TMUA: Sketching and transformations
Graphs of standard functions and their transformations.
- Section 1 Part 1 · MM8 Graphs of functions · spec MM8.1-MM8.4
- Papers 1 and 2
- 7 practice questions
- No calculator
What the specification covers
- Graphs of standard functions including modulus and square root
- af(x), f(x) + a, f(x + a), f(ax) and compositions
- Effect of m, c and of a, b, c in a(x + b)² + c
Key ideas
- \(y=af(x)\): vertical stretch by \(a\). \(y=f(x)+a\): shift up \(a\). \(y=f(x+a)\): shift left \(a\). \(y=f(ax)\): horizontal stretch by \(\tfrac1a\).
- A point \((p,q)\) on \(y=f(x)\) maps to \((p+h,\;aq+k)\) on \(y=af(x-h)+k\).
- In \(y=a(x+b)^2+c\) the vertex is \((-b,c)\) and the graph opens upwards when \(a>0\).
Common mistakes
- \(f(x-3)\) moves the graph right, not left.
- Order matters with composite transformations: apply the one inside the bracket to \(x\) first.
Exam tip
Track one point through the transformation rather than the whole graph.
Worked example
Worked example
The point \((2,5)\) lies on the graph of \(y=f(x)\). Which point must lie on the graph of \(y=2f(x-3)+1\)?
- \((-1, 6)\)
- \((5, 11)\)
- \((-1, 11)\)
- \((5, 12)\)
- \((5, 6)\)
Answer: B: \((5, 11)\)
At \(x=5\), \(2f(5-3)+1=2f(2)+1=11\). So \((5,11)\) is on the new graph.
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
The point \((-3,1)\) lies on the graph of \(y=f(x)\). Which point must lie on the graph of \(y=3f(x-1)-1\)?
- \((-2, 2)\)
- \((-2, 4)\)
- \((-4, 2)\)
- \((-2, 0)\)
- \((-4, 0)\)
Show the answer and solution
Answer: A: \((-2, 2)\)
At \(x=-2\): \(3f(-2-1)-1=3f(-3)-1=3(1)-1=2\). So \((-2,2)\) is on the new graph.
Question 2
The point \((4,5)\) lies on the graph of \(y=f(x)\). Which point must lie on the graph of \(y=3f(x-3)-3\)?
- \((1, 12)\)
- \((7, 18)\)
- \((7, 12)\)
- \((1, 6)\)
- \((7, 6)\)
Show the answer and solution
Answer: C: \((7, 12)\)
At \(x=7\): \(3f(7-3)-3=3f(4)-3=3(5)-3=12\). So \((7,12)\) is on the new graph.
Question 3
The point \((-3,-3)\) lies on the graph of \(y=f(x)\). Which point must lie on the graph of \(y=-2f(x-2)-1\)?
- \((-1, 8)\)
- \((-1, 7)\)
- \((-1, 5)\)
- \((-5, 8)\)
- \((-5, 5)\)
Show the answer and solution
Answer: C: \((-1, 5)\)
At \(x=-1\): \(-2f(-1-2)-1=-2f(-3)-1=-2(-3)-1=5\). So \((-1,5)\) is on the new graph.
Question 4
The point \((3,2)\) lies on the graph of \(y=f(x)\). Which point must lie on the graph of \(y=-2f(x-2)-3\)?
- \((5, -1)\)
- \((5, -7)\)
- \((1, -7)\)
- \((1, 2)\)
- \((5, 2)\)
Show the answer and solution
Answer: B: \((5, -7)\)
At \(x=5\): \(-2f(5-2)-3=-2f(3)-3=-2(2)-3=-7\). So \((5,-7)\) is on the new graph.
Question 5
The point \((3,5)\) lies on the graph of \(y=f(x)\). Which point must lie on the graph of \(y=2f(x-3)-1\)?
- \((6, 9)\)
- \((0, 8)\)
- \((0, 9)\)
- \((6, 8)\)
- \((6, 11)\)
Show the answer and solution
Answer: A: \((6, 9)\)
At \(x=6\): \(2f(6-3)-1=2f(3)-1=2(5)-1=9\). So \((6,9)\) is on the new graph.
Question 6
The point \((4,-1)\) lies on the graph of \(y=f(x)\). Which point must lie on the graph of \(y=3f(x-2)+1\)?
- \((2, -2)\)
- \((6, 0)\)
- \((2, 0)\)
- \((6, -4)\)
- \((6, -2)\)
Show the answer and solution
Answer: E: \((6, -2)\)
At \(x=6\): \(3f(6-2)+1=3f(4)+1=3(-1)+1=-2\). So \((6,-2)\) is on the new graph.
Keep going
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