Transformations of Linear and Quadratic Graphs Revision Notes | GCSE AQA Higher Maths
y = f(x) + a shifts a graph up by a units, while y = f(x - a) translates it right by a units. Free notes with worked examples and 13 questions answered.
Updated 25 Mar 2026 · Revise Right Team
Transformations of Linear and Quadratic Graphs is part of Graphs in GCSE AQA Higher Maths.
3 min
Reading time
39
Lessons in this topic
GCSE AQA Higher
Pathway
Reviewed by Revise Right
Content review
What this lesson covers
y = f(x) + a shifts a graph up by a units, while y = f(x - a) translates it right by a units. Free notes with worked examples and 13 questions answered.
Sketch translations of a given function using function notation such as $f(x+a)$ and $f(x)+a$
Sketch reflections of a given function in the $x$-axis and $y$-axis using $-f(x)$ and $f(-x)$
Identify the effect of each transformation on key features such as intercepts and turning points
Apply combined knowledge of translations and reflections to describe or produce transformed graphs
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Use this lesson to stay inside Graphs while connecting the detail back to the wider Algebra topic.
Translation: A transformation that shifts every point on a graph by the same vector, without changing its shape or orientation. Described using $f(x+a)$ or $f(x)+a$.
Reflection: A transformation that flips a graph about a mirror line. $-f(x)$ reflects in the $x$-axis; $f(-x)$ reflects in the $y$-axis.
Function notation: A compact way of writing transformations: $f(x)$ names the original function, and modifications inside or outside the brackets describe how the graph moves.
Revision notes
Read the core explanations from Transformations of Linear and Quadratic Graphs before testing yourself from memory.
1. Why Transformations Matter
You already know how to sketch graphs like $y = x^2$ or $y = \sin x$. Transformations let you move and flip those graphs without starting from scratch — and Higher-tier examiners test this skill directly.
The key insight is this: every transformation rule is written in function notation, and the position of the change — inside or outside the brackets — tells you exactly what happens to the graph.
2. Translations — Shifting the Graph
A translation moves every point on the graph by the same amount in the same direction. Think of picking up the whole curve and sliding it across the axes — the shape stays identical.
The outside change is intuitive: adding 3 to every output lifts the whole graph 3 units. The horizontal case is where students consistently lose marks, so pay close attention to what follows.
Here is why the direction feels backwards: $f(x + 3) = 0$ is satisfied when $x + 3$ equals whatever made $f$ zero before — so the root moves 3 units to the left, not the right.
Worked example: Sketching a Vertical Translation
The graph of $y = f(x)$ has a turning point at $(2, -3)$ and crosses the $x$-axis at $x = 0$ and $x = 4$. Sketch $y = f(x) + 5$, labelling the new turning point and $x$-intercepts.
$f(x) + 5$ — the $+5$ is outside the brackets — Outside the brackets means a vertical shift. [M1] for recognising this.
Add $5$ to every $y$-value. Graph shifts up 5 units. — Positive value outside brackets → upward shift.
$(2,\ -3) \rightarrow (2,\ -3 + 5) = (2,\ 2)$ — The $x$-coordinate of the turning point is unchanged. [A1] for correct new turning point.
Original roots at $x = 0$ and $x = 4$ had $y = 0$. New $y$-values at those $x$-positions: $0 + 5 = 5
eq 0$. Solve $f(x) + 5 = 0 \Rightarrow f(x) = -5$. The curve no longer crosses the $x$-axis at $x=0$ or $x=4$. — The old intercepts are lifted to $y = 5$. New intercepts exist where $f(x) = -5$, which requires further information about $f$. On a sketch, show the curve crossing the $x$-axis at two new points consistent with the shifted parabola shape.
Draw the same shape as $f(x)$, shifted 5 units upward. Label turning point $(2, 2)$. — The shape is preserved — only the position changes. [A1] for correct shape and labelled turning point.
Answer: Turning point at $(2, 2)$; same curve shape shifted 5 units upward.
Worked example: Sketching a Horizontal Translation
The graph of $y = f(x)$ passes through $(0, 1)$ and has a minimum at $(3, -2)$. Sketch $y = f(x - 4)$, labelling the new minimum and the new $y$-intercept.
$f(x - 4)$ — the $-4$ is inside the brackets — Inside the brackets means a horizontal shift. [M1]
$f(x - 4)$: subtract inside → shift right by 4 — The sign is $-4$, so the graph moves right. Many students flip this — remember: $f(x-a)$ moves right by $a$.
$(3,\ -2) \rightarrow (3 + 4,\ -2) = (7,\ -2)$ — Add 4 to the $x$-coordinate; $y$-coordinate unchanged. [A1]
Set $x = 0$: $y = f(0 - 4) = f(-4)$. Without more information about $f$, mark the $y$-intercept as $f(-4)$ on the sketch. — The original $y$-intercept $(0, 1)$ moves to $(4, 1)$. The new $y$-intercept requires evaluating $f(-4)$.
Draw the same shape as $f(x)$, shifted 4 units to the right. Label minimum $(7, -2)$. — Shape is preserved. [A1] for correct labelled minimum.
Answer: Minimum at $(7, -2)$; original $y$-intercept $(0,1)$ moves to $(4, 1)$.
3. Reflections — Flipping the Graph
Now that translations are clear, reflections follow a similar inside-versus-outside logic. A reflection flips the graph about a mirror line — the shape is preserved but the orientation changes.
There are two reflections you need for Higher-tier: one in the $x$-axis and one in the $y$-axis. The position of the negative sign tells you which one you have.
Worked example: Sketching a Reflection in the $x$-axis
The graph of $y = f(x)$ has a minimum at $(-1, -4)$, crosses the $x$-axis at $x = -3$ and $x = 1$, and has $y$-intercept $(0, -3)$. Sketch $y = -f(x)$, labelling all key features.
$-f(x)$: negative sign is outside the brackets — Outside and negative → reflection in the $x$-axis. [M1]
$(-1,\ -4) \rightarrow (-1,\ -(-4)) = (-1,\ 4)$ — Negate the $y$-coordinate. The minimum becomes a maximum. [A1]
$x = -3$ and $x = 1$ remain unchanged (since $y = 0$ and $-(0) = 0$) — Points on the $x$-axis are fixed under this reflection.
Draw a curve with maximum at $(-1, 4)$, crossing the $x$-axis at $x = -3$ and $x = 1$, passing through $(0, 3)$. — The original U-shape becomes an inverted U-shape (arch). Shape preserved, orientation flipped.
Answer: Maximum at $(-1, 4)$; $x$-intercepts at $x = -3$ and $x = 1$ (unchanged); $y$-intercept at $(0, 3)$.
Worked example: Sketching a Reflection in the $y$-axis
The graph of $y = f(x)$ passes through $(2, 5)$, $(4, 0)$, and $(0, 3)$. Sketch $y = f(-x)$, labelling the images of these three points.
$f(-x)$: negative sign is inside the brackets — Inside and negative → reflection in the $y$-axis. [M1]
$(x, y) \rightarrow (-x, y)$ for each point — Negate the $x$-coordinate; $y$-coordinate is unchanged.
$(2,\ 5) \rightarrow (-2,\ 5)$ — [A1] for correct image.
$(4,\ 0) \rightarrow (-4,\ 0)$ — This was an $x$-intercept; it moves to the other side of the $y$-axis.
$(0,\ 3) \rightarrow (0,\ 3)$ — unchanged — The $y$-intercept is fixed under a reflection in the $y$-axis because $-0 = 0$. [A1]
Answer: Images: $(-2, 5)$, $(-4, 0)$, and $(0, 3)$ (unchanged).
4. Putting It Together — Identifying Transformations from a Description
Examiners often give you a transformed graph and ask you to write the transformation in function notation, or they describe the transformation in words and ask you to write the equation. Both directions require the same understanding — inside versus outside, and the sign rule for horizontal shifts.
A reliable decision process: first check whether the change is to the shape's position vertically or horizontally, then check whether the graph has been flipped. That narrows you to one of the four rules every time.
The blue curve above is $f(x) = x^2 - 2x - 3$. The second curve shows $f(x) + 4$ — shifted 4 units up. The third shows $f(x+3)$ — shifted 3 units to the left. Notice how the shape is identical in all three cases.
5. Exam Technique and Common Mistakes
Transformation questions on Higher-tier carry marks for both the correct shape and correctly labelled key features. A sketch without labels will lose accuracy marks even if the shape is right.
The command word here is sketch — you are not expected to plot exact points from a table of values. You are expected to show the correct shape, the correct position, and labelled intercepts or turning points. Precision matters for the labels, not for the curve itself.
When two transformations are combined — as in $-f(x-2)$ above — apply the one inside the brackets first, then the one outside. This ordering keeps your working consistent with how function notation is read.
Translations and reflections together cover the full Higher-tier requirement for sketching transformations of a given function. The inside-versus-outside rule and the sign direction for horizontal shifts are the two ideas that unlock every question in this topic.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-25.
y = f(x) + a is a vertical translation: adding a outside the brackets shifts the graph up by a units. For example, if f(x) has a turning point at (2, -3), then f(x) + 5 has its turning point at (2, 2), with the same shape.
What does f(x - a) do to a graph?
y = f(x - a) is a horizontal translation right by a units, because a change inside the brackets moves the graph horizontally. For example, a minimum at (3, -2) on f(x) moves to (7, -2) on f(x - 4).
What is the difference between -f(x) and f(-x)?
-f(x) reflects the graph in the x-axis by negating every y-value, so the x-intercepts stay fixed. f(-x) reflects the graph in the y-axis by negating every x-value, so the y-intercept stays fixed.
What happens to a minimum point when a graph is reflected in the x-axis?
Reflecting in the x-axis with -f(x) negates the y-coordinate, so a minimum becomes a maximum. For example, a minimum at (-1, -4) becomes a maximum at (-1, 4), while the x-intercepts at x = -3 and x = 1 stay the same.
How do you apply combined graph transformations?
When two transformations are combined, as in -f(x - 2), apply the one inside the brackets first, then the one outside. Label the key features, such as intercepts and turning points, because a sketch without labels loses accuracy marks.
Which way does f(x + a) move a graph?
f(x + a) moves a graph left by a units when a is positive, the opposite of what the sign suggests. This is because f(x + 3) = 0 when x + 3 equals the old root, so each root moves 3 units to the left.
What stays the same when a graph is reflected in the x-axis?
When a graph is reflected in the x-axis to give -f(x), the x-intercepts stay fixed because negating y = 0 still gives 0. Every other y-value is negated, so a minimum at (-1, -4) becomes a maximum at (-1, 4).
What stays the same when a graph is reflected in the y-axis?
When a graph is reflected in the y-axis to give f(-x), the y-intercept stays fixed because -0 = 0. Every other x-value is negated, so (2, 5) maps to (-2, 5) and (4, 0) maps to (-4, 0).
What happens to the turning point when a graph is translated up?
When a graph is translated up with f(x) + a, the turning point keeps its x-coordinate and a is added to its y-coordinate. For f(x) + 5, a turning point at (2, -3) moves to (2, -3 + 5) = (2, 2).
What happens to the turning point after f(x - 4)?
After f(x - 4), the turning point moves 4 units right, so 4 is added to its x-coordinate and its y-coordinate stays the same. A minimum at (3, -2) moves to (7, -2), and the point (0, 1) moves to (4, 1).
Does a translation change the shape of a graph?
A translation does not change the shape of a graph; it moves every point by the same amount in the same direction. For f(x) = x² - 2x - 3, the curves f(x) + 4 and f(x + 3) have exactly the same shape, just in a new position.
What is the inside versus outside rule for transformations?
A change outside the brackets, such as f(x) + a or -f(x), acts on the y-values and moves or flips the graph vertically. A change inside the brackets, such as f(x + a) or f(-x), acts on the x-values and moves or flips it horizontally.
What should you label when sketching a transformed graph?
A sketch of a transformed graph should show the correct shape and position with labelled intercepts and turning points. A sketch without labels loses accuracy marks even if the shape is right, but exact points from a table of values are not needed.