Grade 12 Maths · Calculus · Paper 1

Stationary Points Grade 12

Real Stationary Points 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: 14 past paper questions
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Question 9.2 · May/June 2025 · Paper 1 · National (DBE) · 4 marks

Given:

f (x) = x^{3} - 6 x^{2} + 9 x - 4 = (x - 4) (x - k)^{2}

9.2 Calculate the coordinates of the turning points of

f

.(4)

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

Stationary Points skills

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More Stationary Points Grade 12 questions and answers

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4. Question 9.2 · May/June 2024 · Paper 1 · National (DBE) · 4 marks
The graph of $f (x) = a x^{3} + b x^{2} + c x - 5$ is drawn below. $E (- 1; 0)$ and $G (5; 0)$ are the $x$ -intercepts of $f$ .
9.2 Calculate the value of x for which f has a local minimum value.(4)
Show the memo (4 ticks)Memo
1. ✓
  Find the derivative of the function $f (x) = x^{3} - 3 x^{2} - 9 x - 5$ $f^{'} (x) = 3 x^{2} - 6 x - 9$
  1 mark · Differentiate using the power rule
2. ✓
  Equate the derivative to zero $3 x^{2} - 6 x - 9 = 0$
  1 mark · Find stationary points
3. ✓
  Factorize the quadratic equation $(x - 3) (x + 1) = 0$
  1 mark · Factorise a trinomial
4. ✓
  Identify the x-value for the local minimum $x = 3$
  1 mark · Find stationary points
5. Question 3.1.5(a) · March 2025 · Provincial · 2 marks
3.1 Given: $f (x) = \frac{2 x + 3}{- x - 3}$

Given: $f (x) = \frac{2 x + 3}{- x - 3}$
3.1.5 Determine the value(s) of $x$ for which:
3.1.5(a) $f$ is decreasing.(2)
Show the memo (2 ticks)Memo
1. ✓
  $x \in R$ $x \in R$
  1 mark · State increasing and decreasing intervals
2. ✓
  $x \neq = - 3$ $x \neq = - 3$
  1 mark · State increasing and decreasing intervals
6. Question 8.3 · September 2024 · Paper 1 · Provincial · 6 marks
8.3 A function $g (x) = a x^{2} + \frac{b}{x}$ has a minimum value at $x = 4$ . The function value at $x = 4$ is 96. Calculate the values of a and b.(6)
Show the memo (6 ticks)Memo
1. ✓
  Derivative is 0 at the minimumMinimum at $x = 4$ , so $g^{'} (4) = 0$
  1 mark · Find stationary points
2. ✓
  Derivative $g (x) = a x^{2} + b x^{- 1}$ $g^{'} (x) = 2 a x - b x^{- 2}$
  1 mark · Differentiate using the power rule
3. ✓
  Equation 1 $2 a (4) - b (4)^{- 2} = 0$ $8 a - \frac{b}{16} = 0$ $128 a - b = 0$ ...(1)
  1 mark · Find stationary points
4. ✓
  Equation 2 $g (4) = 96$ : $16 a + \frac{b}{4} = 96$ $64 a + b = 384$ ...(2)
  1 mark · Substitute into a formula
5. ✓
  Value of a(1) + (2): $192 a = 384$ $a = 2$
  1 mark · Solve simultaneous equations by elimination
6. ✓
  Value of bFrom (1): $b = 128 (2) = 256$
  1 mark · Solve simultaneous equations by substitution
7. Question 9.1.2 · September 2024 · Paper 1 · Provincial · 4 marks
9.1 The graphs of $g^{'} (x) = a x^{2} + b x + c$ and $h (x) = 2 x - 4$ are sketched below. The graph of $g^{'} (x) = a x^{2} + b x + c$ is the derivative graph of a cubic function g. The graphs of h and $g^{'}$ have a common y-intercept at point E. C(-2;0) and D(6;0) are the x-intercepts of the graph of $g^{'}$ . Point A is the x-intercept of h and point B is the turning point of $g^{'}$ . Line AB is parallel to the y-axis.
9.1.2 Determine the equation of the graph of $g^{'}$ in the form $y = a x^{2} + b x + c .$ (4)
Show the memo (4 ticks)Memo
1. ✓
  Substitute roots and point $y = a (x + 2) (x - 6)$ $- 4 = a (2) (- 6)$
  1 mark · Find the equation of a parabola
2. ✓
  Value of a $12 a = 4$ $a = \frac{1}{3}$
  1 mark · Solve a linear equation
3. ✓
  Substitute a and simplify $y = \frac{1}{3} (x + 2) (x - 6) = \frac{1}{3} (x^{2} - 4 x - 12)$
  1 mark · Find the equation of a parabola
4. ✓
  Equation $y = \frac{1}{3} x^{2} - \frac{4}{3} x - 4$
  1 mark · Find the equation of a parabola
8. Question c · Siyavula Grade 12 Mathematics, 6.7 Applications of differential calculus, Worked example 23 · 1 mark
The height (in metres) of a golf ball $t$ seconds after it has been hit into the air, is given by $H (t) = 20 t - 5 t^{2}$ . Determine the following:
c The time at which the vertical velocity is zero.(1)
Show the memo (1 tick)Memo
1. ✓
  v = 0 $v (t) = 0$
  $20 - 10 t = 0$
  $t = 2$
  The velocity is zero after $2 s$ .
  1 mark · Find stationary points

Stationary Points Grade 12 videos

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

Hyperbola Increasing Decreasing | Grade 11 FunctionsKevinmathscience

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