§-Syllabus dot point

WAMath MethodsSyllabus dot point

How do we combine intercepts, stationary points, concavity and asymptotes into a complete and accurate sketch of a curve?

Sketch curves using intercepts, stationary points, their nature, points of inflection, asymptotes and end behaviour

WACE Year 12 Mathematics Methods Unit 3 curve sketching with calculus: a systematic method using intercepts, stationary points and their nature, points of inflection, asymptotes and end behaviour, with a fully worked SCSA-style sketch.

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

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What this dot point is asking
A systematic method
Putting it together
Increasing, decreasing and the sign of $f'$
Asymptotes for non-polynomial curves

What this dot point is asking

SCSA Unit 3 asks you to assemble the calculus tools into one organised process that produces an accurate graph. This dot point is examined in both the calculator-free and calculator-assumed sections; in the calculator-free section the sketch must come entirely from by-hand analysis.

A systematic method

Work through the same checklist every time so no feature is missed.

Putting it together

The features divide the $x$ -axis into intervals on which the curve is increasing or decreasing and concave up or down. Once the key points are plotted, the curve is drawn to match both the gradient sign (from $f'$ ) and the concavity (from $f''$ ) on each interval.

A full curve sketch

We sketch $f(x)=x^{3}-3x^{2}$ by systematically finding its key features before drawing.

Step 1: Find intercepts

The $y$ -intercept is $f(0)=0$ . Setting $f(x)=0$ gives $x^{2}(x-3)=0$ , so the $x$ -intercepts are at $x=0$ (a repeated root, so the curve touches the axis there) and $x=3$ . Identifying the repeated root is important because the curve touches but does not cross the axis at $x=0$ .

Step 2: Find stationary points

Differentiate to get $f'(x)=3x^{2}-6x=3x(x-2)$ . Setting $f'(x)=0$ gives $x=0$ and $x=2$ . The corresponding $y$ -values are $f(0)=0$ and $f(2)=8-12=-4$ , giving candidate stationary points $(0,0)$ and $(2,-4)$ .

Step 3: Classify stationary points

The second derivative is $f''(x)=6x-6$ . At $x=0$ , $f''(0)=-6<0$ , confirming a local maximum at $(0,0)$ . At $x=2$ , $f''(2)=6>0$ , confirming a local minimum at $(2,-4)$ . The sign of $f''$ tells us the concavity and therefore the nature of each turning point.

Step 4: Find the inflection point

Set $f''(x)=6x-6=0$ to get $x=1$ . The sign of $f''$ changes from negative (for $x<1$ ) to positive (for $x>1$ ), confirming a genuine inflection. The $y$ -value is $f(1)=1-3=-2$ , so the point of inflection is $(1,-2)$ .

Step 5: Describe 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 overall rise-fall-rise shape between the two stationary points.

Final answer: The sketch rises from $-\infty$ to the local maximum $(0,0)$ , falls through the inflection $(1,-2)$ to the local minimum $(2,-4)$ , then rises through $(3,0)$ toward $+\infty$ . Markers reward all labelled features and a curve whose shape matches the concavity.

Increasing, decreasing and the sign of $f'$

Between the stationary points the sign of $f'$ tells you whether the curve rises or falls. Where $f'(x)>0$ the curve is increasing; where $f'(x)<0$ it is decreasing. A neat way to organise this is a sign table for $f'$ across the critical $x$ -values, which simultaneously classifies the stationary points and tells you the shape on each interval. For the cubic above, $f'(x)=3x(x-2)$ is positive for $x<0$ , negative for $0<x<2$ and positive again for $x>2$ , confirming the rise-fall-rise pattern.

Asymptotes for non-polynomial curves

Polynomials have no asymptotes, but rational and exponential functions do. For a rational function, a vertical asymptote occurs where the denominator is zero (and the numerator is not), and the horizontal asymptote is found from the behaviour as $x\to\pm\infty$ . For $y=e^{x}$ the line $y=0$ is a horizontal asymptote as $x\to-\infty$ .

Sketching a rational function

Sketch $y = \dfrac{4x}{x^{2}+1}$

Intercepts. Setting $x=0$ gives $y=0$ , and setting $y=0$ gives $x=0$ . The only intercept is the origin.

Asymptote. As $x\to\pm\infty$ , the degree of the denominator exceeds the numerator, so $y\to 0$ . The horizontal asymptote is $y=0$ . The denominator $x^{2}+1$ is never zero, so there is no vertical asymptote.

Stationary points. By the quotient rule,

y'=\frac{4(x^{2}+1)-4x(2x)}{(x^{2}+1)^{2}}=\frac{4-4x^{2}}{(x^{2}+1)^{2}}.

Setting the numerator to zero,

4-4x^{2}=0

gives

x=\pm 1

. Then

y(1)=2

and

y(-1)=-2

Nature. The denominator is always positive, so the sign of $y'$ follows $4-4x^{2}$ : positive for $-1<x<1$ and negative outside. So $(1,2)$ is a local maximum and $(-1,-2)$ is a local minimum.

The curve passes through the origin, rises to the maximum $(1,2)$ , then decreases back toward the asymptote $y=0$ , and is symmetric through the origin (an odd function). Markers reward the asymptote, both stationary points with their nature, and the symmetry observation.

Exam-style practice questions

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

WACE 20228 marksCalculator-free. Consider

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

. (a) Find

f'(x)

and the coordinates of the stationary points. (b) Determine the nature of each stationary point using the second derivative. (c) Find the coordinates of the point of inflection.

Show worked answer →

A standard calculator-free curve-analysis question.

(a) $f'(x)=3x^{2}-6x-9=3(x^{2}-2x-3)=3(x-3)(x+1)$ . Setting $f'(x)=0$ gives $x=3$ and $x=-1$ . Then $f(3)=27-27-27+5=-22$ and $f(-1)=-1-3+9+5=10$ . Stationary points $(3,-22)$ and $(-1,10)$ .

(b) $f''(x)=6x-6$ . At $x=3$ , $f''(3)=12>0$ , so $(3,-22)$ is a local minimum. At $x=-1$ , $f''(-1)=-12<0$ , so $(-1,10)$ is a local maximum.

Markers reward the factorised derivative, both stationary points with $y$ -values, the correct nature, and the inflection with a sign-change justification.

WACE 20245 marksCalculator-assumed. A function is given by

y=\dfrac{4x}{x^{2}+1}

. (a) State the horizontal asymptote and justify it from the end behaviour. (b) Show that the stationary points occur at

x=\pm 1

and classify them.

Show worked answer →

A rational-function sketch question.

(a) As $x\to\pm\infty$ , the denominator $x^{2}+1$ grows faster than the numerator $4x$ , so $y\to 0$ . The horizontal asymptote is $y=0$ .

(b) By the quotient rule, $y'=\dfrac{4(x^{2}+1)-4x(2x)}{(x^{2}+1)^{2}}=\dfrac{4-4x^{2}}{(x^{2}+1)^{2}}$ . Setting the numerator to zero gives $4-4x^{2}=0$ , so $x=\pm 1$ . At $x=1$ , $y=2$ (a maximum, since $y'$ changes $+$ to $-$ ); at $x=-1$ , $y=-2$ (a minimum, since $y'$ changes $-$ to $+$ ).

Markers reward the asymptote justification, the quotient-rule derivative, both stationary points and a sign-table classification.