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How the graph of \(y = f(x)\) changes under translations, stretches and reflections, and combinations of these. You sketch transformed graphs from a given sketch, showing where turning points, intercepts and asymptotes move to, typically for 3 to 6 marks.
Key facts
\(f(x) + a\): translation \(\begin{pmatrix} 0 \\ a \end{pmatrix}\)
\(f(x + a)\): translation \(\begin{pmatrix} -a \\ 0 \end{pmatrix}\)
\(af(x)\): stretch parallel to the \(y\)-axis, scale factor \(a\)
\(f(ax)\): stretch parallel to the \(x\)-axis, scale factor \(\dfrac{1}{a}\)
\(-f(x)\): reflect in the \(x\)-axis; \(f(-x)\): reflect in the \(y\)-axis
\((p, q)\) under \(y = af(x) + b\) \(\to (p, aq + b)\)
\(y = f(x) + a\): translation by \(\begin{pmatrix} 0 \\ a \end{pmatrix}\), i.e. \(a\) units up. A horizontal asymptote \(y = k\) becomes \(y = k + a\)
\(y = f(x + a)\): translation by \(\begin{pmatrix} -a \\ 0 \end{pmatrix}\), i.e. \(a\) units left. So \(f(x - 3)\) moves the graph 3 units right. A vertical asymptote \(x = k\) becomes \(x = k - a\)
With f(x) = x²: f(x) + 2 moves the curve 2 up; f(x − 3) moves it 3 to the right (the opposite of the sign).
Stretches and reflections
diagram
\(y = af(x)\): stretch parallel to the \(y\)-axis, scale factor \(a\). Every \(y\)-coordinate is multiplied by \(a\); points on the \(x\)-axis stay put.
\(y = f(ax)\): stretch parallel to the \(x\)-axis, scale factor \(\dfrac{1}{a}\). Every \(x\)-coordinate is divided by \(a\); points on the \(y\)-axis stay put.
With f(x) = 4 − x²: 2f(x) doubles every y-coordinate; f(2x) halves every x-coordinate.
\(y = -f(x)\): reflection in the \(x\)-axis. \(y = f(-x)\): reflection in the \(y\)-axis.
Rule of thumb: changes outside \(f\) affect \(y\) and do what you expect; changes inside the bracket affect \(x\) and do the opposite.
Combinations
Do one transformation at a time, tracking key points: turning points, intercepts and asymptotes.
Vertical: for \(y = af(x) + b\), stretch first, then translate: \((p, q) \to (p, aq + b)\)
Example: under \(y = 3f(x + 2) - 1\), the point \((4, 5)\) moves to \((2, 14)\)
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Sketch a transformed graph from a given one
4 marksGrade C
Identify the transformation for each part.
Move each key point and each asymptote.
Sketch the same basic shape and label the new coordinates and asymptote equations.
Example. The curve \(y = f(x)\) has a maximum point at \((2, 6)\), crosses the \(x\)-axis at \((-1, 0)\) and \((5, 0)\), and has the horizontal asymptote \(y = -4\). On separate diagrams, sketch (a) \(y = 2f(x)\) (b) \(y = f\left(\dfrac{1}{2}x\right)\), showing the maximum point, the points where the curve crosses the \(x\)-axis and the asymptote.
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(a) Same shape with maximum \((2, 12)\)
(B1)
Crosses at \((-1, 0)\) and \((5, 0)\), asymptote \(y = -8\)
(B1)
(b) Same shape with maximum \((4, 6)\)
(B1)
Crosses at \((-2, 0)\) and \((10, 0)\), asymptote \(y = -4\)
(B1)
Describe a transformation fully
2 marksGrade C
Compare the new equation with \(y = f(x)\) to see what has changed.
Name the transformation.
Give its vector, or its direction and scale factor, or its mirror line.
Example. Describe fully the single transformation that maps the curve \(y = \dfrac{1}{x}\) onto the curve \(y = \dfrac{1}{x + 2} - 5\)
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Translation
(B1)
by the vector \(\begin{pmatrix} -2 \\ -5 \end{pmatrix}\)
(B1)
Find a turning point after combined transformations
3 marksGrade A
Apply the horizontal change to the \(x\)-coordinate.
Apply the vertical change to the \(y\)-coordinate (stretch before translate).
A negative multiplier outside \(f\) turns a minimum into a maximum.
Example. The curve \(y = f(x)\) has a minimum point at \((-2, 5)\). Find the coordinates of the turning point of each curve, and state whether it is a maximum or a minimum: (a) \(y = f(2x + 4)\) (b) \(y = 3 - 2f(x)\)
Moving the wrong way: \(f(x + 3)\) is 3 units left, not right.
Using scale factor 2 for \(f(2x)\) instead of \(\dfrac{1}{2}\)
Forgetting to transform the asymptotes, or not labelling them.
Vague descriptions such as 'moves it' or 'makes it bigger' instead of naming the transformation with its details.
More tips
Memory tricks
Inside the bracket: do the opposite and divide. Outside: do what it says and multiply.
To find where a feature moves under \(f(x - 3)\), solve \(x - 3 = \) its old \(x\)-coordinate.
Vertical asymptotes only move under changes to \(x\); horizontal asymptotes only under changes to \(y\)
A negative multiplier outside \(f\) swaps maxima and minima.
Exam technique
'Describe fully' needs the type of transformation and every detail (vector; direction and scale factor; or mirror line).
Label every point the question mentions with its new coordinates, and every asymptote with its equation.
For combinations, write the image of one key point step by step before you sketch.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Translate a graph vertically and horizontally\(f(x) + a\) moves the graph up by \(a\); \(f(x + a)\) moves it left by \(a\)
Grade D
Stretch a graph parallel to either axis\(af(x)\): stretch parallel to the \(y\)-axis, scale factor \(a\);\(f(ax)\): parallel to the \(x\)-axis, scale factor \(\dfrac{1}{a}\)
Grade D
Reflect a graph in either axis\(-f(x)\) reflects the graph in the \(x\)-axis; \(f(-x)\) reflects it in the \(y\)-axis.
Grade C
Find where points and asymptotes moveApply the transformation to coordinates, e.g. under \(y = 2f(x)\), \((3, 4)\) moves to \((3, 8)\)
Grade C
Describe a single transformation fullyName it and give every detail: the vector, or the direction and scale factor, or the mirror line.
Grade A
Apply combinations of transformations in ordere.g. \(y = 2f(x - 1) + 3\): move right 1, stretch vertically by factor 2, then move up 3.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
The diagram shows a sketch of the curve with equation \(y = \mathrm{f}(x)\). The curve has a maximum point \(P(-1, 8)\) and a minimum point \(Q(2, -19)\).
[object Object]
(a) Write down the coordinates of the minimum point of the curve with equation \(y = 3\mathrm{f}(x)\).[2]
(b) Write down the coordinates of the maximum point of the curve with equation \(y = \mathrm{f}(2x)\).[2]
(c) Write down the coordinates of the maximum point of the curve with equation \(y = 3\mathrm{f}(x)\).[2]
Show the answer and mark scheme
(a)Answer: \(\left(2, -57\right)\)
B1 for the \(x\)-coordinate \(2\)
B1 for the \(y\)-coordinate \(-57\)
Worked solution: The minimum point comes from \(Q(2, -19)\). The \(x\)-coordinate is unchanged. The \(y\)-coordinate becomes \(3 \times (-19) = -57\). So the point is \(\left(2, -57\right)\).
(b)Answer: \(\left(-\frac{1}{2}, 8\right)\)
B1 for the \(x\)-coordinate \(-\frac{1}{2}\)
B1 for the \(y\)-coordinate \(8\)
Worked solution: The maximum point comes from \(P(-1, 8)\). Solving \(2x = -1\) gives \(x = -\frac{1}{2}\). The \(y\)-coordinate is unchanged. So the point is \(\left(-\frac{1}{2}, 8\right)\).
(c)Answer: \(\left(-1, 24\right)\)
B1 for the \(x\)-coordinate \(-1\)
B1 for the \(y\)-coordinate \(24\)
Worked solution: The maximum point comes from \(P(-1, 8)\). The \(x\)-coordinate is unchanged. The \(y\)-coordinate becomes \(3 \times 8 = 24\). So the point is \(\left(-1, 24\right)\).
Question 2Medium6 marks
The diagram shows a sketch of the curve with equation \(y = \mathrm{f}(x)\). The curve has a maximum point \(P(-1, 10)\) and a minimum point \(Q(1, 2)\).
[object Object]
(a) Write down the coordinates of the maximum point of the curve with equation \(y = \mathrm{f}(x + 1) + 1\).[2]
(b) Write down the coordinates of the maximum point of the curve with equation \(y = \mathrm{f}(2x) + 1\).[2]
(c) Write down the coordinates of the minimum point of the curve with equation \(y = 2\mathrm{f}(x + 3)\).[2]
Show the answer and mark scheme
(a)Answer: \(\left(-2, 11\right)\)
B1 for the \(x\)-coordinate \(-2\)
B1 for the \(y\)-coordinate \(11\)
Worked solution: The maximum point comes from \(P(-1, 10)\). Solving \(x + 1 = -1\) gives \(x = -2\). The \(y\)-coordinate becomes \(10 + 1 = 11\). So the point is \(\left(-2, 11\right)\).
(b)Answer: \(\left(-\frac{1}{2}, 11\right)\)
B1 for the \(x\)-coordinate \(-\frac{1}{2}\)
B1 for the \(y\)-coordinate \(11\)
Worked solution: The maximum point comes from \(P(-1, 10)\). Solving \(2x = -1\) gives \(x = -\frac{1}{2}\). The \(y\)-coordinate becomes \(10 + 1 = 11\). So the point is \(\left(-\frac{1}{2}, 11\right)\).
(c)Answer: \(\left(-2, 4\right)\)
B1 for the \(x\)-coordinate \(-2\)
B1 for the \(y\)-coordinate \(4\)
Worked solution: The minimum point comes from \(Q(1, 2)\). Solving \(x + 3 = 1\) gives \(x = -2\). The \(y\)-coordinate becomes \(2 \times 2 = 4\). So the point is \(\left(-2, 4\right)\).
Question 3Hard9 marks
The diagram shows a sketch of the curve with equation \(y = \mathrm{f}(x)\). The curve has a maximum point \(P(1, 10)\) and a minimum point \(Q(-1, 2)\).
[object Object]
(a) The curve with equation \(y = a\mathrm{f}(x) + b\), where \(a\) and \(b\) are non-zero constants, has a maximum point at \((-1, -4)\) and a minimum point at \((1, -36)\). Find the value of \(a\) and the value of \(b\).[5]
(b) The curve with equation \(y = \mathrm{f}(x + c)\), where \(c\) is a constant, has its minimum point on the \(y\)-axis. Find the value of \(c\).[2]
(c) The curve with equation \(y = \mathrm{f}(kx)\), where \(k\) is a positive constant, has a maximum point at \(\left(\frac{1}{3}, 10\right)\). Find the value of \(k\).[2]
Show the answer and mark scheme
(a)Answer: \(a = -4, \ b = 4\)
B1 for deducing that the maximum comes from \(Q\) (it has the same \(x\)-coordinate, \(-1\)), so \(a \lt 0\)
M1 for forming an equation from the maximum point: \(2a + b = -4\)
M1 for forming an equation from the minimum point: \(10a + b = -36\), and solving simultaneously
A1 for \(a = -4\)
A1 for \(b = 4\)
Worked solution: The \(x\)-coordinates are unchanged by \(y = a\mathrm{f}(x) + b\). The maximum is at \(x = -1\), the \(x\)-coordinate of \(Q\), so the minimum \(Q\) has become a maximum: the curve has been reflected in the \(x\)-axis and \(a \lt 0\). So \(2a + b = -4\) and \(10a + b = -36\). Subtracting: \(-8a = 32\), so \(a = -4\), and then \(b = 4\).
(b)Answer: \(c = -1\)
M1 for recognising that the minimum point moves to \((-1 - c, 2)\)
A1 for \(c = -1\)
Worked solution: The graph of \(y = \mathrm{f}(x + c)\) is a translation by \(\begin{pmatrix} -c \\ 0 \end{pmatrix}\), so the minimum point is \((-1 - c, 2)\). It lies on the \(y\)-axis when \(-1 - c = 0\), i.e. \(c = -1\).
(c)Answer: \(k = 3\)
M1 for \(k \times \frac{1}{3} = 1\) (a stretch parallel to the \(x\)-axis with scale factor \(\frac{1}{k}\))
A1 for \(k = 3\)
Worked solution: The graph of \(y = \mathrm{f}(kx)\) is a stretch parallel to the \(x\)-axis with scale factor \(\frac{1}{k}\), so \(P\) moves to \(\left(\frac{1}{k}, 10\right)\). So \(\frac{1}{k} = \frac{1}{3}\) and \(k = 3\).