VCE Mathematical Methods (Year 12) · Functions, Relations and Graphs · 19m read

Polynomial Functions

Ever wondered why some graphs look like smooth hills and valleys while others spike dramatically? Polynomial functions—especially cubics and quartics—let you predict exactly where a graph crosses the x-axis, turns around, and heads off to infinity, all from a simple factored equation. You'll learn to sketch these curves confidently and find the turning points (called stationary points) that examiners love to ask about, making these graphing skills essential for acing your Methods assessments and understanding real-world patterns in physics and engineering.

Learning objective: Sketch cubic and quartic polynomials using factored form, identify stationary points from the graph, and solve polynomial equations.

By the end of this lesson you can:

I can identify the degree, leading coefficient and long-run behaviour of a polynomial.
I can factor a polynomial using the factor theorem and sketch from factored form.
I can identify x-intercepts and their multiplicities from a factored form.

Polynomials are rollercoasters. Degree sets the number of hills; roots are where the track meets the ground.

Reading a polynomial from its factored form

A polynomial P(x) = a(x − r₁)(x − r₂)...(x − rₙ) has degree n and leading coefficient a. The roots r_i are the x-intercepts. The multiplicity of a root determines how the graph behaves at that intercept: multiplicity 1 the graph crosses cleanly, multiplicity 2 the graph touches and turns (looks like a parabola vertex), multiplicity 3 the graph flattens then crosses (a stationary point of inflection).

End behaviour depends on the degree and sign of a. For odd degree: opposite ends (one up, one down). For even degree: both ends go the same direction. If a > 0 the right end goes to +∞; if a < 0 the right end goes to −∞.

Polynomial sketching rules

Factor theorem: If P(r) = 0, then (x − r) is a factor of P(x).
Multiplicity 1: Graph crosses the x-axis cleanly.
Multiplicity 2: Graph touches and turns (local min or max).
Multiplicity 3: Stationary point of inflection — flattens and crosses.
End behaviour: Odd degree: ends opposite; even degree: ends same. Sign of a sets +∞ or −∞.

Quick reference

| Feature | Rule |
|---|---|
| Factor theorem | P(a) = 0 ⇒ (x − a) factor |
| Repeated root (even) | Touches axis |
| Repeated root (odd, ≥3) | Inflection on axis |
| End behaviour | Sign of leading term |
| Stationary points | P′(x) = 0 |

Worked example (Medium)

Sketch y = (x + 2)(x − 1)²(x − 3). State the x-intercepts and their behaviour.

Degree and leading coefficient. Expand mentally: x·x²·x = x⁴ with coefficient +1. Degree 4, a = +1
End behaviour. Even degree, a positive. Both ends go to +∞
Intercepts. List each root and its multiplicity. x = −2 (cross), x = 1 (touch), x = 3 (cross)
Sketch. Start high on the left, cross at −2, come down, touch at 1 and return, cross at 3, end high.

Answer: Intercepts: (−2, 0) crosses, (1, 0) touches and turns, (3, 0) crosses. Y-intercept at (0, +2·1·−3) = (0, −6).

Worked example (Hard)

Show that $x = 2$ is a root of $P(x) = x^3 - 7x + 6$ and hence factorise $P(x)$ fully.

Factor theorem check. Substitute $x = 2$. P(2) = 8 - 14 + 6 = 0 \,\checkmark
Divide $P(x)$ by $(x - 2)$. Polynomial long division (or inspection). P(x) = (x - 2)(x^2 + 2x - 3)
Factor the quadratic. Find numbers that multiply to $-3$ and add to $2$ ($+3$ and $-1$). x^2 + 2x - 3 = (x + 3)(x - 1)
Final factorisation. Combine. P(x) = (x - 2)(x - 1)(x + 3)

Answer: $P(x) = (x - 2)(x - 1)(x + 3)$, so the roots are $x = 2, 1, -3$.

Common mistakes

Ignoring the leading coefficient when determining end behaviour Students identify the degree correctly but forget that the sign of the leading coefficient flips the direction. A negative leading coefficient on a quartic sends both ends to $-\infty$, not $+\infty$. Fix: After finding the degree, always check the sign of the leading coefficient. State: "degree $n$, leading coefficient $a$" and deduce end behaviour from both.
Confusing multiplicity 2 (touch) with multiplicity 3 (inflection crossing) Both create a "flat" appearance at the intercept, but multiplicity 2 means the graph turns around (local extremum), while multiplicity 3 means the graph flattens then crosses (stationary point of inflection). Fix: Multiplicity 2 → touch and turn (parabola-like). Multiplicity 3 → flatten and cross (cubic-like). Sketch a small local picture at each intercept to confirm.
Forgetting the $y$-intercept when sketching VCAA marking schemes typically award a separate mark for the $y$-intercept. Omitting it loses that mark even if the rest of the sketch is perfect. Fix: Always substitute $x = 0$ into the factored form to find the $y$-intercept and label it on the sketch.

The Fundamental Theorem of Algebra and polynomial decomposition

The Fundamental Theorem of Algebra guarantees that every polynomial of degree $n$ with real (or complex) coefficients has exactly $n$ roots counted with multiplicity in $\mathbb{C}$. For real polynomials, complex roots always occur in conjugate pairs $a \pm bi$, so any real polynomial can be factored into a product of real linear factors and irreducible real quadratics (those with negative discriminant).

This explains why a cubic always has at least one real root — with three roots total, at most one conjugate pair can be complex, leaving at least one real. A quartic, however, can have zero, two, or four real roots.

In VCE Methods you factor using the Factor Theorem: test integer candidates from the divisors of the constant term (Rational Root Theorem). Once one root $r$ is found, polynomial long division or synthetic division reduces the degree by one. For a cubic, this yields a quadratic that can be solved with the quadratic formula, completing the factorisation. Understanding this cascade — test, divide, repeat — is the systematic approach examiners expect.

Practice questions

The polynomial P(x) = x³ − 6x² + 11x − 6 has roots 1, 2 and k. Find k.
1. 3
2. −3
3. 6
4. 1
(2 marks)
At x = 2 the graph of y = (x − 2)² (x + 1) has which behaviour?
1. Touches and turns (local min)
2. Crosses cleanly
3. Stationary point of inflection
4. Asymptote
(3 marks)
Find the y-intercept of y = −2(x − 1)(x + 3)(x − 2).
(3 marks)
Sketch and state the y-intercept of $y = -(x + 1)^2 (x - 2)$.
(3 marks)

Summary

Inverse $f^{-1}$: swap $x$ and $y$, solve for $y$. Domain of $f^{-1}$ = range of $f$.
Graphs of $f$ and $f^{-1}$ are reflections across $y = x$.
Composite: $(f \circ g)(x) = f(g(x))$. Apply $g$ first, then $f$.
Domain of $f \circ g$: $x$ values in domain of $g$ where $g(x)$ is in domain of $f$.