§-Syllabus dot point

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How do we use the first and second derivatives to sketch an accurate graph of a function?

Curve sketching combines intercepts, stationary points, concavity and end behaviour to produce an accurate graph.

A systematic method for sketching curves using intercepts, stationary points from the first derivative, and concavity and inflections from the second derivative.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The systematic checklist
Worked example
Using a sign diagram
Sketching the derivative from the original (and vice versa)
Working with restricted domains
Why it matters

What this dot point is asking

This dot point pulls together the first and second derivatives into a single skill: producing an accurate, fully labelled sketch of a function.

The systematic checklist

Work through these steps in order:

Intercepts. Find the $y$ -intercept by evaluating $f(0)$ , and $x$ -intercepts by solving $f(x)=0$ where practical.
Stationary points. Solve $f'(x)=0$ . Each solution is a stationary point; find its $y$ -coordinate.
Classify each stationary point as a local maximum, local minimum or stationary inflection - using either the second derivative test or a sign diagram of $f'$ .
Concavity and inflections. Solve $f''(x)=0$ and check for a sign change to locate points of inflection. Note where the curve is concave up or down.
End behaviour. Describe what happens as $x\to\pm\infty$ .
Plot and join the key points smoothly, consistent with the increasing/decreasing and concavity information.

Worked example

Sketching

f(x)=x^3-3x^2

Sketch $f(x)=x^3-3x^2$ , labelling all intercepts, stationary points, and points of inflection.

Step 1: Find the intercepts

Evaluating $f(0)=0$ gives the $y$ -intercept at the origin. To find the $x$ -intercepts, solve $f(x)=0$ . Factorising gives $x^3-3x^2=x^2(x-3)=0$ , so the curve meets the $x$ -axis at $x=0$ and $x=3$ .

Step 2: Find the stationary points

A stationary point occurs where the gradient is zero, so differentiate and set $f'(x)=0$ :

f'(x)=3x^2-6x=3x(x-2)=0.

This gives

x=0

and

x=2

. Substituting back to find the

y

-coordinates:

f(0)=0

, giving the point

(0,0)

; and

f(2)=8-12=-4

, giving the point

(2,-4)

Step 3: Classify each stationary point using the second derivative

Differentiating $f'(x)$ gives the second derivative $f''(x)=6x-6$ , which tells us the concavity at each stationary point.

At $x=0$ : $f''(0)=6(0)-6=-6<0$ , so the curve is concave down and $(0,0)$ is a local maximum.

At $x=2$ : $f''(2)=6(2)-6=6>0$ , so the curve is concave up and $(2,-4)$ is a local minimum.

Step 4: Locate the point of inflection

A point of inflection is where concavity changes, so solve $f''(x)=0$ :

6x-6=0 \implies x=1.

Check that

f''

changes sign: for

x<1

we have

f''<0

(concave down), and for

x>1

we have

f''>0

(concave up), confirming a genuine inflection. The

y

-coordinate is

f(1)=1-3=-2

, so the point of inflection is

(1,-2)

Step 5: Determine end behaviour

Because the leading term is $x^3$ with a positive coefficient, $f(x)\to\infty$ as $x\to\infty$ and $f(x)\to-\infty$ as $x\to-\infty$ . This confirms the curve rises on the right and falls on the left.

Step 6: Sketch the curve

Plot all key points: intercepts at $(0,0)$ and $(3,0)$ , local maximum at $(0,0)$ , local minimum at $(2,-4)$ , and point of inflection at $(1,-2)$ . The curve rises from the lower left to the local maximum at the origin, falls through the inflection point to the local minimum at $(2,-4)$ , then rises to cross the $x$ -axis again at $x=3$ and continues upward.

Final answer: local maximum $(0,0)$ , local minimum $(2,-4)$ , point of inflection $(1,-2)$ , with $x$ -intercepts at $x=0$ and $x=3$ .

Using a sign diagram

When the second derivative test is inconclusive (i.e. $f''=0$ at a stationary point), build a sign diagram of $f'(x)$ : pick test points either side of each stationary point and record the sign. A change from $+$ to $-$ is a maximum; $-$ to $+$ is a minimum; no change is a stationary point of inflection.

Sketching the derivative from the original (and vice versa)

SACE examiners frequently give you the graph of $f(x)$ and ask for the shape of $f'(x)$ , or the reverse. The translation is mechanical once you remember what each curve encodes. Where $f$ has a turning point, $f'$ crosses the $x$ -axis. Where $f$ is increasing, $f'$ is positive (above the axis); where $f$ is decreasing, $f'$ is negative. Where $f$ has a point of inflection, $f'$ has a turning point of its own. Reading these correspondences in both directions is one of the most reliable sources of marks in Topic 1, because the question tests understanding rather than algebra.

Working with restricted domains

When a function is only defined on a closed interval, say $-2 \le x \le 4$ , the endpoints become candidate maximum or minimum points even though they are not stationary. Evaluate $f$ at every stationary point and at both endpoints, then compare. The global maximum on the interval is simply the largest of those $y$ -values, and the global minimum the smallest. Marking the endpoints with their coordinates on the sketch is expected.

Why it matters

Curve sketching is where all of Topic 1 comes together, and it underpins later work: an accurate graph makes optimisation problems intuitive, helps you set up areas under curves in integral calculus, and lets you interpret models. The discipline of the checklist - intercepts, stationary points, concavity, end behaviour - is what turns a rough idea into a precise, mark-earning diagram.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20234 marksCalculator-assumed. The graph of the derivative

y = f'(x)

crosses the

x

-axis at the origin

O

and at

x = b

, and has a local maximum at point

A

where

x = a

(with

0 < a < b

). State whether

f'(x)

and

f''(x)

are positive, negative, or zero when

x = a

and when

x = b

Show worked answer →

Read the information off the graph of $f'(x)$ . Remember $f'(x)$ is the $y$ -value of the plotted derivative curve, and $f''(x)$ is the slope of that derivative curve.

At $x = a$ : $A$ is a local maximum of the derivative curve, sitting above the $x$ -axis, so $f'(a) > 0$ . Because $A$ is a turning point of $f'(x)$ , the slope of $f'(x)$ is zero there, so $f''(a) = 0$ .

At $x = b$ : the derivative curve crosses the $x$ -axis, so $f'(b) = 0$ . To the left of $b$ the derivative is positive and to the right it is negative, so the derivative curve is decreasing through $b$ , giving a negative slope, so $f''(b) < 0$ .

Marks: one each for $f'(a) > 0$ , $f''(a) = 0$ , $f'(b) = 0$ , $f''(b) < 0$ . The key idea is distinguishing the height of the $f'(x)$ curve (which gives $f'$ ) from its slope (which gives $f''$ ).

SACE 20225 marksCalculator-assumed. Consider

f(x) = x^3 - 3x^2 - 9x + 5

. Find the coordinates and nature of all stationary points, and hence sketch the curve showing these features.

Show worked answer →

Differentiate: $f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)$ . Setting $f'(x) = 0$ gives $x = 3$ and $x = -1$ .

Classify with $f''(x) = 6x - 6$ . At $x = -1$ , $f''(-1) = -12 < 0$ , so a local maximum; $f(-1) = -1 - 3 + 9 + 5 = 10$ , giving $(-1, 10)$ . At $x = 3$ , $f''(3) = 12 > 0$ , so a local minimum; $f(3) = 27 - 27 - 27 + 5 = -22$ , giving $(3, -22)$ .

The sketch rises to the maximum $(-1, 10)$ , falls through the inflection at $x = 1$ to the minimum $(3, -22)$ , then rises; as $x \to \pm\infty$ the cubic goes to $\pm\infty$ .

Marks: one for $f'(x)$ factorised, two for both stationary points with coordinates, one for classifying their nature, and one for a labelled sketch consistent with the features.