Algebra - Graphs

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Overview

A graph transformation changes where a curve sits without changing its shape. In this Higher-tier topic, candidates translate graphs horizontally or vertically and reflect them in the coordinate axes. The original curve is called (y=f(x)), but its formula may never be given: the graph and a few labelled points can provide everything needed.

This matters because a correct transformation can secure marks quickly, whereas one sign error can move every point to the wrong place. Typical questions ask you to sketch, write down coordinates, or describe fully a transformation. The most reliable approach is to track points, including a turning point, and then reproduce the entire curve through their images.

The topic connects algebraic substitution to quadratic roots and turning points, coordinate geometry, translation vectors and trigonometric graphs. Understanding inputs and outputs explains why horizontal movement reverses the sign inside the brackets. It also helps you check answers rather than relying on a memorised picture. Work through the examples, cover each rule before recalling it, and use the podcast to practise explaining your reasoning aloud. Aim for accurate coordinates, an unchanged shape and precise mathematical language.

Specification: Pearson Edexcel GCSE Mathematics 1MA1, A13, “sketch translations and reflections of a given function”. A13 is Higher-only. Coordinate reading is prerequisite knowledge, not itself Higher-only. Pearson limits these function transformations to translations and reflections; stretches are outside this topic's specification scope. 1 2

Key Concepts

1. Function notation: separate the input from the output

The expression (f(x)) means the output of a function when its input is (x). It does not mean (f\times x). If ((2,5)) lies on (y=f(x)), then (f(2)=5): input 2 produces output 5.

Write a general original point as ((p,q)), with (f(p)=q). This is your anchor. An outside change acts on the output (q); an inside change acts on the input expression, so you solve for the new horizontal coordinate. Candidates do not need to invent an equation for an unfamiliar curve.

Cover and recall: If (f(-3)=7), what point lies on the graph? Answer before checking: ((-3,7)), not ((7,-3)).

2. Vertical translations: (y=f(x)+a)

Adding (a) outside the function adds (a) to every output. Therefore ((p,q)\mapsto(p,q+a)). For positive (a), the curve moves up; for negative (a), it moves down. The horizontal coordinate is unchanged.

Example: Under (y=f(x)+3), a maximum at ((1,5)) moves to ((1,8)). A different point ((-2,-1)) moves to ((-2,2)). Both move exactly three units upwards, and the maximum remains a maximum.

A full description is “translation by vector (\begin{pmatrix}0\3\end{pmatrix})”. A translation has a vector, not a mirror line. Think of lifting a rigid tracing of the original curve: nothing is stretched, and every point moves equally.

Why does this work? At the same input (p), the new output is (f(p)+a=q+a). This explanation is more secure than “plus always means up”, because plus inside the function behaves differently.

3. Horizontal translations: (y=f(x+a))

An inside addition moves the graph horizontally in the opposite directionto its sign. The rule is ((p,q)\mapsto(p-a,q)). Thus (f(x+2)) moves left two units, while (f(x-2)) moves right two units. The height is unchanged.

To explain the apparent reversal, preserve the old input. If the original point has input (p=1), then for (f(x+2)) the new coordinate must satisfy (x+2=1). Hence (x=-1). The function still receives 1 and produces the same old output, but it now does so two units further left.

Example:A maximum at ((1,5)) moves to ((-1,5)) under (f(x+2)), and to ((3,5)) under (f(x-2)). The corresponding vectors are (\begin{pmatrix}-2\0\end{pmatrix}) and (\begin{pmatrix}2\0\end{pmatrix}).

Check with substitution: If (f(x)=x^2), then (f(x-2)=(x-2)^2), whose minimum is at (x=2). It cannot be a move left. Apply the same reasoning when (a) is negative: subtracting a negative coordinate change reverses its sign.

4. Reflection in the x-axis: (y=-f(x))

A minus sign outside negates every output: ((p,q)\mapsto(p,-q)). This is a reflection in thex-axis, the line (y=0). Horizontal positions remain unchanged, while points above the axis move the same distance below it.

Example:((1,5)\mapsto(1,-5)), and ((-2,-3)\mapsto(-2,3)). A maximum becomes a minimum. Every x-intercept stays fixed because (-0=0).

A reflection is not a downward translation. A point at height 5 moves ten units to height -5, while one at height 2 moves four units to height -2. Their distances from the mirror line, rather than their displacement vectors, determine the image. Credit for a full description requires the correct reflection and mirror line, subject to the individual mark scheme.

5. Reflection in the y-axis: (y=f(-x))

A minus sign inside reverses the horizontal coordinate: ((p,q)\mapsto(-p,q)). This reflects the curve in they-axis, the line (x=0). Heights are unchanged. A maximum remains a maximum, although its position may change.

Example:((1,5)\mapsto(-1,5)), while ((-4,-2)\mapsto(4,-2)). Every point already on the y-axis is fixed.

For (f(x)=x^2), (f(-x)=(-x)^2=x^2), so the image looks unchanged: the original graph is symmetric about the y-axis. The transformation rule still applies. An asymmetric graph makes the reflection easier to recognise. Do not assume that an unchanged-looking graph means no reflection occurred.

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The dashed dark curve is the original; purple is its image. The atlas uses an asymmetric illustrative function with a local minimum at (1, -2), so left translation and y-axis reflection can be distinguished by the rest of the curve.

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6. Sketch the whole image, not only the turning point

Choose at least three informative points: usually a turning point and two well-separated points. For a complicated curve, three points are only a starting point. Track every turning point, visible endpoint and branch needed to reproduce the shape.

Read the axis scales first. A square may represent two units, so a shift of two coordinate units might mean one square. Transform each chosen ordered pair using one consistent rule, plot its image, and draw the same kind of curve through the new points. Use a smooth line for a smooth original, not straight segments. Retain corners if the original has corners.

For the parabola (f(x)=(x-1)^2-4), use ((-1,0)), ((1,-4)) and ((3,0)). Under (f(x-2)), their images are ((1,0)), ((3,-4)) and ((5,0)). The entire parabola moves right two units. Label the image of the turning point and the transformed equation.

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Check that each original point moves right two units with no change in height. Cover the purple curve and predict the three image coordinates before looking again.

7. Describe fully and reverse the process

For a translation, state its name and vector, with horizontal movement above vertical movement. If corresponding points are ((p,q)) and ((r,s)), the vector is (\begin{pmatrix}r-p\s-q\end{pmatrix}). Check another pair: a translation must give the same displacement for every point.

For a reflection, state its name and the equation of the mirror line. “Reflection in (y=0)” and “reflection in the x-axis” are equivalent. Do not confuse (x=0), a vertical line, with (y=0), a horizontal line.

Reverse example:A curve moves left four and down one. Its equation becomes (y=f(x+4)-1), and its vector is (\begin{pmatrix}-4\-1\end{pmatrix}). The positive inside sign records a leftward movement. A named point alone does not generally determine an entire transformation; check the curve or additional corresponding points.

8. Combine changes in the order the equation requires

For (y=-f(x-2)+3), move right two, reflect in the x-axis, then move up three. The direct point rule is ((p,q)\mapsto(p+2,3-q)). A maximum at ((1,5)) becomes a minimum at ((3,-2)).

The order of the outside operations matters. “Reflect, then move up three” gives (-f(x)+3). “Move up three, then reflect” gives (-(f(x)+3)=-f(x)-3). Their outputs differ by six.

Horizontal and vertical translations can be tracked independently, but do not invent a universal rule that all transformations commute. With a reflection and a translation in the same coordinate direction, order can change the result. Coordinate mapping or substitution settles the question accurately.

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9. Turning points, intercepts and asymptotes

Translations preserve whether a turning point is a maximum or minimum. An x-axis reflection exchanges maxima and minima; a y-axis reflection preserves their type.

An image of an intercept is not necessarily an intercept. Under a sideways translation, an original y-intercept moves off the y-axis. Under a vertical translation, an original x-intercept moves off the x-axis. For (g(x)=f(x+a)+b), the new y-intercept is ((0,f(a)+b)); its roots satisfy (f(x+a)=-b).

If a reciprocal graph has an asymptote, transform that line too. Translating (y=1/x) right two and up one gives (y=1/(x-2)+1), with asymptotes (x=2) and (y=1). Do not connect branches across an asymptote. This is an application of the same translations, not an extra stretch rule.

Mathematical Relationships

All transformation and coordinate rules below are Must memorise. They are not supplied on Pearson's 2026 Higher-tier formula sheet. No formula from that sheet is needed for the four core transformations. 5

(y=f(x)+a:\quad(p,q)\mapsto(p,q+a)), vector (\begin{pmatrix}0\a\end{pmatrix}).

(y=f(x+a):\quad(p,q)\mapsto(p-a,q)), vector (\begin{pmatrix}-a\0\end{pmatrix}).

(y=-f(x):\quad(p,q)\mapsto(p,-q)), reflection in (y=0).

(y=f(-x):\quad(p,q)\mapsto(-p,q)), reflection in (x=0).

(y=f(x-h)+k:\quad(p,q)\mapsto(p+h,q+k)), vector (\begin{pmatrix}h\k\end{pmatrix}).

For combined reflections and translations, (y=\varepsilon f(\delta(x-h))+k) maps ((p,q)) to ((h+\delta p,k+\varepsilon q)), where (\varepsilon,\delta\in{$-1,1}). This compact rule is Must memorise if used, but it is optional: the four basic rules and step-by-step tracking are sufficient.$

Here (p,q) are original coordinates; (a,h,k) are signed constants. The symbols (\varepsilon) and (\delta$) choose whether the output and horizontal coordinate are reversed. Also Must memorise: (f(p)=q); vector ((r-p,s-q)), written as a column; and, for (g(x)=f(x+a)+b), (g(0)=f(a)+b). These relationships explain coordinate reading, descriptions and intercept checks.$

Exam Focus: marks, timing and command words

Each 1MA1 paper lasts 90 minutes for 80 marks, giving an overall budget of (90/80=1.125) minutes per mark. Allow approximately 3-4 minutes for 3 marks, 4-5 for 4 marks, and 6-7 for 6 marks, adjusting to difficulty and leaving checking time. This is revision guidance, not a Pearson timing instruction. A13 can appear on the non-calculator paper or either calculator paper. 1

Official examples: Pearson's teaching guidance reproduces New SAMs Paper 2H Q19: a sketch of (f(-x)) for 1 mark, and a sketch of (-f(x)+3) for 1 mark. It also includes a 2-mark coordinate question for (-f(x+3)). 2

In June 2024 Paper 2H Q21, the official scheme awards C2 for the fully correct transformation; C1 is available for specified partial work, including one correct component or at least two correctly transformed final points. This does not guarantee a direction-only mark in every question. Individual schemes take precedence. 3 4

Sketch means show the transformed curve with its defining features. Write down means provide the requested coordinates or expression directly. Describe fully means transformation plus vector or mirror line. Explain needs a mathematical reason: for example, (x+2=p), so the new horizontal coordinate is (p-2). Do not replace mathematical calculation with generic essay advice about “both sides”.

The worked and practice questions in this resource are original Edexcel-style questions. Their stated allocations are indicative teaching schemes, not official Pearson mark schemes.

Practical Applications and Synoptic Understanding

A fixed additional charge changes a cost graph from (C(x)) to (C(x)+a): every cost increases equally. Delaying a time-dependent signal by two seconds changes it to (s(t-2)): the same event occurs later, a rightward translation. Changing a signed height convention from above-ground positive to below-ground positive negates the output, illustrating an x-axis reflection.

For trigonometric graphs, a shift of (\sin(x)) to (\sin$(x-30^\circ)) moves the graph right 30 degrees, not 30 grid squares. GCSE trigonometric graph arguments are in degrees. Follow labelled units; a 30-second delay is 0.5 minutes if the horizontal axis uses minutes. Pure coordinate questions normally have no physical units, so do not attach centimetres to an ordered pair. 1$

There is no required practical for this GCSE Mathematics topic. Graphing software or tracing paper can help you check a practice sketch, but the exam answer must demonstrate the required graph or coordinates.

Listen, Retrieve and Check

Ten-minute female-tutor podcast: 00:00 introduction; 01:00 core concepts; 06:00 exam tips; 08:00 recall quiz; 09:00 summary. Pause whenever you need longer to answer.

Close the guide and write all four point rules from memory. Then apply each to ((1,5)), check your answers and explain any error. Revisit tomorrow, three days later and one week later. Mix examples rather than practising only one rule at a time. Explaining why the inside sign reverses direction is a stronger check than recognising a familiar diagram.

References

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