Sketch the graph of a polynomial in factored form using the behaviour of the leading term for large x and the multiplicity of each zero, deciding where the curve crosses, touches or has a horizontal inflection, and locate a zero between integers from a table of values

What this dot point is asking

This is your first time sketching a polynomial from its factors, and NESA wants a sketch built from reasoning, not from plotting dozens of points. Three pieces of information fix the shape of the curve. The leading term tells you where the two ends go as $x \to \pm\infty$. The zeroes tell you where the curve meets the x-axis. And the multiplicity of each zero, the power on its factor, tells you whether the curve crosses straight through, just touches and turns, or flattens into a horizontal inflection as it crosses. Put those together and a smooth curve almost draws itself. A separate skill rounds out the dot point: when a polynomial will not factor, you locate a real zero between two integers by spotting a sign change in a table of values.

This is the Year 11 first-contact version. We reason entirely from the factors and the multiplicity rule; the calculus that proves why the flattening happens, and the machinery of stationary points away from the axis, belong to the Year 12 page on graphing polynomials and multiplicity. Here the goal is to read the shape off the factored form with confidence.

The answer

End behaviour from the leading term

For very large positive or negative $x$, the leading term $a_n x^n$ dwarfs every other term, so the polynomial behaves exactly like that single term. Two things decide the picture: whether the degree $n$ is even or odd, and the sign of the leading coefficient $a_n$.

• Odd degree, positive leading coefficient: down to $-\infty$ on the left, up to $+\infty$ on the right (the two ends go opposite ways, like $y = x^3$).
• Odd degree, negative leading coefficient: up on the left, down on the right.
• Even degree, positive leading coefficient: up to $+\infty$ at both ends (like $y = x^2$).
• Even degree, negative leading coefficient: down to $-\infty$ at both ends.

The shortcut: odd degrees send the arms to diagonally opposite corners; even degrees send both arms the same way. Because an odd-degree curve runs from $-\infty$ to $+\infty$ and is continuous, it must cross the x-axis somewhere, so every odd-degree polynomial has at least one real zero. An even-degree curve need not have any real zero at all (for example $y = x^2 + 1$).

Reading the zeroes off the factored form

If a polynomial is fully factored into linear factors (and possibly an irreducible quadratic), its zeroes are read off instantly: each factor $(x - \alpha)$ gives a zero at $x = \alpha$, and the curve meets the x-axis there. An irreducible quadratic factor, one with negative discriminant such as $x^2 + x + 1$, contributes no x-intercept at all, but it still affects the sign of the polynomial between the real zeroes (it is always positive when its leading coefficient is positive).

The y-intercept is just $P(0)$, found by substituting $x = 0$, which for a factored polynomial means multiplying the constant in each bracket. It is a fast, independent check on your sketch.

Multiplicity: cross, touch, or horizontal inflection

The power on a factor is the heart of this dot point. If

then $x = \alpha$ is a zero of multiplicity $m$. A zero of multiplicity $1$ is a simple zero; multiplicity $2$ or more is a multiple or repeated zero (double, triple, and so on). The multiplicity decides the local shape:

The sign-change rule is the reasoning behind all three cases. As $x$ passes through $\alpha$, the factor $(x - \alpha)^m$ changes sign only when $m$ is odd, because an even power is never negative. So an odd power flips the sign of $P(x)$ (the curve moves from one side of the axis to the other, a crossing), while an even power keeps the sign the same (the curve returns to the side it came from, a touch). The extra flattening at $m \ge 2$ happens because near $\alpha$ a high power like $(x - \alpha)^3$ is extremely small, hugging the axis, exactly as $y = x^3$ does at the origin.

A stage-by-stage sketch from the factors

The reliable method is to build the curve up in four passes, never trying to draw it all at once. Take

a cubic with three distinct simple zeroes, as the model.

Stage 1, end behaviour. Expanded, the leading term is $x \cdot x \cdot x = x^3$: degree $3$, positive leading coefficient. The left arm falls to $-\infty$ and the right arm rises to $+\infty$. Sketch those two arrows first (the figure above). They anchor the whole picture.

Stage 2, intercepts. The factors give zeroes at $x = -2$, $x = 1$ and $x = 3$, each a simple zero (multiplicity $1$). The y-intercept is

so the curve passes through $(0, 6)$. Mark these four points.

Stage 3, behaviour at each zero. Every zero here is simple, so the curve crosses straight through each one, changing sign as it goes. Reading the sign of $P(x)$ between the zeroes (or just remembering that a simple zero flips the sign), the curve is below the axis for $x < -2$, above between $-2$ and $1$, below between $1$ and $3$, and above for $x > 3$. Drawing a short slanted crossing stroke at each intercept records this before you commit to the full curve.

Stage 4, join into one smooth curve. Now connect the left arm, the three crossings and the point $(0, 6)$ into a single smooth curve, letting it turn over between consecutive zeroes. A cubic with three simple zeroes has exactly two turning points, one local maximum between $x = -2$ and $x = 1$ and one local minimum between $x = 1$ and $x = 3$. The finished sketch is below.

A double root: the curve touches the axis

When a factor is squared the zero is a double root, and the curve touches the axis there and turns back, making a turning point on the axis rather than a crossing. Consider the quartic

Its leading term is $x^2 \cdot x \cdot x = x^4$: degree $4$, positive leading coefficient, so both arms go up to $+\infty$. The zeroes are $x = -2$ (simple, cross), $x = 1$ (double, touch) and $x = 3$ (simple, cross), and the y-intercept is $P(0) = (-1)^2(2)(-3) = -6$. At the double root $x = 1$ the curve comes down, just kisses the axis, and lifts back off on the same side, never changing sign there.

A triple root: the curve flattens into a horizontal inflection

When a factor is cubed the zero has multiplicity $3$, which is odd, so the curve crosses, but the high power makes it hug the axis first: it flattens onto the axis, becoming momentarily horizontal, and then crosses through, exactly like $y = x^3$ at the origin. This is a horizontal inflection. Take

a quartic with leading term $x^3 \cdot x = x^4$ (degree $4$, both arms up to $+\infty$). The zeroes are $x = -1$ (triple) and $x = 3$ (simple), and the y-intercept is $P(0) = (1)^3(-3) = -3$. At $x = -1$ the curve flattens onto the axis and crosses; at $x = 3$ it makes an ordinary simple crossing.

Locating a zero between integers from a value table

Not every polynomial factors. When it does not, you can still pin down where a real zero lies using one idea: a polynomial is continuous, so if it is negative at one value of $x$ and positive at another, it must equal zero somewhere in between. Build a table of values at successive integers and look for a sign change from one column to the next; a zero lies between those two integers. You can then halve the interval, testing the midpoint, to trap the zero more tightly. (This is the reasoning that becomes the bisection method and, with calculus, Newton's method later on.)

Take $P(x) = x^3 - 4x + 1$. Its table of integer values is

There are three sign changes, between $-3$ and $-2$, between $0$ and $1$, and between $1$ and $2$, so this cubic has a real zero in each of $(-3, -2)$, $(0, 1)$ and $(1, 2)$, accounting for all three zeroes a cubic can have. To narrow the zero in $(0, 1)$, test the midpoint: $P(0.5) = 0.5^3 - 4(0.5) + 1 = 0.125 - 2 + 1 = -0.875$, which is negative, while $P(0) = 1$ is positive, so that zero lies in the left half $(0, 0.5)$.

How exam questions ask about graphing polynomials

The wording is predictable once you map it to the three pieces of information.

• "Sketch the graph of $P(x) = \dots$ (factored), showing all intercepts." Do the four stages: end behaviour from the leading term, x-intercepts with multiplicities, behaviour (cross/touch/flatten) at each, then the y-intercept, then join. Label the intercepts; markers want the touch-versus-cross distinction made correctly.
• "Describe the behaviour of the graph at $x = a$," given a factor like $(x - a)^4$ or $(x + 2)^3$. Read the multiplicity off the power and quote the rule: even means touch and turn, odd at least $3$ means flatten and cross, $1$ means cross at an angle.
• "Between which two integers does a zero of $P(x)$ lie?" Build a value table and find a sign change; the zero sits between the two integers where $P$ changes sign. Halve the interval if asked to refine.
• "How many times does the graph cross the x-axis?" Count the zeroes of odd multiplicity (those are crossings); even-multiplicity zeroes touch but do not cross, and irreducible quadratic factors contribute no x-intercept.
• "What is the smallest degree of a polynomial with this graph?" Add the multiplicities you can see (each touch is at least $2$, each flatten-and-cross is at least $3$, each plain crossing is at least $1$), and match the end behaviour to an even or odd degree.