformed point is $(- 2, - 3)$ .

4. Combining Multiple Transformations★★★★☆⏱ 20 min

When applying more than one transformation, the order of application changes the final result. Follow this simple rule: always apply stretches and reflections first, then shifts. For transformations inside the function, factor the argument to correctly identify the order and size of each transformation.

📐 Worked Example

Starting from $y = x^{2}$ , write the equation of the graph after: vertical stretch by factor 2, reflection over the $x$ -axis, shift up 1 unit, shift left 3 units.

1
1. Start with the original function, define $f (x) = x^{2}$ .
2
1. Apply vertical stretch by factor 2: $y = 2 f (x) = 2 x^{2}$ .
3
1. Apply reflection over $x$ -axis: multiply by $- 1$ : $y = - 2 x^{2}$ .
4
1. Apply shift up 1 unit: add 1 outside the function: $y = - 2 x^{2} + 1$ .
5
1. Apply shift left 3 units: replace $x$ with $(x + 3)$ : $y = - 2 (x + 3)^{2} + 1$ .
6
If you shifted first then stretched, you get the incorrect result $y = - 2 (x + 3)^{2} + 2$ , which has the wrong vertical shift.

✓ Quick check

Check your understanding of order:

What is the correct order to get $y = 2 f (x + 1) - 3$ from $y = f (x)$ ?
- Shift left 1, stretch vertically by 2, shift down 3
- Stretch vertically by 2, shift left 1, shift down 3
- Shift down 3, shift left 1, stretch vertically by 2
Reveal answer
1 —
Correct! The rule is: stretches/reflections first, then shifts, work from inside out, so horizontal shift before vertical transformations outside the bracket.

5. Common Pitfalls

Wrong move:

Shifting $f (x)$ right by 2 units to get $f (x + 2)$

Why:

Horizontal shifts are opposite to the sign of the change inside the bracket.

Correct move:

Shifting right by 2 units gives $f (x - 2)$ .

Wrong move:

Writing $f (2 x - 6)$ as $f (2 (x - 6))$ and shifting right 6 units

Why:

You must factor out the coefficient of x to find the correct shift size.

Correct move:

Factor to get $f (2 (x - 3))$ , so the shift is right 3 units.

Wrong move:

Applying vertical shift before vertical stretch, getting $y = 2 f (x) + 6$ for 'stretch by 2, shift up 3'

Why:

Stretches must always be applied before shifts to avoid scaling the shift.

Correct move:

Apply stretch first: $y = 2 f (x)$ , then add shift: $y = 2 f (x) + 3$ .

Wrong move:

Confusing reflections: writing $y = f (- x)$ for reflection over the x-axis

Why:

Negative sign outside affects y (x-axis reflection), negative inside affects x (y-axis reflection).

Correct move:

Reflection over x-axis is $y = - f (x)$ , reflection over y-axis is $y = f (- x)$ .

Wrong move:

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

Why:

The coefficient of x is the reciprocal of the horizontal scale factor.

Correct move:

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

6. Quick Reference Cheatsheet

Transformation	Algebraic Form	Effect
Vertical shift up $k$	$y = f (x) + k$	$(a, b) \to (a, b + k)$
Vertical shift down $k$	$y = f (x) - k$	$(a, b) \to (a, b - k)$
Horizontal shift left $k$	$y = f (x + k)$	$(a, b) \to (a - k, b)$
Horizontal shift right $k$	$y = f (x - k)$	$(a, b) \to (a + k, b)$
Vertical stretch by $k$	$y = k f (x)$	All $y$ -values $\times k$
Horizontal stretch by $k$	$y = f (\frac{x}{k})$	All $x$ -values $\times k$
Reflect over x-axis	$y = - f (x)$	Flip over x-axis
Reflect over y-axis	$y = f (- x)$	Flip over y-axis

7. Frequently Asked

Does the order of transformations matter?

Yes, order matters when combining transformations on the same axis, or when mixing stretches and shifts. Always apply stretches/reflections before shifts, and work inside-out from the function brackets.

Why is horizontal shift opposite to the sign?

For $f (x + a)$ with $a > 0$ , we replace $x$ with $x + a$ , so the graph reaches the same output $y$ $a$ units earlier than the original, hence shifting left by $a$ units, not right.

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.

2022 · 1
Find transformed quadratic equation
2023 · 2
Sketch transformed exponential graph
2021 · 1
Combine reflection and stretch

What's Next

Mastering transformations of graphs is a foundational skill you will use across the entire IB AA SL course. You will apply it when working with composite and inverse functions, writing quadratic functions in vertex form, and graphing transformed trigonometric functions. It also makes sketching unfamiliar functions for calculus problems much faster, and helps you interpret how changing parameters changes a function's behavior. This topic builds directly on core function concepts, and leads into more advanced graphing skills you will need for exam questions across both papers.