IB Maths AA HL Topic 2 — Functions Paper 1 & 2 HL only ~9 min read

Graphs & Roots of Polynomial Functions

A polynomial graph is fully determined by four things: the y-intercept, the roots, the multiplicity of each root, and the end behaviour. Once you have these, sketching is mechanical. The Fundamental Theorem of Algebra guarantees a real polynomial of degree n has exactly n zeros (counting complex pairs and repeats), which is what lets you back out an equation from a graph or finish factorising once one zero is known.

📘 What you need to know

Y-intercept: substitute x = 0 to get y = a₀ (the constant term).
Roots: solve P(x) = 0 — usually by factorising.
Multiplicity: if (x − k)^m is a factor, then x = k is a root of multiplicity m. The graph behaves differently at each value of m.
End behaviour is set by the degree (even/odd) and the sign of the leading coefficient.
Fundamental Theorem of Algebra: a real polynomial of degree n has exactly n zeros (counting multiplicity and complex pairs).
Conjugate root pairs: if a + bi is a zero of a real polynomial, so is a − bi.
Every odd-degree polynomial has at least one real zero — its graph must cross the x-axis somewhere.
At most n − 1 stationary points for degree n; at least one turning point between any two distinct roots.

Multiplicity — what each one looks like

The multiplicity of a root tells you how the curve behaves at the x-axis at that point. Four cases cover everything:

Mult 1

crosses

straight pass-through

Mult 2

touches

turning point on the axis

Mult odd ≥ 3

crosses

stationary point of inflection

Mult even ≥ 4

touches

turning point on the axis (flatter)

Three multiplicity types at a root

Quick mental rule: odd multiplicity → crosses (the curve goes through); even multiplicity → touches (the curve bounces off). Higher multiplicity just makes the curve flatter at the root.

End behaviour from the leading term

For very large |x|, only the leading term a_nxⁿ matters — every other term is negligible by comparison. So the ends are determined by two things: parity of n, and sign of a_n.

a_n > 0, n even

↖ ↗

both ends to +∞

a_n < 0, n even

↙ ↘

both ends to −∞

a_n > 0, n odd

↙ ↗

−∞ on left, +∞ on right

a_n < 0, n odd

↖ ↘

+∞ on left, −∞ on right

Mnemonic: even degrees behave like a parabola — both ends go the same way. Odd degrees behave like a cubic — the ends go opposite ways. The sign of the leading coefficient flips the picture vertically.

The sketching recipe

🧭 Recipe — sketching a polynomial

Find the y-intercept: substitute x = 0.
Find the roots and their multiplicities: factorise (or use given factored form).
Plot each root as either a “crosses” or “touches” point based on multiplicity.
Determine the end behaviour from the sign and parity of the leading term.
Connect with a smooth curve, ensuring at least one turning point between consecutive distinct roots.
Label all intercepts (with coordinates) and indicate the shape clearly.

Solving polynomial equations

The Fundamental Theorem of Algebra guarantees that a real polynomial of degree n has exactly n zeros (in ℂ, counting multiplicity). Two facts make these searchable:

Real linear factor

x = k real → (x − k) is a factor

factor theorem

Conjugate pair

a ± bi roots → (x−a)² + b² is a factor

non-real roots come in pairs

Odd degree

at least one real zero

graph must cross the x-axis

🧭 Recipe — find all zeros given one

Write the corresponding factor from the given zero (real → linear; complex → quadratic from the conjugate pair).
Divide the polynomial by that factor (long division or comparing coefficients).
Solve the resulting quotient — typically a quadratic, finished off with the formula.
List all zeros: the original given one(s) plus the new ones from the quotient.

Worked examples

WE 1

Sketch a polynomial from its factorised form

Sketch the graph of f(x) = (x + 1)(x − 2)²(x − 4), labelling all intercepts.

Step 1: y-intercept — substitute x = 0 f(0) = (1)(−2)²(−4) = (1)(4)(−4) = −16 → (0, −16) Step 2: Roots and multiplicities x = −1 (mult 1) → crosses; x = 2 (mult 2) → touches x = 4 (mult 1) → crosses Step 3: End behaviour — leading term x · x² · x = x⁴; positive coefficient, even degree both ends → +∞ x-ints (−1, 0), (2, 0) [touch], (4, 0); y-int (0, −16); both ends to +∞ curve enters from top-left, dips through (−1, 0), comes back up through (0, −16), kisses (2, 0), dips again, then rises through (4, 0) up to top-right

WE 2

Find a polynomial from its graph

A polynomial graph touches the x-axis at (−2, 0) with a stationary point of inflection, crosses the x-axis at (1, 0), and passes through (0, 16). The graph tends to −∞ as x → ±∞. Find a possible equation for the polynomial.

Step 1: Identify factors from multiplicities stationary inflection at (−2, 0) → multiplicity 3 → (x + 2)³ crosses at (1, 0) → multiplicity 1 → (x − 1) Step 2: Set up form with unknown leading coefficient y = a(x + 2)³(x − 1) Step 3: Use y-intercept to find a y(0) = a · (2)³ · (−1) = −8a = 16 → a = −2 Step 4: Verify end behaviour leading term: −2 · x³ · x = −2x⁴ (negative, even degree) → both ends → −∞ ✓ y = −2(x + 2)³(x − 1) “stationary point of inflection” is the giveaway for multiplicity 3 — without it, you’d think (x + 2) was a simple factor

WE 3

Describe the behaviour at each root

For f(x) = −(x + 5)(x − 1)²(x + 2)³, state: (a) the degree, (b) the y-intercept, (c) the behaviour at each root.

(a) Degree — sum of multiplicities 1 + 2 + 3 = 6 (a) degree 6 (b) y-intercept: f(0) f(0) = −(5)(1)²(2)³ = −(5)(1)(8) = −40 (b) (0, −40) (c) Multiplicities x = −5: mult 1 → crosses x = 1: mult 2 → touches (turning point on axis) x = −2: mult 3 → crosses with stationary inflection (c) crosses at −5; touches at 1; crosses with flat tangent at −2 leading term: −x¹·x²·x³ = −x⁶ (negative, even) → both ends to −∞

WE 4

State the number and nature of zeros

A real polynomial f(x) has degree 5. It has a real zero at x = 2 with multiplicity 2 and a real zero at x = −1 with multiplicity 1. State the number and nature of any further zeros.

Step 1: Total zeros from FTA degree 5 → 5 zeros (counting multiplicity) Step 2: Real zeros accounted for x = 2 (mult 2) + x = −1 (mult 1) = 3 zeros Step 3: Remaining zeros 5 − 3 = 2 zeros left f is real → any complex zeros come in conjugate pairs → these 2 are a conjugate pair two non-real zeros forming a complex conjugate pair two further real zeros are also possible in principle, but the question wording “any further zeros” + odd-degree set-up makes a complex pair the standard answer

WE 5

Find all zeros given one

Given that x = 3 is a zero of f(x) = x³ − 5x² + 11x − 15, find all three zeros of f.

Step 1: x = 3 is a zero → (x − 3) is a factor Step 2: Divide f by (x − 3) — compare coefficients x³ − 5x² + 11x − 15 = (x − 3)(x² + ax + b) expand: x³ + ax² + bx − 3x² − 3ax − 3b match x²: a − 3 = −5 → a = −2 match const: −3b = −15 → b = 5 quotient: x² − 2x + 5 Step 3: Solve x² − 2x + 5 = 0 x = (2 ± √(4 − 20))/2 = (2 ± √(−16))/2 = (2 ± 4i)/2 x = 1 ± 2i zeros: x = 3, x = 1 + 2i, x = 1 − 2i cubic has three zeros total — one real and one complex conjugate pair, matching the FTA

WE 6

Construct a polynomial from given zeros

A real polynomial f(x) of degree 4 has zeros at x = −2, x = 5, and x = 3 + i. The leading coefficient is 1. Find f(x) in expanded form.

Step 1: Use conjugate root theorem f real → 3 − i is also a zero four zeros: −2, 5, 3 + i, 3 − i ✓ Step 2: Combine the complex pair into a real quadratic (x − (3 + i))(x − (3 − i)) = (x − 3)² − i² = (x − 3)² + 1 = x² − 6x + 10 Step 3: Combine the real linear factors (x + 2)(x − 5) = x² − 3x − 10 Step 4: Multiply the two quadratics f(x) = (x² − 3x − 10)(x² − 6x + 10) = x⁴ − 6x³ + 10x² − 3x³ + 18x² − 30x − 10x² + 60x − 100 = x⁴ − 9x³ + 18x² + 30x − 100 f(x) = x⁴ − 9x³ + 18x² + 30x − 100 always pair complex conjugates first — they multiply to a clean real quadratic, simplifying the expansion

💡 Top tips

Factorise before sketching: a polynomial in factored form gives roots and multiplicities immediately.
Memorise the four end-behaviour cases. Two parities × two signs = four pictures — quick mental reference.
Multiplicity rule of thumb: odd → crosses; even → touches. Higher multiplicities just flatten the curve at that root.
Complex zeros come in conjugate pairs for real polynomials — never alone. If you’ve found one, you’ve actually found two.
Pair complex conjugates first when constructing or factorising — they collapse to a real quadratic and remove the imaginary work.
Use the y-intercept to nail the leading coefficient when reverse-engineering an equation from a graph.
Check the sign and parity of your final leading term against the graph’s end behaviour. If they don’t match, the leading coefficient is the wrong sign.

⚠ Common mistakes

Treating a multiplicity-2 root as multiplicity-1. The graph touches the axis instead of crossing — easy to spot when you know to look.
Missing the stationary inflection at a multiplicity-3 root. The graph crosses but with a flat tangent.
Reading end behaviour from a non-leading term. Only a_nxⁿ matters as |x| → ∞.
Forgetting to count complex zeros when applying the FTA. Degree n means n zeros total — real and complex combined.
Listing only one of a complex pair. If 2 + 5i is a zero, then 2 − 5i is too (for real polynomials).
Setting the leading coefficient to 1 when reverse-engineering from a graph. Use the y-intercept (or another given point) to determine it.
Forgetting placeholders when dividing by a quadratic factor formed from a complex pair — same multiple-missing-term issue as in the division note.

You can now read a polynomial from its graph and back. The next note, Sum & Product of Roots of Polynomials, exposes a shortcut: the sum and product of all the zeros are encoded in just two coefficients of the polynomial. Useful when you have partial information about the roots and need to find the rest without fully factorising.

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