Function Transformations — SL 2.7 Transformations of functions (shifts, stretches, reflections) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Long before graphing calculators existed, mathematicians needed efficient ways to understand how changing an equation affected the shape and position of its curve. Rather than plotting each new function point by point, they discovered that certain algebraic changes — adding a constant, multiplying by a factor, or negating a variable — produced predictable geometric changes on the graph. These observations became the foundation of function transformations, one of the most powerful tools in all of mathematics.

1637 • Descartes and the Coordinate Plane

René Descartes published La Géométrie, introducing the coordinate system that lets us plot algebraic equations as curves — the essential prerequisite for studying transformations visually.

1748 • Euler Formalises Functions

Leonhard Euler defined the modern concept of a function in Introductio in Analysin Infinitorum, establishing the notation y = f(x) and enabling systematic study of how modifying f affects its graph.

1800s • Symmetry and Group Theory

Mathematicians like Évariste Galois and Felix Klein studied symmetry operations — translations, reflections, and rotations — as abstract algebraic structures, giving transformations a rigorous theoretical backbone.

1960s–Today • Computer Graphics & Digital Imaging

Function transformations became the mathematical engine behind computer graphics, animation, and image processing. Every time you zoom, rotate, or resize a digital image, transformation rules are at work.

The central question that function transformations answer is deceptively simple: if you already know the graph of y = f(x), how can you quickly sketch y = 2f(x − 3) + 1 without plotting dozens of individual points? Mastering this skill saves enormous time on IB exams and builds deep intuition about how equations and graphs are connected.

SECTION 2

Core Principles of Function Transformations

Every transformation you will encounter in IB SL 2.7 falls into one of three categories: translations (shifts), stretches and compressions (scaling), and reflections. The key insight is that each transformation corresponds to a specific algebraic change inside the equation y = f(x). Once you learn the pattern, you can apply it to any function — parabolas, cubics, exponentials, or trigonometric curves.

Translations (Shifts)

Adding or subtracting a constant moves the entire graph without changing its shape. Horizontal shifts are controlled inside the function argument; vertical shifts are controlled outside.

Stretches & Compressions

Multiplying by a constant scales the graph by stretching or compressing it. A factor outside f(x) scales vertically; a factor inside the argument scales horizontally.

Reflections

Negating the output (−f(x)) reflects the graph across the x-axis. Negating the input (f(−x)) reflects the graph across the y-axis.

Inside vs. Outside Rule

Changes inside the function argument affect the x-direction and work in the opposite way you might expect. Changes outside affect the y-direction and work in the intuitive direction.

✦

KEY TAKEAWAY

Think of the original function f(x) as a rubber stamp. Translations slide the stamp to a new position on the paper. Stretches enlarge or shrink the stamp. Reflections flip the stamp over. In every case, the stamp's basic design (the shape of the curve) stays recognisable — only its position, size, or orientation changes.

SECTION 3

Visualising Translations

The diagram below shows how a simple parent function y = x² is affected by vertical and horizontal translations. Notice how the shape of the parabola remains identical — only its position on the coordinate plane changes. The cyan curve is the original, the amber curve shows a vertical shift, and the pink curve shows a horizontal shift.

The cyan parabola y = x² is the parent function. Shifting vertically by +3 (amber) moves the vertex up. Shifting horizontally by replacing x with (x − 3) (pink) moves the vertex right. Note that the horizontal direction is opposite to the sign inside the bracket.

The most common mistake students make is with horizontal translations. When you see y = f(x − 3), you might assume the graph moves left because of the minus sign. In fact, the graph moves right by 3. The reasoning is straightforward: for the output to match the original f(0), you now need x = 3 rather than x = 0. Everything that used to happen at x = 0 now happens at x = 3 — hence the graph shifts right. This "opposite" behaviour for horizontal changes is one of the most important patterns to internalise.

SECTION 4

Mathematical Framework

The IB syllabus requires you to work with the general transformed function in the form shown below. Each parameter controls a specific geometric change, and the order in which you apply them matters. Let's define each parameter carefully.

GENERAL TRANSFORMATION

y = a · f(b(x − h)) + k

a = vertical stretch factor (|a| > 1 stretches, 0 < |a| < 1 compresses; a < 0 reflects across x-axis). b = horizontal stretch factor (graph is scaled by 1/|b|; b < 0 reflects across y-axis). h = horizontal translation (positive h shifts right). k = vertical translation (positive k shifts up).

VERTICAL TRANSLATION

y = f(x) + k

Every point (x, y) on the original graph moves to (x, y + k). If k > 0 the graph shifts up; if k < 0 it shifts down.

HORIZONTAL TRANSLATION

y = f(x − h)

Every point (x, y) moves to (x + h, y). If h > 0 the graph shifts right; if h < 0 it shifts left. Remember: the direction is opposite to the sign you see inside the bracket.

REFLECTIONS

y = −f(x) reflects across x-axis ; y = f(−x) reflects across y-axis

For −f(x), every y-coordinate is negated: (x, y) → (x, −y). For f(−x), every x-coordinate is negated: (x, y) → (−x, y).

💡

IB Exam Tip

On IB papers, transformations are often described using column vector notation. A translation by the vector (h, k) means the graph shifts h units horizontally and k units vertically. Always state whether the graph moves right/left and up/down.

SECTION 5

Stretches, Compressions & Reflections in Detail

While translations simply slide a graph, stretches and compressions change its proportions. A vertical stretch by factor a multiplies every y-value by a, pulling the graph away from (or squeezing it toward) the x-axis. A horizontal stretch by factor 1/b does the same in the x-direction. Reflections are a special case where the stretch factor is −1.

Comparing four transformations of y = x²: a horizontal compression (amber) makes the parabola narrower, a vertical compression (violet) makes it wider, and a reflection in the x-axis (red) flips it upside down.

Transformation	Equation Form	Effect on Points	Visual Effect
Vertical stretch by a	y = a·f(x), \|a\| > 1	(x, y) → (x, ay)	Graph pulled away from x-axis
Vertical compression by a	y = a·f(x), 0 < \|a\| < 1	(x, y) → (x, ay)	Graph squeezed toward x-axis
Horizontal stretch by 1/b	y = f(bx), 0 < \|b\| < 1	(x, y) → (x/b, y)	Graph pulled away from y-axis
Horizontal compression by 1/b	y = f(bx), \|b\| > 1	(x, y) → (x/b, y)	Graph squeezed toward y-axis
Reflection in x-axis	y = −f(x)	(x, y) → (x, −y)	Graph flipped vertically
Reflection in y-axis	y = f(−x)	(x, y) → (−x, y)	Graph flipped horizontally

Summary of stretch, compression, and reflection transformations

SECTION 6

Worked Example

Let's walk through a complete IB-style problem. Suppose the graph of y = f(x) passes through the point (2, 5). We want to find the corresponding point on the graph of y = 3f(2(x − 1)) + 4.

Finding the Image of a Point Under Combined Transformations

1
Step 1 — Identify the Transformations

Compare y = 3f(2(x − 1)) + 4 with the general form y = a·f(b(x − h)) + k. We can read off: a = 3 (vertical stretch by 3), b = 2 (horizontal compression by factor 1/2), h = 1 (shift right 1), and k = 4 (shift up 4).
2
Step 2 — Apply the Horizontal Transformations to the x-coordinate

Start with the original x-value: x = 2. For a horizontal compression by factor 1/b = 1/2, divide by b: x' = 2/2 = 1. Then apply the horizontal translation h = 1: x' = 1 + 1 = 2.

New x-coordinate: x' = 2
3
Step 3 — Apply the Vertical Transformations to the y-coordinate

Start with the original y-value: y = 5. For a vertical stretch by factor a = 3: y' = 3 × 5 = 15. Then apply the vertical translation k = 4: y' = 15 + 4 = 19.

New y-coordinate: y' = 19
4
Step 4 — State the Image Point

The point (2, 5) on y = f(x) maps to (2, 19) on y = 3f(2(x − 1)) + 4.

Image point: (2, 19)
5
Step 5 — Verify (optional but recommended)

To check, substitute x = 2 into the transformed equation: y = 3f(2(2 − 1)) + 4 = 3f(2 × 1) + 4 = 3f(2) + 4 = 3(5) + 4 = 19. ✓ This confirms our answer.

⚠

Order Matters!

Always apply stretches and reflections before translations. For the x-coordinate: divide by b, then add h. For the y-coordinate: multiply by a, then add k. Reversing the order will give an incorrect result.

SECTION 7

Common Pitfalls & Comparison of Transformations

Students often lose marks on IB exams not because they don't understand transformations in principle, but because they mix up the details. The table below highlights the most frequent errors and how to avoid them.

Pitfall	What Students Do Wrong	Correct Approach
Horizontal direction	Shift f(x − 3) to the left because of the minus sign	f(x − 3) shifts right by 3. Think: x must be 3 to get f(0).
Horizontal stretch factor	Multiply x-values by b for y = f(bx)	Divide x-values by b. The scale factor is 1/b, not b.
Order of operations	Translate before stretching	Always stretch/reflect first, then translate.
Confusing −f(x) and f(−x)	Using reflection in x-axis when y-axis is intended	−f(x) flips y-values (x-axis reflection). f(−x) flips x-values (y-axis reflection).
Invariant points	Forgetting that some points don't move	Points on the axis of reflection or with y = 0 (for vertical stretches) stay fixed.

Common IB exam pitfalls with function transformations

💡

THE INSIDE–OUTSIDE MNEMONIC

Here's a handy way to remember: changes inside the function argument (next to x) affect the horizontal direction and are counterintuitive (opposite sign or reciprocal factor). Changes outside the function affect the vertical direction and are intuitive (they do what they look like they should do).

SECTION 8

Connection to Advanced Topics

The transformation ideas you learn in SL 2.7 are not just exam content — they lay the groundwork for more advanced mathematical concepts. Understanding how functions change when you modify their equations is foundational to topics like trigonometric modelling (SL 2.11), where you use transformations to fit sine and cosine curves to real-world periodic data, and calculus, where understanding shifts and stretches helps you interpret derivatives and integrals graphically.

SL 2.7 Concept	Where It Leads
Vertical translation y = f(x) + k	Integration: adding a constant of integration C shifts the antiderivative vertically
Horizontal translation y = f(x − h)	Trigonometry: phase shift in y = sin(x − c) models time delays in waves
Vertical stretch y = af(x)	Trigonometry: amplitude in y = a sin(x) controls the height of oscillations
Horizontal stretch y = f(bx)	Trigonometry: period = 2π/b controls how fast a wave oscillates
Reflections y = −f(x), y = f(−x)	Even and odd functions: f(−x) = f(x) means symmetry about the y-axis; f(−x) = −f(x) means rotational symmetry about the origin

How SL 2.7 transformations connect to higher-level IB topics

In HL Mathematics, transformations extend to composite transformations expressed as matrices, which can handle rotations, shears, and projections in addition to the shifts, stretches, and reflections covered here. If you continue to university-level mathematics, you'll encounter linear algebra, where transformation matrices become one of the central objects of study.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why the graph of y = f(x − 4) shifts to the right by 4 units, even though the equation contains a "minus 4". Why does the horizontal direction seem to go opposite to the sign?

Show answer

In y = f(x − 4), the function f receives the input (x − 4) rather than x. For the transformed function to produce the same output as the original f(0), we need x − 4 = 0, which means x = 4. In other words, everything that used to happen at x = 0 now happens at x = 4, so the entire graph has shifted 4 units to the right. The "opposite" direction arises because the subtraction happens to the input x before the function acts on it — you need a larger x to compensate for the subtraction. This is the counterintuitive inside-the-function rule.

Question 2: BASIC CALCULATION

The point (3, −2) lies on the graph of y = g(x). Find the coordinates of the corresponding point on the graph of y = g(x) + 7.

Show answer

The transformation y = g(x) + 7 is a vertical translation upward by 7 units. Only the y-coordinate changes: the new y-value is −2 + 7 = 5. The x-coordinate stays unchanged. Therefore the corresponding point is (3, 5). This is a straightforward application of the rule (x, y) → (x, y + k) with k = 7.

Question 3: INTERMEDIATE

The graph of y = f(x) has a maximum point at (4, 6). Find the coordinates of the maximum point on the graph of y = −2f(x − 3) + 1. State whether this new point is a maximum or minimum.

Show answer

Apply transformations in order. Start with the point (4, 6). The horizontal translation x − 3 shifts right by 3, so the x-coordinate becomes 4 + 3 = 7. The vertical stretch by −2 multiplies the y-coordinate: −2 × 6 = −12. Then the vertical translation +1 adds 1 to the y-coordinate: −12 + 1 = −11. The new point is (7, −11). Because the factor a = −2 is negative, the graph has been reflected in the x-axis, so the original maximum becomes a minimum. The answer is: minimum point at (7, −11).

Question 4: APPLIED

A Ferris wheel's height above the ground (in metres) is modelled by h(t) = 10sin(t) + 12 for time t in radians. The operator wants the wheel to: (a) spin twice as fast, and (b) be mounted on a 5-metre platform. Write the new equation and identify each transformation.

Show answer

Spinning twice as fast means the period is halved, which requires a horizontal compression by factor 2. This is achieved by replacing t with 2t inside the sine function. Mounting the wheel 5 metres higher is a vertical translation up by 5. The new equation is h(t) = 10sin(2t) + 17. The transformation y = f(2t) is a horizontal compression by factor 1/2 (the wheel completes its cycle in half the original time), and the +17 (which is 12 + 5) is the combined vertical translation. The amplitude (10 m) remains unchanged, meaning the wheel's diameter hasn't changed.

Question 5: CRITICAL THINKING

The function y = f(x) satisfies f(−x) = f(x) for all x (it is an even function). If the graph of y = f(x) is shifted 3 units to the right to produce y = f(x − 3), explain whether the resulting function is still even. Justify your answer with a specific test or example.

Show answer

A function g(x) is even if g(−x) = g(x) for all x. Let g(x) = f(x − 3). Then g(−x) = f(−x − 3). For g to be even, we would need f(−x − 3) = f(x − 3) for all x. Since f is even, f(−x − 3) = f(x + 3) (replacing the argument −x − 3 with its negation using the even property). So the condition becomes f(x + 3) = f(x − 3) for all x. This would mean the function has period 6, which is not generally true for an arbitrary even function. For example, let f(x) = x². Then g(x) = (x − 3)² and g(−x) = (−x − 3)² = (x + 3)². Since (x − 3)² ≠ (x + 3)² in general (try x = 1: 4 ≠ 16), the shifted function is not even. The horizontal translation destroys the symmetry about the y-axis.

SUMMARY

Lesson Summary

Summary

Function transformations allow you to take the graph of any parent function y = f(x) and produce a new graph through predictable algebraic changes. The general form y = a·f(b(x − h)) + k encodes four transformations: a vertical stretch by factor |a| (with reflection if a < 0), a horizontal compression by factor 1/|b| (with reflection if b < 0), a horizontal translation by h units, and a vertical translation by k units.

The critical insight is the inside–outside rule: changes inside the function argument affect the x-direction and work opposite to what you might expect (subtract → shift right; multiply → compress), while changes outside affect the y-direction intuitively. Always apply stretches and reflections before translations. These principles extend directly to trigonometric modelling, calculus, and beyond.