Grade 11 Maths · Functions · Paper 1

Graph Interpretation Grade 11

Real Graph Interpretation past paper questions and answers from 2024 to 2025, with each memo split into the ticks examiners give. Try one, mark yourself, and see exactly which skill cost you marks.

past paper questions: 36 past paper questions
skills: 6 skills
videos: 6 videos

Try a Graph Interpretation exam question

No sign-up. Takes 3 minutes.

Question 5.5 · November 2025 · Paper 1 · Provincial · 2 marks

The graphs of

f (x) = \frac{a}{x + p} + q

and

g (x) = \frac{x}{3} + c

are drawn below. C

(3; - 1)

is the point of intersection between the asymptotes of

f

. A and B are

x

- and

y

-intercepts of

f

and

g

respectively. E

(4; 2)

is a point on

f

in the first quadrant. $Diagram for this question$

5.5 Determine the value of

x

, for which

f (x) \cdot g (x) < 0

(2)

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

Graph Interpretation skills

Each skill is one thing a memo gives a tick for. Tap one for its own questions, notes and videos.

More Graph Interpretation Grade 11 questions and answers

Try each one on paper before you open the memo.

4. Question 5.5 · November 2025 · Paper 2 · Provincial · 2 marks
Given: $f (x) = a sin (\frac{x}{2}) - 1$ and $g (x) = cos x + q$ , $x \in [- 90°; 270°]$

The graphs of $f$ and $g$ are drawn below.

Use the graphs to answer the following questions where $x \in [- 90°; 270°]$ : $Diagram for this question$
5.5 For which values of $x$ is $f (x) . g (x) \leq 0$ ?(2)
Show the memo (1 tick)Memo
1. ✓
  Correct answer (✓✓) $- 90° \leq x \leq 90°$ (also $x = 180°$ and $x = 270°$ , where one of the graphs is zero)
  2 marks · Solve product inequality from graphs
5. Question 1.5 · June 2024 · Paper 2 · Provincial · 4 marks
$PQRS$ is a quadrilateral with vertices $P (- 5; 0)$ , $Q (0; t)$ , $R (3; - 2)$ and $S (- 2; - 5)$ .
$P$ is on the x-axis and $Q$ is on the y-axis.
1.5 Calculate the size of $P \hat{Q} R$ .(4)
Show the memo (4 ticks)Memo
1. ✓
  METHOD 1: Substitution for the gradient of QR $m_{QR} = \frac{- 2 - 3}{3 - 0}$
  1 mark · Calculate gradient between two points
2. ✓
  Gradient of QR $m_{QR} = - \frac{5}{3}$
  1 mark · Calculate gradient between two points
3. ✓
  Product of the gradients (method) $m_{PQ} \times m_{QR} = \frac{3}{5} \times (- \frac{5}{3}) = - 1$
  1 mark · Use parallel and perpendicular gradients
4. ✓
  Answer $PQ ⊥ QR$ , so $P \hat{Q} R = 9 0^{\circ}$
  1 mark · Use parallel and perpendicular gradients
6. Question 2.5 · June 2024 · Paper 2 · Provincial · 5 marks
In the diagram below, $A (- 2; 4)$ , $B (- 2; - 1)$ and $C (p; - 4)$ are the vertices of $Δ A BC$ . $Q$ is the x-intercept of the line $A C$ .
$θ$ is the angle of inclination of $A C$ and $B \hat{A} C = 3 7^{\circ}$ .
2.5 Determine the equation of the line parallel to $BC$ and passing through point $A$ .(5)
Show the memo (5 ticks)Memo
1. ✓
  Substitution of B and C $m_{BC} = \frac{- 4 + 1}{4 + 2}$
  1 mark · Calculate gradient between two points
2. ✓
  Gradient of BC $m_{BC} = - \frac{1}{2}$
  1 mark · Calculate gradient between two points
3. ✓
  Gradient of the new line $m_{n e w} = m_{BC} = - \frac{1}{2}$
  1 mark · Use parallel and perpendicular gradients
4. ✓
  Substitution of point A $y - 4 = - \frac{1}{2} (x + 2)$
  1 mark · Find the equation of a line
5. ✓
  Equation $y = - \frac{1}{2} x - 1 + 4$ $y = - \frac{1}{2} x + 3$
  1 mark · Find the equation of a line
7. Question 3.1 · November 2024 · Paper 2 · Provincial · 2 marks
In the diagram, P $(x; y)$ , Q $(- 2; 2)$ and R $(2; 3)$ are the vertices of triangle PQR. Line PR passes through the point of origin at O. The equation of line PQ is given as $y = 6 x + 14$ . S is the $x$ -intercept of line QP.

Note: the question text gives R $(2; 3)$ but the diagram labels R $(3; 4)$ . The official memo marks both versions: method 1 uses R $(3; 4)$ , method 2 uses R $(2; 3)$ . $Diagram for this question$
3.1 Calculate the gradient of PR.(2)
Show the memo (2 ticks)Memo
1. ✓
  METHOD 1 (R(3;4)): Correct substitution $m_{PR} = \frac{4 - 0}{3 - 0}$
  1 mark · Calculate gradient between two points
2. ✓
  METHOD 1 (R(3;4)): Answer $= \frac{4}{3}$
  1 mark · Calculate gradient between two points
8. Question 4.1.2 · November 2024 · Paper 2 · Provincial · 2 marks
In the diagram, $Δ$ ABC is drawn with vertices A $(- 7; 0)$ , B $(3; 8)$ and C. P is the midpoint of line AB. Q is a point on line AC. AQC is a line on the $x$ -axis. E $(- \frac{1}{2}; 2)$ is a point on line PQ. PQ $∥$ BC $Diagram for this question$
4.1 Calculate the:
4.1.2 Gradient of AB(2)
Show the memo (2 ticks)Memo
1. ✓
  Substitution $m = \frac{8 - 0}{3 + 7}$
  1 mark · Calculate gradient between two points
2. ✓
  Answer $= \frac{4}{5}$
  1 mark · Calculate gradient between two points
9. Question 5.6 · November 2024 · Paper 1 · Provincial · 3 marks
The lines $y = - x + 4$ and $y = x - 2$ are the axes of symmetry of the function $f (x) = \frac{- 3}{x + p} + q$ .
5.6 For which values of $x$ will $x . f (x) \leq 0$ ?(3)
Show the memo (2 ticks)Memo
1. ✓
  $x \leq 0$ $x \leq 0$
  1 mark · Solve product inequality from graphs
2. ✓
  $3 < x \leq 6$ $3 < x \leq 6$
  2 marks · Solve product inequality from graphs
10. Question 6.5 · November 2024 · Paper 1 · Provincial · 5 marks
Sketched below is the graph of $f (x) = x^{2} - 4 x - 5$ for $x \in (- \infty; 6]$ .

A and B are the $x$ -intercepts and C the $y$ -intercept of $f$ . $Diagram for this question$
6.5 Suppose $h$ is a tangent, to the curve $f$ at a certain point, with a gradient of $- 2$ . Calculate the $y$ -intercept of $h$ .(5)
Show the memo (5 ticks)Memo
1. ✓
  Equating $f (x) = h (x)$ $h (x) = - 2 x + c$
  $f (x) = x^{2} - 4 x - 5$
  $x^{2} - 4 x - 5 = - 2 x + c$
  1 mark · Find where two graphs intersect
2. ✓
  Simplification $x^{2} - 2 x - 5 - c = 0$
  1 mark · Write a quadratic in standard form
3. ✓
  $Δ = 24 + 4 c$ $Δ = b^{2} - 4 a c$
  $= (- 2)^{2} - 4 (1) (- 5 - c)$
  $= 4 + 20 + 4 c$
  $= 24 + 4 c$
  1 mark · Calculate the discriminant
4. ✓
  $Δ = 0$ Tangent $\Rightarrow$ one solution
  $∴ Δ = 0$
  1 mark · Use the discriminant to find k
5. ✓
  $c = - 6$ $24 + 4 c = 0$
  $4 c = - 24$
  $c = - 6$
  Therefore, $y$ -int $= - 6$
  1 mark · Use the discriminant to find k

Graph Interpretation Grade 11 videos

Short lessons from South African teachers first. Videos load only when you tap play, to save your data.

Straight line graphs grade 9: Gradient Part 1Kevinmathscience

Vertical Distance Grade 11 FunctionsKevinmathscience

Parabola + Straight Line| Grade 11 FunctionsKevinmathscience

f(x) bigger than 0 | Grade 11 FunctionsKevinmathscience

For Which Values of K | Grade 11 FunctionsKevinmathscience

for which values of x is f(x).g(x)≥0. functions grade 12Benit mathematics

Free on Android

Know exactly which skill is costing you marks.

Mark past papers tick by tick in the NavyBlue app. It remembers every mark you lose, shows you the skills to fix first, and picks your next questions.

The NavyBlue app showing a learner's weak skills