Grade 10 Maths · Functions

Solve an inequality from a graph

Use a graph to find the x-values where it is above or below the x-axis or another graph (for example where f(x) > 0).

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f(x) bigger than 0 | Grade 11 FunctionsKevinmathscience

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Question 6.4 · May/June 2024 · Paper 1 · National (DBE) · 2 marks

The graphs of

f (x) = x^{2} - 2 x - 3

and

g (x) = m x + c

are drawn below. D and E are the

x

-intercepts and P is the

y

-intercept of

f

. The turning point of

f

is T

(1; - 4)

. The graphs of

f

and

g

intersect at P and E. Diagram for this question

6.4 Write down the values of x for which

f (x) - g (x) > 0

.(2)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: solve an inequality from a graph

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3. Question 7.4.2 · May/June 2025 · Paper 2 · National (DBE) · 3 marks
In the diagram, the graphs of $f (x) = 2 cos x + 1$ and $g (x) = sin 2 x$ are drawn for the interval $x \in [- 18 0^{\circ}; 18 0^{\circ}]$ .
7.4 Use the graphs to determine the values of x, in the interval $x \in [- 18 0^{\circ}; 18 0^{\circ}]$ for which:
7.4.2 $cos x \leq - \frac{1}{2}$ (3)
Show the memo (3 ticks)Memo
1. ✓
  Relate the inequality to the given graph f(x) $cos x \leq - \frac{1}{2}$
  Multiply by 2:
  $2 cos x \leq - 1$
  Add 1 to both sides:
  $2 cos x + 1 \leq 0$
  This means we are looking for where $f (x) \leq 0$ .
  1 mark · Solve an inequality from a graph
2. ✓
  Find the x-intercepts of f(x) $2 cos x + 1 = 0$
  $cos x = - \frac{1}{2}$
  Reference angle $= 6 0^{\circ}$
  $x = 18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$
  $x = - 18 0^{\circ} + 6 0^{\circ} = - 12 0^{\circ}$
  1 mark · Find the x-intercepts
3. ✓
  Determine the intervals where the graph is below or on the x-axis $x \in [- 18 0^{\circ}; - 12 0^{\circ}] \cup [12 0^{\circ}; 18 0^{\circ}]$
  
  OR
  
  $- 18 0^{\circ} \leq x \leq - 12 0^{\circ}$ or $12 0^{\circ} \leq x \leq 18 0^{\circ}$
  1 mark · Solve an inequality from a graph
4. Question 4.4 · May/June 2025 · Paper 1 · National (DBE) · 4 marks
The graph of $f (x) = \frac{4}{x - 3} + 4$ is drawn below. M is the point where the asymptotes of $f$ intersect. C and D are the x- and y-intercepts respectively of $f$ . A is the point on $f$ that is closest to M.
4.4 Determine the values of $x$ for which $f (x) \leq 0$ .(4)
Show the memo (2 ticks)Memo
1. ✓
  Find the x-intercept C by equating f(x) to 0 $\frac{4}{x - 3} + 4 = 0$
  $\frac{4}{x - 3} = - 4$
  $4 = - 4 (x - 3)$
  $- 1 = x - 3$
  $x = 2$
  $C (2; 0)$
  2 marks · Find the x-intercepts
2. ✓
  Determine interval from the graphFrom the graph, $f (x) \leq 0$ occurs between the x-intercept and the vertical asymptote.
  $2 \leq x < 3$
  2 marks · Solve an inequality from a graph
5. Question 4.1.2 · March 2025 · Provincial · 2 marks
4.1 The graph of $f^{- 1} (x) = 3 x$ , $x \geq 0$ is drawn. $f^{- 1}$ passes through A(3; 3).
4.1.2 For which values of $x$ will $f (x) \leq f^{- 1} (x)$ ?(2)
Show the memo (1 tick)Memo
1. ✓
  Identify interval $0 \leq x \leq 3$ (Intersection is at origin and (3;3))
  2 marks · Solve an inequality from a graph
6. Question 4.1.3 · June 2025 · Paper 1 · Provincial · 2 marks
4.1 Given: $f (x) = - x^{2} + 4$
4.1.3 Determine the value(s) of $x$ for which $f (x) > 0$ .(2)
Show the memo (2 ticks)Memo
1. ✓
  Interval between the $x$ -intercepts $- 2$ and $2$ : the graph is above the $x$ -axis between them
  1 mark · Solve an inequality from a graph
2. ✓
  Notation $- 2 < x < 2$ OR $x \in (- 2; 2)$ (strict signs or round brackets, because $f (x) > 0$ , not $\geq 0$ )
  1 mark · Solve an inequality from a graph

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