NCEA Level 2Turning pointsAS 91262

Turning points: 7 exam questions solved step by step

What this lesson covers

  • At a turning point the gradient is zero.
  • Whether it is a maximum or a minimum is proved by the second derivative or the gradients either side.
Key idea

7 questions, one method

  • 1. Differentiate and set f′(x) = 0.
  • 2. Solve for every x.
  • 3. Find each y from f, not f′.
  • 4. Prove max or min, with a reason.
Your turn

Question 1 · Achieved

  • y = x² − 8x + 3
  • Find the turning point and
  • show that it is a minimum.
Worked example

Q1: y = x² − 8x + 3

  • Gradient zero.
  • y″ > 0: a minimum.
  1. dy/dx = 2x − 8differentiate
  2. 2x − 8 = 0, so x = 4gradient zero
  3. y = 4² − 8(4) + 3the height
  4. = 16 − 32 + 3 = −13
  5. y″ = 2 > 0a minimum
  6. minimum at (4, −13)
Your turn

Question 2 · Achieved

  • y = 5 + 6x − 3x²
  • Find the turning point. Use the
  • gradient either side to classify it.
Worked example

Q2: y = 5 + 6x − 3x²

  • Test x = 0 and x = 2.
  • + then −: a maximum.
  1. dy/dx = 6 − 6xdifferentiate
  2. 6 − 6x = 0, so x = 1gradient zero
  3. y = 5 + 6(1) − 3(1)² = 8
  4. x = 0: 6 − 0 = 6 > 0rising
  5. x = 2: 6 − 12 = −6 < 0falling
  6. maximum at (1, 8)
Your turn

Question 3 · Merit

  • y = x³ − 3x² − 9x + 5
  • Find both turning points
  • and the nature of each.
Worked example

Q3: y = x³ − 3x² − 9x + 5

  • Take out 3, factorise.
  • Classify each with y″.
  1. dy/dx = 3x² − 6x − 9
  2. 3x² − 6x − 9 = 0gradient zero
  3. 3(x − 3)(x + 1) = 0take out 3, factorise
  4. x = 3 or x = −1
  5. y(3) = 27 − 27 − 27 + 5 = −22
  6. y(−1) = −1 − 3 + 9 + 5 = 10
  7. y″ = 6x − 6
  8. y″(3) = 12 > 0: min (3, −22)
  9. y″(−1) = −12 < 0: max (−1, 10)
Your turn

Question 4 · Merit

  • y = x⁴ − 2x²
  • Find all the turning points
  • and the nature of each.
  • How many are there?
Worked example

Q4: y = x⁴ − 2x²

  • Never divide by x.
  • Factor it out instead.
  1. dy/dx = 4x³ − 4x
  2. 4x³ − 4x = 0gradient zero
  3. x² = 1÷ 4x loses x = 0
  4. 4x(x² − 1) = 0take out 4x
  5. 4x(x − 1)(x + 1) = 0difference of squares
  6. x = 0, x = 1 or x = −1
  7. y(0) = 0
  8. y(±1) = 1 − 2 = −1
Worked example

Q4: the nature of each point

  • One y″ for each x.
  • State all three.
  1. (0, 0), (1, −1), (−1, −1)
  2. y″ = 12x² − 4differentiate again
  3. x = 0: y″ = −4 < 0max (0, 0)
  4. x = 1: y″ = 12 − 4 = 8 > 0min (1, −1)
  5. x = −1: y″ = 12 − 4 = 8 > 0min (−1, −1)
  6. max (0, 0); min (±1, −1)
Your turn

Question 5 · Merit, in context

  • In a Pukekohe glasshouse,
  • T = 12 + 6t − 0.5t² (°C),
  • t hours after 6 am (0 ≤ t ≤ 12).
  • Find the highest temperature, and when.
Worked example

Q5: the glasshouse

  • The maximum is a T value.
  • Answer with units and time.
  1. dT/dt = 6 − tdifferentiate
  2. 6 − t = 0, so t = 6gradient zero
  3. T = 12 + 6(6) − 0.5(6)²the height
  4. = 12 + 36 − 18 = 30
  5. T″ = −1 < 0a maximum
  6. t = 6: 6 am + 6 h = 12 noon
  7. 30 °C at 12 noon
Your turn

Question 6 · Excellence

  • y = x³ + kx² − 24x + 2 has a
  • turning point at x = 2.
  • Find k, then find the other turning
  • point. Justify the nature of both.
Worked example

Q6: find k

  • Turning point: gradient 0.
  • Put x = 2 into dy/dx.
  1. dy/dx = 3x² + 2kx − 24k is a constant
  2. 3(2)² + 2k(2) − 24 = 0flat at x = 2
  3. 12 + 4k − 24 = 0
  4. 4k = 12, so k = 3
Worked example

Q6: both turning points

  • x = 2 must be a root.
  • Justify both.
  1. y = x³ + 3x² − 24x + 2
  2. dy/dx = 3x² + 6x − 24 = 0
  3. 3(x + 4)(x − 2) = 0take out 3, factorise
  4. x = 2 or x = −4
  5. y(2) = 8 + 12 − 48 + 2 = −26
  6. y(−4) = −64 + 48 + 96 + 2 = 82
  7. y″ = 6x + 6
  8. x = 2: y″ = 18 > 0min (2, −26)
  9. x = −4: y″ = −18 < 0max (−4, 82)
Your turn

Question 7 · Excellence

  • y = x³ − 3x² + 3x + 1
  • Find where the gradient is zero.
  • Is it a turning point? Justify.
Worked example

Q7: flat, but does it turn?

  • y″ = 0 proves nothing.
  • Test either side.
  1. dy/dx = 3x² − 6x + 3 = 0
  2. 3(x − 1)² = 0take out 3, factorise
  3. x = 1: y = 1 − 3 + 3 + 1 = 2
  4. y″ = 6x − 6 = 0 at x = 1no verdict
  5. x = 0: dy/dx = 3 > 0rising
  6. x = 2: 12 − 12 + 3 = 3 > 0rising
  7. (1, 2): flat, not a turning point
Summary

The method that solved all seven

  • Find every point, then prove its nature.
StepHow
Find xsolve f′(x) = 0, all roots
Find ysubstitute into f, not f′
f″ testf″ < 0 max, f″ > 0 min
f″ = 0test the gradient either side
Either side+ then − max, − then + min
State itcoordinates, nature, reason

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