NCEA Level 2Gradient graphsAS 91262

Gradient graphs: 7 exam questions solved step by step

What this lesson covers

  • The gradient graph is zero where the function turns, positive where it rises and negative where it falls.
  • A cubic has a quadratic gradient graph and a quadratic has a straight-line gradient graph.
Key idea

7 questions, one method

  • 1. Turning point: f′ = 0, an x-intercept.
  • 2. Rising: f′ above the axis.
  • 3. Falling: f′ below the axis.
  • 4. Degree drops by one: one smooth curve.
Your turn

Question 1 · Achieved

  • The graph of f(x) = x² + 2x − 3 is shown.
  • Sketch the gradient graph, y = f′(x).
Question 1 · Achieved
Worked example

Q1: f(x) = x² + 2x − 3

  • Turning point first.
  • Then the signs.
  1. f turns at x = −1its minimum
  2. so f′(−1) = 0the x-intercept
  3. x < −1: falling, f′ < 0
  4. x > −1: rising, f′ > 0
  5. quadratic, so f′ is a linedegree drops by one
  6. f′(x) = 2x + 2differentiate
  7. 2x + 2 = 0 at x = −1
  8. f′(0) = 2the y-intercept
  9. line through (−1, 0) and (0, 2)
Worked example

Q1: the sketch

  • Cross at x = −1.
  • Below, then above.
  • One straight line.
Q1: the sketch
Your turn

Question 2 · Achieved, in context

  • A drone takes off, rises, then comes down.
  • Its gradient graph: dh/dt = 6 − 2t (m/s)
  • When is the drone highest?
  • Rising or falling at t = 4 s?
Question 2 · Achieved, in context
Worked example

Q2: the drone

  • Highest: gradient zero.
  • Sign before and after.
  1. dh/dt = 0: 6 − 2t = 0stops rising
  2. t = 3
  3. t < 3: dh/dt > 0, rising
  4. t > 3: dh/dt < 0, falling
  5. + then −: a maximumsign change
  6. highest at t = 3 s
  7. t = 4: 6 − 2 × 4 = −2
  8. falling at 2 m/s when t = 4 s
Your turn

Question 3 · Merit

  • The cubic shown has a minimum at x = −1
  • and a maximum at x = 3.
  • Sketch y = f′(x). Where is it greatest?
Question 3 · Merit
Worked example

Q3: a cubic that ends falling

  • Intercepts, then signs.
  • Do not copy the curve.
  1. f′ = 0 at x = −1 and x = 3turning points
  2. x < −1: falling, f′ < 0
  3. −1 < x < 3: rising, f′ > 0
  4. x > 3: falling, f′ < 0
  5. cubic, so f′ is a paraboladegree drops by one
  6. − + −: it opens downwardupside down
  7. top at x = (−1 + 3) ÷ 2 = 1symmetric
  8. f′ greatest at x = 1
Worked example

Q3: the sketch

  • Under each turning point.
  • Opens downward.
  • Top at x = 1.
Q3: the sketch
Your turn

Question 4 · Merit

  • f(x) = x³ − 6x² + 9x + 1
  • Where does y = f′(x) cross the x-axis?
  • Find its lowest point, then sketch it.
Question 4 · Merit
Worked example

Q4: f(x) = x³ − 6x² + 9x + 1

  • Term by term.
  • Factorise for the zeros.
  1. x³ → 3x², −6x² → −12xpower down, minus 1
  2. 9x → 9, 1 → 0a constant → 0
  3. f′(x) = 3x² − 12x + 9
  4. = 3(x² − 4x + 3)take out 3
  5. = 3(x − 1)(x − 3)factorise
  6. crosses at x = 1 and x = 3
  7. lowest at x = (1 + 3) ÷ 2 = 2
  8. f′(2) = 12 − 24 + 9 = −3
  9. lowest point (2, −3)
Worked example

Q4: the sketch

  • Through x = 1 and x = 3.
  • Lowest at (2, −3).
  • Meets the y-axis at 9.
Q4: the sketch
Your turn

Question 5 · Merit

  • The gradient graph y = f′(x) is shown.
  • Find the turning points of f, and say
  • which is a max and which is a min.
  • Sketch a possible graph of f.
Question 5 · Merit
Worked example

Q5: from f′ back to f

  • Zeros are turning points.
  • Sign change: max or min.
  1. f′ = 0 at x = −2 and x = 1turning points
  2. x < −2: f′ > 0, f rising
  3. −2 < x < 1: f′ < 0, f falling
  4. x > 1: f′ > 0, f rising
  5. x = −2: + then −, a maximum
  6. x = 1: − then +, a minimum
  7. f′ is a parabola, so f is cubic
  8. max at x = −2, min at x = 1
Worked example

Q5: a possible f

  • Rise, fall, rise.
  • Max at −2, min at 1.
  • Any height will do.
Q5: a possible f
Your turn

Question 6 · Excellence

  • f(x) = x³ − 3x² + 3x + 2
  • Sam says f′(1) = 0, so f turns at x = 1.
  • Is Sam right? Sketch y = f′(x).
  • Careful: zero gradient is not always a turn.
Question 6 · Excellence
Worked example

Q6: is it a turning point?

  • Check the sign either side.
  • No sign change, no turn.
  1. f′(x) = 3x² − 6x + 3differentiate
  2. = 3(x² − 2x + 1)take out 3
  3. = 3(x − 1)²a perfect square
  4. f′(1) = 3(0)² = 0stationary
  5. f′(0) = 3(−1)² = 3 > 0rising before
  6. f′(2) = 3(1)² = 3 > 0rising after
  7. turning point at x = 1no sign change
  8. f′ touches the axis at x = 1
  9. f flattens at x = 1, keeps rising
Worked example

Q6: the sketch

  • Touches at x = 1, no crossing.
  • Above the axis everywhere else.
Q6: the sketch
Your turn

Question 7 · Excellence

  • A cubic f has a positive x³ term and
  • turning points at x = a and x = b.
  • Show f′ is lowest at x = (a + b) ÷ 2.
  • Then use it for a = −3 and b = 5.
Question 7 · Excellence
Worked example

Q7: any cubic

  • f′ is zero at a and b.
  • Its turn is midway.
  1. f′(x) = k(x − a)(x − b), k > 0
  2. = k(x² − (a + b)x + ab)expand
  3. k(2x − (a + b)) = 0differentiate f′
  4. 2x = a + b
  5. x = (a + b) ÷ 2midway
  6. k > 0: f′ opens up, a minimum
  7. a = −3, b = 5: x = (−3 + 5) ÷ 2
  8. = 1
  9. f falls most steeply at x = 1
Summary

The method that solved all seven

  • Turns, signs, degree, then one smooth curve.
StepHow
Turnsf′ = 0: crosses the x-axis
Risingf′ above the axis
Fallingf′ below the axis
Degreedrops by one
Steepestfurthest from the axis
Reversesign change: max or min
No changeflat, but not a turn

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