Exam questions · Maths · Functions, Sequences and Rates of Change
Transformations of Graphs
- 6 exam questions
- 18 marks
- 9 quick checks
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1 Write down [2 marks]
The graph of \(y = f(x)\) is shown. The turning point is \((2, -1)\). (a) Write down the coordinates of the turning point of the graph of \(y = f(x) + 2\). (1 mark) (b) Write down the coordinates of the turning point of the graph of \(y = f(x + 1)\). (1 mark)
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Model answer
(a) \((2, 1)\). (b) \((1, -1)\).
Mark scheme
- (a) \((2, 1)\) — B1
- (b) \((1, -1)\) — B1
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2 Write down [2 marks]
The graph of \(y = g(x)\) has a minimum point at \((-1, -3)\). (a) Write down the coordinates of the minimum point of \(y = g(x) + 4\). (1 mark) (b) Write down the coordinates of the maximum point of \(y = -g(x)\). (1 mark)
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Model answer
(a) \((-1, 1)\). (b) \((-1, 3)\).
Mark scheme
- (a) \((-1, 1)\) — B1
- (b) \((-1, 3)\) — B1
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3 Write down [3 marks]
The graph of \(y = x^2\) is transformed. Write down the equation of the new graph after (a) a translation of 3 units to the right, (1 mark) (b) a translation of 2 units up, (1 mark) (c) a reflection in the \(x\)-axis. (1 mark)
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Model answer
(a) \(y = (x - 3)^2\). (b) \(y = x^2 + 2\). (c) \(y = -x^2\).
Mark scheme
- (a) \(y = (x - 3)^2\) — B1
- (b) \(y = x^2 + 2\) — B1
- (c) \(y = -x^2\) — B1
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4 Write down [4 marks]
The graph of \(y = f(x)\) has a maximum point at \((3, 5)\). Write down the coordinates of the maximum point of the graph of (a) \(y = f(x - 2)\) (1 mark) (b) \(y = f(x) - 4\) (1 mark) (c) \(y = -f(x)\) (1 mark) (d) \(y = f(-x)\) (1 mark)
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Model answer
(a) \((5, 5)\). (b) \((3, 1)\). (c) \((3, -5)\), which is now a minimum. (d) \((-3, 5)\).
Mark scheme
- (a) \((5, 5)\) — B1
- (b) \((3, 1)\) — B1
- (c) \((3, -5)\) — B1
- (d) \((-3, 5)\) — B1
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5 Write down [3 marks]
The diagram shows the graph of \(y = \sin x\) and a transformation of it, for \(0^\circ \le x \le 360^\circ\). (a) Write down the equation of the transformed graph. (1 mark) (b) Describe fully the single transformation. (2 marks)
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Model answer
(a) \(y = \sin x + 1\). (b) A translation by the vector \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\), which is 1 unit up.
Mark scheme
- (a) \(y = \sin x + 1\) — B1
- (b) Translation — B1
- (b) Vector \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\), or 1 unit up — B1
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6 Show that [4 marks]
\(f(x) = x^2 - 4x + 3\). Show that \(f(x + 1) = x^2 - 2x\), and hence solve \(f(x + 1) = 0\). (4 marks)
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Model answer
\(f(x + 1) = (x + 1)^2 - 4(x + 1) + 3 = x^2 + 2x + 1 - 4x - 4 + 3 = x^2 - 2x\). Then \(x(x - 2) = 0\), so \(x = 0\) or \(x = 2\).
Mark scheme
- \((x + 1)^2 - 4(x + 1) + 3\) — M1
- \(x^2 - 2x\) with the working shown — A1
- \(x(x - 2) = 0\) — M1
- \(x = 0\) and \(x = 2\) — A1
Quick check
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1
What does \(y = f(x) + 3\) do to the graph of \(y = f(x)\)?
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B: Moves it up 3
Adding to the function moves the graph up.
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2
What does \(y = f(x + 2)\) do to the graph of \(y = f(x)\)?
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A: Moves it left 2
A plus inside the bracket moves the graph left.
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3
What does \(y = -f(x)\) do to the graph of \(y = f(x)\)?
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D: Reflects it in the \(x\)-axis
A minus outside changes the \(y\)-values.
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4
What does \(y = f(-x)\) do to the graph of \(y = f(x)\)?
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C: Reflects it in the \(y\)-axis
A minus inside changes the \(x\)-values.
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5
The maximum of \(y = f(x)\) is at \((3, 5)\). Where is the maximum of \(y = f(x - 2)\)?
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B: \((5, 5)\)
The graph moves right 2.
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6
The maximum of \(y = f(x)\) is at \((3, 5)\). Where is the turning point of \(y = -f(x)\)?
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A: \((3, -5)\)
The \(y\)-coordinate changes sign.
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7
What is the equation of \(y = x^2\) after a translation of 3 units to the right?
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D: \(y = (x - 3)^2\)
Moving right replaces \(x\) with \(x - 3\).
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8
What is the maximum value of \(y = \sin x + 1\)?
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C: 2
The sine graph is moved up by 1, so its maximum is \(1 + 1 = 2\).
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9
Which equation gives the same graph as \(y = \cos x\)?
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B: \(y = \sin(x + 90^\circ)\)
Moving the sine graph left by \(90^\circ\) gives the cosine graph.