Module 1 · L9 of 15~35 min⚡ +50 XP in Learn · +25 to complete
Translations of Functions
When you drag an app icon across your phone screen, nothing about the icon itself changes, it simply moves. The same thing happens with functions. A translation slides the entire graph up, down, left, or right without stretching or flipping it.
Today's hook, The graph of $y = x^2$ has its vertex at $(0, 0)$. If you wanted to move this parabola so its vertex is at $(2, 3)$, what changes would you make to the equation? Would you add or subtract numbers, and where would they go?
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Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Worksheets
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
The graph of $y = x^2$ has its vertex at $(0, 0)$. Imagine you wanted to move this parabola so its vertex is at $(2, 3)$. What changes would you need to make to the equation? Would you add or subtract numbers, and where would they go, inside or outside the squared term?
Key insight: Horizontal shifts are counter-intuitive, $f(x - h)$ moves right by $h$, while $f(x + h)$ moves left by $h$.
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What you'll master
Know
Key facts
$f(x) + k$ shifts the graph vertically
$f(x - h)$ shifts the graph horizontally
Translations do not change the shape of the graph
Understand
Concepts
Why horizontal shifts behave "backwards" from intuition
How translations affect key features (vertex, intercepts, asymptotes)
How to read the translation from an equation in vertex form
Can do
Skills
Sketch translated graphs from their equations
Write the equation of a translated graph
Determine the new coordinates of key points after translation
Find the range and domain of translated functions
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Key terms
Translation
A transformation that slides every point on a graph by the same distance in the same direction.
Vertical Translation
Adding $k$ outside the function, shifts the graph up ($k > 0$) or down ($k < 0$).
Horizontal Translation
Adding $h$ inside the function, shifts the graph right ($h > 0$ in $x - h$) or left.
Vertex Form
$y = a(x - h)^2 + k$, vertex is at $(h, k)$, revealing the translation directly.
Domain
Horizontal shifts change the domain; vertical shifts do not.
Range
Vertical shifts change the range; horizontal shifts do not.
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Establish it yourself with a graphing application
Work through the core explanation before applying it.
04b
Establish it yourself with a graphing application
+10 XP
The syllabus dot-point for this lesson begins "Establish using graphing applications", so this is the activity that earns it rather than being told the rule. Use Desmos, GeoGebra or any graphing calculator.
Task. Graph $y = x^2$, then $y = (x-3)^2$, then $y = (x-3)^2 + 2$, all on the same axes.
Predict before you plot. Before you look: predict which way each change moves the parabola, and by how much. Then check. Pay attention to the sign: does $(x-3)^2$ move it left or right?
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Vertical and horizontal translations
core concept · +3 XP at end
A translation slides every point on a graph by the same distance in the same direction. The shape, size, and orientation of the graph do not change, only its position.
Vertical translations: $y = f(x) + k$
When you add or subtract a constant outside the function, the entire graph moves up or down:
$y = f(x) + 3$ shifts the graph 3 units up
$y = f(x) - 2$ shifts the graph 2 units down
Every point $(x, y)$ on the original graph moves to $(x, y + k)$. The $x$-coordinates stay the same; only the $y$-coordinates change.
Horizontal translations: $y = f(x - h)$
When you add or subtract a constant inside the function, the graph moves left or right. This is where students often make mistakes, because the direction is opposite to what the sign suggests:
$y = f(x - 3)$ shifts the graph 3 units to the right
$y = f(x + 2)$ shifts the graph 2 units to the left
Think of it this way: if you want $f(x - 3)$ to produce the same output as $f(0)$, you need $x = 3$. So the feature that was at $x = 0$ has now moved to $x = 3$, to the right.
The "opposite direction" trap. Many students see $f(x + 2)$ and assume it shifts right by 2. It does not. The $+2$ is inside the brackets with $x$, and it shifts left by 2. A simple memory trick: set the inside equal to zero, $x + 2 = 0$ gives $x = -2$, so the reference point moves to $-2$, which is on the left side.
Combined translations
When both transformations appear together, handle them one at a time:
$$y = f(x - h) + k$$
Shift horizontally by $h$ units (right if $h > 0$, left if $h < 0$)
Shift vertically by $k$ units (up if $k > 0$, down if $k < 0$)
The order of these two shifts does not matter, horizontal and vertical translations commute with each other.
Misconception to fix, $(a + b)^2 \neq a^2 + b^2$. The correct expansion is $(a + b)^2 = a^2 + 2ab + b^2$. The middle term $2ab$ is essential and commonly forgotten when working with vertex form.
Pause, copy the two translation rules: $y = f(x) + k$ shifts vertically ($k > 0$ up), $y = f(x - h)$ shifts horizontally ($h > 0$ right), with the note on which changes domain and which changes range.
Did you get this? True or false: $y = f(x + 3)$ shifts the graph of $f$ to the right by 3 units.
Quick check: Which transformation maps $y = f(x)$ to $y = f(x - 2) + 5$?
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Effect on key features
core concept
We just saw that $y = f(x-h) + k$ shifts a graph $h$ units right and $k$ units up. That raises a question: if the graph moves, exactly what happens to specific features like the turning point, intercepts, and asymptotes? This card answers it → each feature's coordinates shift by exactly $(h, k)$, but $x$-intercepts only shift if there's a horizontal translation.
Translations affect different features in predictable ways:
Vertex / turning point: Moves by the same translation vector $(h, k)$
$y$-intercept: Changes with both horizontal and vertical shifts
$x$-intercept(s): Change with both horizontal and vertical shifts
Domain: Horizontal shifts move the domain left or right; vertical shifts do not change the domain
Range: Vertical shifts move the range up or down; horizontal shifts do not change the range
Vertex form reveals the translation instantly. For a parabola $y = a(x - h)^2 + k$, the vertex is at $(h, k)$. This tells you exactly how far and in what direction the basic parabola $y = ax^2$ has been translated.
Vertex/turning point: add $(h, k)$ to original vertex coordinates; Domain: horizontal shift moves domain; vertical shift does NOT change domain
Pause, copy the feature-shift rule: add $(h, k)$ to the vertex; note that horizontal shifts move the domain but vertical shifts do not into your book.
Fill the blanks: drag each token to the correct blank.
horizontalverticalrangedomain
A ___ shift changes the ___ but not the domain.
A ___ shift changes the ___ but not the range.
Worked examples · reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
Worked example 1 · describing a translation+5 XP on full reveal
Describe the transformation that maps $y = f(x)$ to $y = f(x + 4) - 1$.
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$x + 4 = x - (-4) \Rightarrow h = -4$
Identify the horizontal shift, $h = -4$ means 4 units to the left
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$k = -1$
Identify the vertical shift, $k = -1$ means 1 unit down
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Translation: 4 units left and 1 unit down ✓
State both shifts clearly
Worked example 2 · finding the new vertex+5 XP on full reveal
The vertex of $y = f(x)$ is at $(2, -3)$. Find the vertex of $y = f(x - 5) + 2$.
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Horizontal shift: $h = 5$ (5 units right). Vertical shift: $k = 2$ (2 units up)
Worked example 3 · writing the equation+5 XP on full reveal
The graph of $y = x^2$ is translated 2 units to the right and 4 units down. Write the equation of the transformed graph.
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$y = (x - 2)^2$
Apply horizontal shift, 2 units right means replace $x$ with $x - 2$
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$y = (x - 2)^2 - 4$
Apply vertical shift, 4 units down means subtract 4 from the whole function
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Final answer: $y = (x - 2)^2 - 4$ ✓
Vertex at $(2, -4)$, confirms 2 right and 4 down from $(0, 0)$
Common mistakes · the 4 traps that cost marks
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Dodge the traps, then apply
Meet the mistakes that cost marks, then do it yourself.
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Thinking $f(x + h)$ shifts right
This is the most common transformation error in all of Year 11. Students see $f(x + 3)$ and instinctively think "plus 3 means shift right 3." It does not. It shifts left 3.
✓ Fix: Always set the inside equal to zero. $x + 3 = 0$ gives $x = -3$, so the reference point moves to $-3$, on the left.
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Confusing which constant does which shift
Students sometimes think $f(x - 2) + 3$ shifts left 2 and down 3, or right 2 and down 3, or any other incorrect combination.
✓ Fix: Remember, inside = horizontal, outside = vertical. Minus inside = right, plus outside = up.
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Forgetting that horizontal shifts do not change the range
A common error is to adjust the range when a graph is shifted horizontally. Sliding left or right does not raise or lower the graph, so the $y$-values do not change.
✓ Fix: Only vertical shifts affect the range. Only horizontal shifts affect the domain.
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Writing the transformed equation in the wrong form
When asked for the equation after translation, some students expand the brackets unnecessarily. While expanded form is correct, factorised/vertex form is usually more useful for transformations.
✓ Fix: Leave your answer in the form $y = f(x - h) + k$ or $y = a(x - h)^2 + k$ unless the question asks for expansion.
Activity 1, Describe the translation
For each equation, describe how the graph of $y = f(x)$ has been translated. State the direction and number of units.
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$y = f(x) + 5$
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$y = f(x - 4)$
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$y = f(x + 2) - 3$
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$y = f(x - 1) + 6$
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Match: Which equation produces a shift of 2 units left and 3 units down?
Quick-fire practice · 5 reps +2 XP per reveal
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Drill it, then lock it in
Run the quick drill and copy the summary into your book.
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Describe the translation: $y = f(x) + 5$.
5 units up. Adding 5 outside the function is a vertical shift.
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Describe the translation: $y = f(x - 4)$.
4 units right. $x - 4$ inside the brackets shifts right by 4.
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The vertex of $f$ is at $(1, 4)$. Find the vertex after $y = f(x + 2) - 5$.
Write the equation of $y = x^2$ translated 3 units to the right and 4 units up.
$y = (x - 3)^2 + 4$. Right 3: replace $x$ with $x - 3$. Up 4: add 4 outside.
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Does shifting $y = \sqrt{x}$ horizontally by 4 units change its range?
No. Horizontal shifts only affect the domain (the set of valid $x$-values), not the range. The range remains $[0, \infty)$. The domain shifts from $[0, \infty)$ to $[4, \infty)$ (if shifted right 4).
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Revisit your thinking
Earlier you were asked: How would you move $y = x^2$ so its vertex is at $(2, 3)$? What changes would you make to the equation?
To move the vertex from $(0, 0)$ to $(2, 3)$, you need to shift the graph 2 units to the right and 3 units up. The new equation is $y = (x - 2)^2 + 3$. The $-2$ inside the brackets causes the horizontal shift to the right, and the $+3$ outside causes the vertical shift up. This is the vertex form of a parabola, and it is one of the most useful equations in all of Year 11 mathematics because it lets you read the vertex directly from the equation.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
UnderstandBand 32 marks
Q8. Explain in words why $y = f(x - 3)$ represents a shift 3 units to the right, even though the number $-3$ appears inside the function. Use the idea of inputs and outputs in your explanation. (2 marks)
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ApplyBand 33 marks
Q9. The vertex of $y = f(x)$ is at $(-1, 2)$. Write the coordinates of the vertex after each transformation: (a) $y = f(x) + 4$ (b) $y = f(x - 3)$ (c) $y = f(x + 2) - 5$ (3 marks)
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AnalyseBand 54 marks
Q10. A parabola with equation $y = x^2$ is translated so that its vertex moves to $(4, -2)$. (a) Write the equation of the translated parabola. (b) State the domain and range of the translated parabola. (c) Explain why the domain changed or did not change, and why the range changed or did not change. (4 marks)
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Comprehensive answers (click to reveal)
Multiple choice, drill bank
MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.
1. A, $f(x) + 3$ shifts 3 units up.
2. D, $f(x - 2)$ shifts 2 units right.
3. B, Vertical shift up: $(1, 4) \to (1, 6)$.
4. B, Horizontal shift left: $(3, 5) \to (2, 5)$.
5. A, 3 right, 4 up gives vertex $(3, 4)$.
Activity 1, Describe the Translation model answers
1. 5 units up
2. 4 units right
3. 2 units left and 3 units down
4. 1 unit right and 6 units up
Short answer model answers
Q8 (2 marks): To get the same output from $f(x - 3)$ as you would from $f(0)$, you need $x - 3 = 0$, which means $x = 3$ [1]. So the point that was at $x = 0$ has moved to $x = 3$, which is 3 units to the right [1].
(a) $y = (x - 4)^2 - 2$ [1] (b) Domain: $(-\infty, \infty)$; Range: $[-2, \infty)$ [1] (c) The domain did not change because there was no horizontal restriction on the original parabola, and horizontal translations do not introduce restrictions [1]. The range changed because the vertical translation of 2 units down shifted the minimum $y$-value from $0$ to $-2$ [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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quiz
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