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Polynomial, power and modulus functions: VCE Math Methods Unit 3

Syllabus dot point

“Graphs of polynomial functions and key features including stationary points and points of inflection, intercepts, asymptotes, end behaviour, and the graphs of power functions f(x)=xnf(x) = x^n for n∈Qn \in Q and the modulus function f(x)=∣x∣f(x) = |x|”

9 min read

Quick answer

Polynomial, power and modulus functions form the algebraic graph families of VCE Math Methods Unit 3: cubics and quartics from factor-theorem analysis, power functions xnx^n for rational nn with their characteristic shapes (square root, cube root, hyperbola), and the V-shaped modulus graph.

Jump to a section
  1. What this dot point is asking
  2. Polynomial functions
  3. Power functions
  4. The modulus function
  5. Examples in context
  6. Try this
  7. Exam-style questions

What this dot point is asking

VCAA wants the standard graph shapes for polynomial functions up to quartic, the rational-power family f(x)=xnf(x) = x^n for n∈Qn \in Q (including the square root and cube root), and the modulus function. You need to identify key features (intercepts, turning points, end behaviour, asymptotes if any) and sketch transformed versions cleanly.

Polynomial functions

A polynomial of degree nn has the form f(x)=anxn+an−1xn−1+⋯+a1x+a0f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 with an≠0a_n \neq 0. The end behaviour is dictated by the leading term anxna_n x^n.

Quadratics

f(x)=ax2+bx+cf(x) = a x^2 + b x + c with one turning point at x=−b2ax = -\frac{b}{2a}. The discriminant Δ=b2−4ac\Delta = b^2 - 4ac tells you the number of real roots: two if Δ>0\Delta > 0, one (repeated) if Δ=0\Delta = 0, none if Δ<0\Delta < 0. Standard forms: general ax2+bx+ca x^2 + b x + c, factored a(x−p)(x−q)a(x - p)(x - q), turning-point a(x−h)2+ka(x - h)^2 + k.

Cubics

f(x)=ax3+bx2+cx+df(x) = a x^3 + b x^2 + c x + d has up to two stationary points and at least one real root. Standard factored forms:

  • Three distinct real roots: f(x)=a(x−p)(x−q)(x−r)f(x) = a(x - p)(x - q)(x - r).
  • A repeated root: f(x)=a(x−p)2(x−q)f(x) = a(x - p)^2 (x - q). The graph touches the x-axis at x=px = p (without crossing) and crosses at x=qx = q.
  • One real root: f(x)=a(x−p)(x2+bx+c)f(x) = a(x - p)(x^2 + b x + c) with the quadratic factor irreducible.

For a positive leading coefficient, the cubic rises from bottom left to top right. For a negative leading coefficient, it falls from top left to bottom right.

Quartics

f(x)=ax4+…f(x) = a x^4 + \dots has up to three stationary points and 0, 2 or 4 real roots. Common factored form f(x)=a(x−p)2(x−q)2f(x) = a(x - p)^2 (x - q)^2 has a positive-leading "W" shape if a>0a > 0, touching the x-axis at x=px = p and x=qx = q.

Key features to identify

For any polynomial sketch, mark all of:

  • x-intercepts (solve f(x)=0f(x) = 0 via the factor theorem)
  • y-intercept (f(0)f(0))
  • Stationary points (f′(x)=0f'(x) = 0, classify with f′′f'' or sign analysis)
  • End behaviour from the leading term

Power functions

A power function has the form f(x)=xnf(x) = x^n for some rational nn. The shape depends on the sign and form of nn.

Positive integer powers

f(x)=xnf(x) = x^n for n∈{1,2,3,… }n \in \{1, 2, 3, \dots\}. Even powers (x2x^2, x4x^4) are symmetric in the y-axis with a minimum at the origin. Odd powers (x3x^3, x5x^5) are symmetric about the origin with a stationary point of inflection at the origin.

Negative integer powers

f(x)=x−n=1xnf(x) = x^{-n} = \frac{1}{x^n} for n∈{1,2,3,… }n \in \{1, 2, 3, \dots\}. These have a vertical asymptote at x=0x = 0 and a horizontal asymptote at y=0y = 0. Maximal domain is R∖{0}\mathbb{R} \setminus \{0\}.

For n=1n = 1: f(x)=1xf(x) = \frac{1}{x}, hyperbola in quadrants 1 and 3.

For n=2n = 2: f(x)=1x2f(x) = \frac{1}{x^2}, both branches in the upper half-plane.

Fractional powers (roots)

f(x)=x1/nf(x) = x^{1/n} for n∈{2,3,4,… }n \in \{2, 3, 4, \dots\}.

Square root f(x)=x=x1/2f(x) = \sqrt{x} = x^{1/2}. Maximal domain [0,∞)[0, \infty), range [0,∞)[0, \infty). The graph starts at the origin and curves up and right.

Cube root f(x)=x3=x1/3f(x) = \sqrt[3]{x} = x^{1/3}. Maximal domain R\mathbb{R}, range R\mathbb{R}. Passes through the origin with a vertical tangent. Odd-symmetric (rotation symmetric about the origin).

General fractional powers f(x)=xp/qf(x) = x^{p/q} with gcd⁡(p,q)=1\gcd(p, q) = 1 are defined on [0,∞)[0, \infty) when qq is even and on R\mathbb{R} when qq is odd.

The modulus function

f(x)=∣x∣f(x) = |x| is defined piecewise:

∣x∣={xif x≥0−xif x<0|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

The graph is a V-shape with vertex at the origin, range [0,∞)[0, \infty).

Standard transformations follow the usual form y=a∣b(x−h)∣+ky = a|b(x - h)| + k. For example, y=∣x−2∣−1y = |x - 2| - 1 is the V-shape translated 2 right and 1 down, with vertex at (2,−1)(2, -1).

A useful identity: ∣x∣2=x2|x|^2 = x^2, and x2=∣x∣\sqrt{x^2} = |x| (not xx).

Worked example: sketching a cubic with a repeated root

Step 1: Determine the end behaviour

The leading term is −2x3-2 x^3, so the cubic falls from top-left to bottom-right. Identifying the leading term first fixes the overall shape of the sketch and prevents the most common orientation error.

Step 2: Find the x-intercepts

x=−1x = -1 is a repeated root (the factor (x+1)2(x+1)^2), so the graph touches the x-axis at (−1,0)(-1, 0) without crossing; x=2x = 2 is a simple root, so the graph crosses the x-axis at (2,0)(2, 0). The multiplicity of each root determines whether the graph bounces or passes through the axis.

Step 3: Find the y-intercept

f(0)=−2(1)2(−2)=4f(0) = -2 (1)^2 (-2) = 4, giving the y-intercept (0,4)(0, 4). Substituting x=0x = 0 directly into the factored form is quicker than expanding; this point anchors the curve between the two x-intercepts.

Step 4: Locate the stationary points

f′(x)=−2⋅[2(x+1)(x−2)+(x+1)2]=−2(x+1)[2(x−2)+(x+1)]=−2(x+1)(3x−3)=−6(x+1)(x−1)f'(x) = -2 \cdot [2(x+1)(x-2) + (x+1)^2] = -2(x+1)[2(x-2) + (x+1)] = -2(x+1)(3x - 3) = -6(x+1)(x-1). So f′(x)=0f'(x) = 0 at x=−1x = -1 and x=1x = 1. The point (−1,0)(-1, 0) lies on the x-axis (confirming the repeated root is a stationary point of inflection), and f(1)=−2(2)2(−1)=8f(1) = -2 (2)^2 (-1) = 8, so (1,8)(1, 8) is a local maximum. Using the product rule on the factored form avoids expanding a cubic before differentiating.

Common traps
Confusing the shape at a repeated root
A double root means the graph touches the x-axis without crossing. A triple root means the graph crosses with a stationary point of inflection.
Forgetting domain restrictions on rational power functions
x\sqrt{x} is defined only for x≥0x \geq 0; x3\sqrt[3]{x} is defined for all real xx.
Treating ∣x∣|x| like xx
∣x∣|x| is never negative. x2=∣x∣\sqrt{x^2} = |x|, not xx, when xx could be negative.
Mixing up end behaviour
For an odd-degree polynomial with negative leading coefficient, the graph falls from top left to bottom right (opposite to the positive case).

Examples in context

Example 1. Reading a factored cubic model. A profit model is P(x)=−(x−1)2(x−5)P(x) = -(x - 1)^2(x - 5) (thousand dollars) for production level xx. The roots are x=1x = 1 (double, where the curve touches the axis) and x=5x = 5 (single, where it crosses). The yy-intercept is P(0)=−(1)(−5)=5P(0) = -(1)(-5) = 5. The negative leading term means the cubic falls from top left to bottom right, so profit is positive for small xx and turns negative beyond x=5x = 5.

Example 2. Domain of a square-root model. The time for an object to fall a distance dd is t=d5t = \sqrt{\frac{d}{5}} seconds (a power function with n=12n = \tfrac12). Its domain is d≥0d \ge 0 and range t≥0t \ge 0. Doubling the drop height does not double the time: from d=20d = 20 (t=2t = 2 s) to d=80d = 80 (t=4t = 4 s) requires four times the height, reflecting the square-root shape.

Try this

Q1. State the xx-intercepts and their type for y=(x+3)(x−2)2y = (x + 3)(x - 2)^2. [2 marks]

  • Cue. x=−3x = -3 (single, crosses); x=2x = 2 (double, touches and turns).

Q2. State the maximal domain and range of f(x)=x+4f(x) = \sqrt{x + 4}. [2 marks]

  • Cue. Domain [−4,∞)[-4, \infty); range [0,∞)[0, \infty).

Q3. The graph of y=∣x∣y = |x| is transformed to y=∣x−1∣+2y = |x - 1| + 2. (a) State the vertex. (b) Find yy when x=4x = 4. [2+1 marks]

  • Cue. (a) Vertex (1,2)(1, 2). (b) ∣4−1∣+2=5|4 - 1| + 2 = 5.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2023 VCAA-style (Exam 1)3 marks
Sketch the graph of y=(x−1)2(x+2)y = (x - 1)^2 (x + 2), showing all axis intercepts and stationary points.
Show worked answer →

Factored form already gives x-intercepts. y=0y = 0 when x=1x = 1 (double root, touches the axis) or x=−2x = -2 (single root, crosses).

y-intercept: y(0)=(−1)2(2)=2y(0) = (-1)^2 (2) = 2.

Differentiate to locate stationary points. Expand or use the product rule: y=(x−1)2(x+2)y = (x-1)^2(x+2), so y′=2(x−1)(x+2)+(x−1)2=(x−1)[2(x+2)+(x−1)]=(x−1)(3x+3)=3(x−1)(x+1)y' = 2(x-1)(x+2) + (x-1)^2 = (x-1)[2(x+2) + (x-1)] = (x-1)(3x+3) = 3(x-1)(x+1).

y′=0y' = 0 at x=1x = 1 and x=−1x = -1. At x=1x = 1, y=0y = 0 (the repeated root, a local minimum). At x=−1x = -1, y=(−2)2(1)=4y = (-2)^2 (1) = 4 (a local maximum).

Sketch: cubic shape rising from bottom left, crossing at x=−2x = -2, peaking at (−1,4)(-1, 4), touching the x-axis at (1,0)(1, 0), then rising to the top right.

Markers reward correct intercepts, correct stationary points with classification, and the right end behaviour for a positive-leading cubic.

2024 VCAA-style (Exam 2)2 marks
State the maximal domain and range of f(x)=x−3f(x) = \sqrt{x - 3}.
Show worked answer →

The square root is defined only when the radicand is non-negative.

Maximal domain: x−3≥0x - 3 \geq 0, so x∈[3,∞)x \in [3, \infty).

Range: x−3≥0\sqrt{x - 3} \geq 0, so the range is [0,∞)[0, \infty).

Markers want interval notation, not inequalities, and both endpoints stated.

Practise this

Sources & how we know this

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