Home   >  Maths quizzes   >  Grade 8+ Quizzes   >  Transforming graphs y=f(x)

MathsLinks Logo Transforming graphs y=f(x) Worksheets

Transforming graphs y=f(x) Quiz

Grade 8+ • 8B • 10 Questions (22 Marks)

Grade 1Grade 2Grade 3Grade 4Grade 5Grade 6Grade 7Grade 8/9
Score: 0 / 10
Transforming graphs y=f(x) — Notes & Worked Examples Show / Hide

Explanation:

Changes outside the bracket affect the y-coordinates as expected, whereas changes inside the bracket affect the x-coordinates in the opposite direction:

• y = f(x) + a translates by vector 0a (up by a);  y = f(x + a) translates by vector −a0 (left by a).
• y = −f(x) reflects in the x-axis (y changes sign);  y = f(−x) reflects in the y-axis (x changes sign).

Example 1: The curve y = f(x) has a turning point at P(3, −4). Find the coordinates of the turning point on: (a) y = f(x − 2) + 5  (b) y = −f(x).

f(x) P(3, −4) −f(x) (3, 4) (5, 1) f(x−2)+5

Solution:

(a) Inside (x − 2) shifts right by 2: x = 3 + 2 = 5
Outside (+ 5) shifts up by 5: y = −4 + 5 = 1 ⇒ (5, 1)
(b) y = −f(x) reflects in the x-axis (multiply y by −1):
Answer: (a) (5, 1)   (b) (3, 4)

Example 2: Given f(x) = x2 − 4x, find the equation of the curve after a reflection in the y-axis followed by a translation of 0−3.

Solution:

Reflection in y-axis replaces x with −x:
f(−x) = (−x)2 − 4(−x) = x2 + 4x
Translation by 0−3 subtracts 3 from the function:
Answer: y = x2 + 4x − 3
Question 1
[1 mark]
The curve y = f(x) has a minimum turning point at (4, -5).
Write down the coordinates of the minimum turning point of y = f(x - 5) + 4.
Worked Solution & Answer
Translation by (54) maps (4, -5) → (9, -1). B1
Answer: (9, -1)
Question 2
[1 mark]
The curve y = f(x) has a minimum turning point at (6, -2).
Write down the coordinates of the minimum turning point of y = f(x - 5) + 1.
Worked Solution & Answer
Translation by (51) maps (6, -2) → (11, -1). B1
Answer: (11, -1)
Question 3
[2 marks]
Here is the graph of y = f(x)
xy–6–4–2246–4–224O
(a) On the grid below, draw the graph of y = f(x) - 4 [1 mark]
xy–6–4–2246–4–224O
(b) On the grid below, draw the graph of y = f(-x) [1 mark]
xy–6–4–2246–4–224O
Worked Solution & Answer
(a)
Translation by (0, –4) B1
xy–6–4–2246–4–224O
Answer (a): Trapezium translated 4 units down with vertices (–3, –4), (–2, –3), (–1, –3), (0, –4)
(b)
Reflection in the y-axis B1
xy–6–4–2246–4–224O
Answer (b): Trapezium reflected in the y-axis with vertices (0, 0), (1, 1), (2, 1), (3, 0)
Question Paper (PDF) • Mark Scheme (PDF)
Question 4 AQA Jun 2025 1H Q21
[2 marks]
The graph of y = f(x) for 0 ≤ x ≤ 3 is shown.
–4–4–3–3–2–2–1–111223344Oxy
(a) On the grid below, draw the graph of y = -f(x) for 0 ≤ x ≤ 3
The graph of y = f(x) is shown to help you. [1 mark]
–4–4–3–3–2–2–1–111223344Oxy
(b) On the grid below, draw the graph of y = f(x) + 1 for 0 ≤ x ≤ 3
The graph of y = f(x) is shown to help you. [1 mark]
–4–4–3–3–2–2–1–111223344Oxy
Worked Solution & Answer
(a)
y = –f(x) reflects the graph in the x-axis B1
–4–4–3–3–2–2–1–111223344Oxy
Answer (a): Triangle with vertices (0, 0), (1, –3), (3, 0)
(b)
y = f(x) + 1 translates the graph 1 unit up B1
–4–4–3–3–2–2–1–111223344Oxy
Answer (b): Triangle with vertices (0, 1), (1, 4), (3, 1)
Question Paper (PDF) • Mark Scheme (PDF)
Question 5 IGCSE Jun 2022 1H Q20
[2 marks]
The diagram shows a sketch of part of the curve with equation y = f(x)
yxO(s, t)y = f(x)
There is one maximum point on this curve.
The coordinates of this maximum point are (s, t)
Find, in terms of s and t, the coordinates of the maximum point on the curve with equation
(i) y = f(x - 2) [1 mark]
(ii) y = 3f(x) [1 mark]
Worked Solution & Answer
(i)
Translation 2 units right: (s + 2, t) B1
Answer (i): (s + 2, t)
(ii)
Vertical stretch scale factor 3: (s, 3t) B1
Answer (ii): (s, 3t)
Question Paper (PDF) • Mark Scheme (PDF)
Question 6
[2 marks]
The graph of y = f(x) is shown on the grid below.
xyO–4–2246–22y = f(x)
(a) On the grid above, sketch the graph of y = f(x - 2) [1 mark]
(b) Write down the equation of graph **B**. [1 mark]
xyO–4–224–22graph Agraph B
y =
Worked Solution & Answer
(a)
Translation by vector (2, 0) B1
Answer (a): Parabola translated 2 units right: passes through (1, 3), (2, 0), minimum (3, –1), (4, 0), (5, 3)
(b)
Reflection in the y-axis maps y = g(x) to y = g(–x) B1
Answer (b): y = y = g(-x)
Question Paper (PDF) • Mark Scheme (PDF)
Question 7
[2 marks]
The graph of y = f(x) is shown on the grid below.
xy–5–4–3–2–112345–4–3–2–112345O
(a) On the grid above, sketch the graph of y = f(x + 2) [1 mark]
(b) Write down an equation of graph **B**. [1 mark]
xy–4–3–2–11234–3–2–1123Ograph Agraph B
y =
Worked Solution & Answer
(a)
Translation by (–2, 0) B1
Answer (a): Curve translated 2 units left, crossing x-axis at (–3, 0), (–1, 0), (0, 0) and passing through (–2, 2)
(b)
Reflection in x-axis gives y = –g B1
Answer (b): y = y = -g(x)
Question Paper (PDF) • Mark Scheme (PDF)
Question 8
[3 marks]
The graph of y = f(x) is shown on the grid below.
xyOy = f(x)
(a) On the grid above, sketch the graph of y = f(-x) [1 mark]
(b) The graph of y = \tan x° is translated to give the graph of y = g(x)
Following the translation the point Q, shown on the graph above, moves to point R.
Point R has coordinates (90, -5)
Find an expression for g(x) in terms of x. [2 marks]
xyOQ
Worked Solution & Answer
(a)
Reflect in the y-axis B1
Answer (a): Graph reflected in the y-axis through (–2, –4), (0, 0), (1, 1), (3, 0), (5, 4)
(b)
Q is at (360, 0) and moves to (90, –5), which is a translation of 270° left and 5 units down M1
⇒ g(x) = tan(x + 270)° – 5 A1
Answer (b): \tan(x + 270)° - 5
Question Paper (PDF) • Mark Scheme (PDF)
Question 9
[3 marks]
The graph of y = f(x) is shown on the grid.
xy–6–5–4–3–2–1123456–6–4–2246Oy = f(x)A(–2, 1)
(a) On the grid, draw the graph with equation y = f(x + 1) - 3 [2 marks]
(b) Write down the coordinates of point B. [1 mark]
Worked Solution & Answer
(a)
Shift 1 unit left and 3 units down M1 A1
Answer (a): Graph translated by vector (–1, –3) with vertices at (–5, 0), (–4, 0), (–3, –2), (0, 1), (2, 1)
(b)
Reflection in the y-axis maps (–2, 1) to (2, 1) B1
Answer (b): (2, 1)
Question Paper (PDF) • Mark Scheme (PDF)
Question 10 IGCSE Jan 2021 1H Q21
[4 marks]
A curve has equation y = f(x)
The coordinates of the minimum point on this curve are (-9, 15)
(a) y = f(x + 3) [1 mark]
(b) y = 13f(x) [1 mark]
(b) Given that a > 0 and that 0 < b < 360
find the value of a and the value of b. [2 marks]
–2–112O180360
Worked Solution & Answer
(a)
Shift left 3 -> (–12, 15) B1
Answer (a): (-12, 15)
(b)
Vertical scale by 1/3 -> (–9, 5) B1
Answer (b): (-9, 5)
(b)
Amplitude a = 2 M1
cos(x + 90)° shifts minimum to x = 90°, so b = 90 A1
Answer (b): a = 2, b = 90
Question Paper (PDF) • Mark Scheme (PDF)
Score: 0 / 10 (0 / 22 marks)
Transforming graphs y=f(x) Worksheets Next: Algebraic proof →

Grade 8+ Quizzes

14 Topics
8A - Quad. sim. equations8B - Transforming graphs y=f(x)
8C - Algebraic proof8D - Completing the square
8E - Quadratic inequalities8F - Quadratic sequences
8G - Velocity time graphs8H - Circles and tangents
8J - Circle theorem proof8K - Vector proof
8L - Probability equation questions8M - Invariant points
8N - Spheres, cones and cylinders8P - Volume of a frustum