Edexcel A level Maths (9MA0) · Pure mathematics › Algebra and functions
Practise Transformations of graphs. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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 \(P(2, 20)\) and a minimum point \(Q(-1, -7)\).
[object Object]
(a) Write down the coordinates of the minimum point of the curve with equation \(y = 2\mathrm{f}(x)\).[2]
(b) Write down the coordinates of the minimum point of the curve with equation \(y = 2\mathrm{f}(x)\).[2]
(c) Write down the coordinates of the maximum point of the curve with equation \(y = -\mathrm{f}(x)\).[2]
Show the answer and mark scheme
(a)Answer: \(\left(-1, -14\right)\)
B1 for the \(x\)-coordinate \(-1\)
B1 for the \(y\)-coordinate \(-14\)
Worked solution: The minimum point comes from \(Q(-1, -7)\). The \(x\)-coordinate is unchanged. The \(y\)-coordinate becomes \(2 \times (-7) = -14\). So the point is \(\left(-1, -14\right)\).
(b)Answer: \(\left(-1, -14\right)\)
B1 for the \(x\)-coordinate \(-1\)
B1 for the \(y\)-coordinate \(-14\)
Worked solution: The minimum point comes from \(Q(-1, -7)\). The \(x\)-coordinate is unchanged. The \(y\)-coordinate becomes \(2 \times (-7) = -14\). So the point is \(\left(-1, -14\right)\).
(c)Answer: \(\left(-1, 7\right)\)
B1 for the \(x\)-coordinate \(-1\)
B1 for the \(y\)-coordinate \(7\)
Worked solution: The maximum point comes from \(Q(-1, -7)\). The \(x\)-coordinate is unchanged. The \(y\)-coordinate becomes \(-(-7) = 7\). So the point is \(\left(-1, 7\right)\).
Question 2Medium6 marks
The diagram shows a sketch of the curve with equation \(y = \mathrm{f}(x)\). The curve has a maximum point \(P(-1, 4)\) and a minimum point \(Q(-3, -4)\).
[object Object]
(a) Write down the coordinates of the minimum point of the curve with equation \(y = \mathrm{f}(x + 3) + 2\).[2]
(b) Write down the coordinates of the maximum point of the curve with equation \(y = 2\mathrm{f}(x - 3)\).[2]
(c) Write down the coordinates of the minimum point of the curve with equation \(y = \mathrm{f}(3x) + 1\).[2]
Show the answer and mark scheme
(a)Answer: \(\left(-6, -2\right)\)
B1 for the \(x\)-coordinate \(-6\)
B1 for the \(y\)-coordinate \(-2\)
Worked solution: The minimum point comes from \(Q(-3, -4)\). Solving \(x + 3 = -3\) gives \(x = -6\). The \(y\)-coordinate becomes \(-4 + 2 = -2\). So the point is \(\left(-6, -2\right)\).
(b)Answer: \(\left(2, 8\right)\)
B1 for the \(x\)-coordinate \(2\)
B1 for the \(y\)-coordinate \(8\)
Worked solution: The maximum point comes from \(P(-1, 4)\). Solving \(x - 3 = -1\) gives \(x = 2\). The \(y\)-coordinate becomes \(2 \times 4 = 8\). So the point is \(\left(2, 8\right)\).
(c)Answer: \(\left(-1, -3\right)\)
B1 for the \(x\)-coordinate \(-1\)
B1 for the \(y\)-coordinate \(-3\)
Worked solution: The minimum point comes from \(Q(-3, -4)\). Solving \(3x = -3\) gives \(x = -1\). The \(y\)-coordinate becomes \(-4 + 1 = -3\). So the point is \(\left(-1, -3\right)\).
Question 3Hard9 marks
The diagram shows a sketch of the curve with equation \(y = \mathrm{f}(x)\). The curve has a maximum point \(P(1, 5)\) and a minimum point \(Q(-1, -3)\).
[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 \((-1, 8)\) and a minimum point at \((1, -24)\). 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 \(\left(\frac{1}{3}, 5\right)\). 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: \(-3a + b = 8\)
M1 for forming an equation from the minimum point: \(5a + b = -24\), 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 \(-3a + b = 8\) and \(5a + b = -24\). Subtracting: \(-8a = 32\), so \(a = -4\), and then \(b = -4\).
(b)Answer: \(c = -1\)
M1 for recognising that the minimum point moves to \((-1 - c, -3)\)
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 \((-1 - c, -3)\). 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 \(\left(\frac{1}{k}, 5\right)\). So \(\frac{1}{k} = \frac{1}{3}\) and \(k = 3\).