Sketching Cubic Graphs (Grade 12)

Using the derivative to find stationary points, deciding which is a maximum and which a minimum, locating the point of inflection, and putting it all together in a sketch.

1What to find, in order

Intercepts — y-intercept by letting x = 0; x-intercepts by factorising.
Stationary points — solve f′(x) = 0.
Nature of each — maximum or minimum.
Point of inflection — solve f″(x) = 0.

2Worked example

Worked ExampleSketch f(x) = x³ − 3x² − 9x + 5

1
y-intercept: f(0) = 5.
Let x = 0.
2
f′(x) = 3x² − 6x − 9.
Differentiate term by term.
3
3(x² − 2x − 3) = 0 → 3(x − 3)(x + 1) = 0.
Take out the 3 first — it makes the factorising trivial.
4
x = 3 or x = −1.
The two stationary points.
5
f(−1) = −1 − 3 + 9 + 5 = 10 → (−1 ; 10).
6
f(3) = 27 − 27 − 27 + 5 = −22 → (3 ; −22).
7
Since a > 0 the graph rises, so (−1 ; 10) is the maximum and (3 ; −22) the minimum.
For a positive cubic the LEFT stationary point is always the maximum.
8
f″(x) = 6x − 6 = 0 → x = 1, and f(1) = −6.
Point of inflection at (1 ; −6) — exactly halfway between the turning points.

💡

The point of inflection of a cubic always sits midway between the two stationary points. Here −1 + 32 = 1. Use it as a free check on your work.

3Deciding maximum or minimum

f″(x) < 0 ⇒ maximum (concave down)
f″(x) > 0 ⇒ minimum (concave up)

For f″(x) = 6x − 6: at x = −1, f″ = −12 < 0, so it is a maximum. At x = 3, f″ = 12 > 0, so it is a minimum. ✓

4Where is the graph increasing?

f is increasing where f′(x) > 0 and decreasing where f′(x) < 0. For this graph it increases for x < −1 and x > 3, and decreases for −1 < x < 3.

⚠️

'Concave up' is about f″, not f′. The graph is concave up for x > 1 — that is where f″(x) = 6x − 6 is positive — even though it is still decreasing between 1 and 3.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Determine the y-intercept of f(x) = x³ − 3x² − 9x + 5.
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f(0) = 5
2
Determine the coordinates of the stationary points of f(x) = x³ − 3x² − 9x + 5.
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f′(x) = 3(x−3)(x+1) = 0 → (−1 ; 10) and (3 ; −22)
3
State, with a reason, which stationary point is the local maximum.
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(−1 ; 10), because f″(−1) = −12 < 0 (concave down).
4
Determine the coordinates of the point of inflection.
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f″(x) = 6x − 6 = 0 → x = 1 → (1 ; −6)
5
For which values of x is f decreasing?
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−1 < x < 3
6
For which values of x is f concave up?
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f″(x) > 0 → x > 1
7
Where does the point of inflection of a cubic always lie relative to the turning points?
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Exactly midway between them.
8
Determine the stationary points of g(x) = x³ − 12x.
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g′ = 3x² − 12 = 0 → x = ±2 → (2 ; −16) and (−2 ; 16)