f(0)=0, so the y-intercept is 0. Solve x^3-3x^2=0 x^2(x-3)=0, giving x-intercepts at x=0 and x=3.

What are stationary points?

f'(x)=3x^2-6x=3x(x-2). Setting f'(x)=0 gives x=0 and x=2. - f(0)=0(0,0).

f''(x)=6x-6. - f''(0)=-6 0, so (2,-4) is a local minimum.

What is end behaviour?

As x→∞, f(x)→∞; as x→-∞, f(x)→-∞ (cubic with positive leading coefficient).

SAMath MethodsSyllabus dot point

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 4.78 min answerUpdated 2026-05-29

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

Worked example

Sketch $f(x)=x^3-3x^2$ .

Intercepts. $f(0)=0$ , so the $y$ -intercept is $0$ . Solve $x^3-3x^2=0\Rightarrow x^2(x-3)=0$ , giving $x$ -intercepts at $x=0$ and $x=3$ .

Stationary points. $f'(x)=3x^2-6x=3x(x-2)$ . Setting $f'(x)=0$ gives $x=0$ and $x=2$ .

$f(0)=0\Rightarrow(0,0)$ .
$f(2)=8-12=-4\Rightarrow(2,-4)$ .

Classify. $f''(x)=6x-6$ .

$f''(0)=-6<0$ , so $(0,0)$ is a local maximum.
$f''(2)=6>0$ , so $(2,-4)$ is a local minimum.

Inflection. $f''(x)=6x-6=0$ at $x=1$ . Sign changes from negative to positive, so $(1,f(1))=(1,-2)$ is a point of inflection.

End behaviour. As $x\to\infty$ , $f(x)\to\infty$ ; as $x\to-\infty$ , $f(x)\to-\infty$ (cubic with positive leading coefficient).

The curve rises to the local max $(0,0)$ , falls through the inflection $(1,-2)$ to the local min $(2,-4)$ , then rises through the $x$ -intercept at $x=3$ .

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.

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.

2018 SACE Stage 24 marksThe 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 required 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 (positive). Because A is a turning point of f'(x), the slope of f'(x) is zero there, so f''(a) = 0 (zero).

At x = b: the derivative curve crosses the x-axis, so f'(b) = 0 (zero). 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 (negative).

Marks: one each for the four entries 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'').