Grade 10 Maths · Functions

Shift a graph

Slide a graph left, right, up or down and write its new equation (for example moving y = x² two units right gives y = (x − 2)²).

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Functions : Horizontal Translations - Horizontal Shift - y = f(x - c) Examples and ExplanationRadford Mathematics

Grade 12 Functions Student Question Find equation of p(x) if f(x) is shifted 2 units left | NTENTE \ New Technological Education

Horizontal Translation of Quadratic Functions - NerdstudyNerdstudy

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Question 7.5 · May/June 2025 · Paper 2 · National (DBE) · 2 marks

In the diagram, the graphs of

f (x) = 2 cos x + 1

and

g (x) = sin 2 x

are drawn for the interval

x \in [- 18 0^{\circ}; 18 0^{\circ}]

7.5 Graph g is shifted

4 5^{\circ}

to the right to obtain a new graph h.

Determine the equation of h in its simplest form.(2)

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More past paper questions: shift a graph

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3. Question 3.2.1 · March 2025 · Provincial · 2 marks
Given: $f (x) = (x - 1)^{2} - 9$
3.2 Consider the function $g (x) = x^{2}$ .
3.2.1 What transformation did the graph of $f (x) = (x - 1)^{2} - 9$ undergo in order to obtain the graph of $g$ .(2)
Show the memo (2 ticks)Memo
1. ✓
  Describe vertical shift9 units upwards
  1 mark · Shift a graph
2. ✓
  Describe horizontal shift1 unit to the left
  1 mark · Shift a graph
4. Question 3.2.3 · March 2022 · Provincial · 2 marks
3.2 The graph of $g (x) = - 2 cos 2 x$ for $x \in [- 18 0^{\circ}; 18 0^{\circ}]$ is drawn below.
3.2.3 For which value(s) of $x$ is $f (x + 3 0^{\circ}) - g (x + 3 0^{\circ}) = 0$ for $x \in [- 18 0^{\circ}; 18 0^{\circ}]$ ?(2)
Show the memo (2 ticks)Memo
1. ✓
  Interpret equation $f (x + 3 0^{\circ}) = g (x + 3 0^{\circ})$ implies intersection points shifted left by $3 0^{\circ}$ .
  Intersection points of original graphs from 3.1 ( $2 sin x - 1 = - 2 cos 2 x$ ... actually 3.1 was $4 sin x + 2 cos 2 x = 2 \to 2 sin x + cos 2 x = 1 \to cos 2 x = 1 - 2 sin x$ . Check graph equation: $f (x) = 2 sin x - 1, g (x) = - 2 cos 2 x$ . Intersection: $2 sin x - 1 = - 2 cos 2 x \to 2 sin x + 2 cos 2 x = 1$ . This is different from 3.1 equation $4 sin x + 2 cos 2 x = 2$ which simplifies to $2 sin x + cos 2 x = 1$ . Ah, $2 sin x - 1 = - 2 cos 2 x \Leftrightarrow 2 sin x + 2 cos 2 x = 1$ . The equation in 3.1 is $4 sin x + 2 cos 2 x = 2 \Rightarrow 2 sin x + cos 2 x = 1$ . So intersections are at $x$ values from 3.1: $- 18 0^{\circ}, 0^{\circ}, 9 0^{\circ}, 18 0^{\circ}$ .
  0 marks · Shift a graph
2. ✓
  Apply shiftWe need $x$ such that $(x + 3 0^{\circ})$ is a solution to $f (θ) = g (θ)$ .
  $θ \in {- 18 0^{\circ}; 0^{\circ}; 9 0^{\circ}; 18 0^{\circ}}$
  $x + 3 0^{\circ} = 0^{\circ} ⟹ x = - 3 0^{\circ}$
  $x + 3 0^{\circ} = 9 0^{\circ} ⟹ x = 6 0^{\circ}$
  $x + 3 0^{\circ} = 18 0^{\circ} ⟹ x = 15 0^{\circ}$
  2 marks · Shift a graph
5. Question 4.2.4 · March 2025 · Provincial · 3 marks
4.2 Consider $g (x) = (\frac{1}{3})^{x}$ .
4.2.4 Describe the translation from $h$ to $g$ .(3)
Show the memo (3 ticks)Memo
1. ✓
  Analyze equations $h (x) = \frac{1}{3} (\frac{1}{3})^{x} + 7 = (\frac{1}{3})^{x + 1} + 7$
  $g (x) = (\frac{1}{3})^{x}$
  1 mark · Use the laws of exponents
2. ✓
  Describe horizontal shift1 unit to the right (from $x + 1$ to $x$ )
  1 mark · Shift a graph
3. ✓
  Describe vertical shift7 units downwards (from $+ 7$ to $0$ )
  1 mark · Shift a graph
6. Question 6.4.2 · November 2025 · Paper 1 · Provincial · 3 marks
The graphs of $f (x) = - x^{2} - 6 x$ and $g (x) = a x^{2} + b x$ are drawn below. $g$ is the reflection of $f$ along the $x$ -axis. A and B $(- 5; n)$ are symmetrical on $f$ along $x = - p$ , i.e. the axes of symmetry. $Diagram for this question$
6.4 Determine the value(s) of $k$ for which:
6.4.2 $f (x + k)$ will have two real roots that have different signs.(3)
Show the memo (2 ticks)Memo
1. ✓
  $x$ -values $- x^{2} - 6 x = 0$
  $x^{2} + 6 x = 0$
  $x (x + 6) = 0$
  $x = 0$ or $x = - 6$
  1 mark · Find the x-intercepts
2. ✓
  Answer $∴ - 6 < k < 0$
  2 marks · Shift a graph

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