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 \((3, -4)\). (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 \(y = f(x - 2)\). [1 mark]
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Model answer
(a) \((3, -6)\). (b) \((5, -4)\).
Mark scheme
- (a) \((3, -6)\) — B1
- (b) \((5, -4)\) — B1
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2 Write down [2 marks]
The graph of \(y = g(x)\) has a minimum point at \((-2, -5)\). (a) Write down the coordinates of the maximum point of \(y = -g(x)\). [1 mark] (b) Write down the coordinates of the minimum point of \(y = g(-x)\). [1 mark]
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Model answer
(a) \((-2, 5)\). (b) \((2, -5)\).
Mark scheme
- (a) \((-2, 5)\) — B1
- (b) \((2, -5)\) — 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 4 units to the left, [1 mark] (b) a translation of 3 units down, [1 mark] (c) a translation by the vector \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\). [1 mark]
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Model answer
(a) \(y = (x + 4)^2\). (b) \(y = x^2 - 3\). (c) \(y = (x - 2)^2 - 3\).
Mark scheme
- (a) \(y = (x + 4)^2\) — B1
- (b) \(y = x^2 - 3\) — B1
- (c) \(y = (x - 2)^2 - 3\) — B1
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4 Write down [4 marks]
The graph of \(y = f(x)\) has a maximum point at \((-2, 6)\). Write down the coordinates of the maximum point of the graph of (a) \(y = f(x + 3)\) [1 mark] (b) \(y = f(x) + 1\) [1 mark] (c) \(y = -f(x)\) [1 mark] (d) \(y = f(-x)\) [1 mark]
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Model answer
(a) \((-5, 6)\). (b) \((-2, 7)\). (c) \((-2, -6)\), which is now a minimum. (d) \((2, 6)\).
Mark scheme
- (a) \((-5, 6)\) — B1
- (b) \((-2, 7)\) — B1
- (c) \((-2, -6)\) — B1
- (d) \((2, 6)\) — 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) \(y = \sin(x - 90^\circ)\). (b) A translation by the vector \(\begin{pmatrix} 90 \\ 0 \end{pmatrix}\), which is \(90^\circ\) to the right.
Mark scheme
- (a) \(y = \sin(x - 90^\circ)\) — B1
- (b) Translation — B1
- (b) Vector \(\begin{pmatrix} 90 \\ 0 \end{pmatrix}\), or \(90^\circ\) to the right — B1
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6 Work out [4 marks]
\(f(x) = x^2 - 6x + 5\). Solve \(f(x - 2) = 0\). [4 marks]
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Model answer
\(f(x) = (x - 1)(x - 5)\), so \(f(x - 2) = (x - 3)(x - 7)\). So \(x = 3\) or \(x = 7\).
Mark scheme
- \(f(x) = (x - 1)(x - 5)\) — B1
- \(f(x - 2) = (x - 2 - 1)(x - 2 - 5)\) or \((x - 3)(x - 7)\) — M1
- \(x = 3\) — A1
- \(x = 7\) — 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.