Determining the rule for the graph of a polynomial
Theory
Determining the rule of a polynomial from its graph reverses the sketching process: each \(x\)-intercept gives a factor, a repeated factor shows where the curve touches or flattens on the axis, and one further point together with the end behaviour fixes the leading coefficient \(a\). The rule is written in factor form \(y=a(x-p)(x-q)\cdots\) and expanded only if a polynomial in \(x\) is required.
The graph of a polynomial meets the \(x\)-axis at its real zeros. Reading a rule from a graph reverses the usual process: an intercept at \(x=p\) contributes a factor \((x-p)\), so a curve crossing at \(x=p\), \(x=q\) and \(x=r\) has the form \(y=a(x-p)(x-q)(x-r)\). The constant \(a\) is the leading coefficient; it controls the vertical stretch and the direction of the tails.
The way the curve meets the axis reveals the multiplicity of each factor. A simple crossing is a single factor \((x-p)\); where the curve touches the axis and turns back it has an even multiplicity, most often a squared factor \((x-p)^2\); where it flattens and crosses — a stationary point of inflection — it has an odd multiplicity greater than one, typically a cubed factor \((x-p)^3\). Counting factors with multiplicity gives the degree.
Two pieces of information fix \(a\). The end behaviour — whether the tails rise or fall — sets the sign of \(a\) (positive if a cubic runs bottom-left to top-right or a quartic opens upward, negative if reflected). One further point on the curve, usually the \(y\)-intercept, then gives the exact value of \(a\) when substituted into the factor form.
Factor form — one factor per simple \(x\)-intercept, with a leading coefficient \(a\):
Repeated factors encode how the curve meets the axis — a touch is a squared factor, a stationary point of inflection is a cubed factor:
Find the leading coefficient by substituting one more point \((x_0,y_0)\) that lies on the curve:
Building a polynomial rule from its graph
- Read the \(x\)-intercepts. For each intercept at \(x=p\), write a factor \((x-p)\) — mind the sign, so \(x=-2\) gives \((x+2)\).
- Account for the behaviour at each intercept. A crossing is a single factor; a touch is a squared (even) factor; a flatten-and-cross (stationary point of inflection) is a cubed (odd, \(>1\)) factor. The multiplicities should sum to the degree.
- Write \(y=a(x-p)(x-q)\cdots\) with an unknown leading coefficient \(a\).
- Find \(a\). Substitute one more known point (often the \(y\)-intercept) and solve for \(a\).
- Check and finish. Confirm the sign of \(a\) matches the end behaviour, then expand only if the polynomial form in \(x\) is required.
| \(y\) | \(=\) | \(a(x+1)(x-2)(x-3)\) |
| \(12\) | \(=\) | \(a(0+1)(0-2)(0-3)\) |
| \(12\) | \(=\) | \(a(1)(-2)(-3)=6a\) |
| \(a\) | \(=\) | \(2\) |
| \(y\) | \(=\) | \(a(x+2)^{2}(x-1)\) |
| \(4\) | \(=\) | \(a(0+2)^{2}(0-1)\) |
| \(4\) | \(=\) | \(a(4)(-1)=-4a\) |
| \(a\) | \(=\) | \(-1\) |
| \(y\) | \(=\) | \(a(x+2)(x+1)(x-1)(x-2)\) |
| \(-4\) | \(=\) | \(a(2)(1)(-1)(-2)\) |
| \(-4\) | \(=\) | \(4a\) |
| \(a\) | \(=\) | \(-1\) |
| \(y\) | \(=\) | \(a(x-1)^{3}\) |
| \(2\) | \(=\) | \(a(0-1)^{3}=a(-1)\) |
| \(a\) | \(=\) | \(-2\) |
| \(y\) | \(=\) | \(-2(x^{3}-3x^{2}+3x-1)\) |
| \(=\) | \(-2x^{3}+6x^{2}-6x+2\) |
Common pitfalls
Frequently asked questions
How do you find the rule of a polynomial from its graph?
Read the \(x\)-intercepts and write a factor for each one: an intercept at \(x=p\) gives a factor \((x-p)\). Multiply the factors, include an unknown leading coefficient \(a\), then substitute one more known point to find \(a\). The rule is \(y=a(x-p)(x-q)\cdots\).
What does it mean when a graph touches the x-axis instead of crossing it?
A graph that touches the axis and turns back without crossing has a repeated factor there. A touch corresponds to an even power, most commonly a squared factor \((x-p)^2\); the curve meets the axis but does not change sign at that intercept.
How do you find the leading coefficient a?
Write the rule in factor form with an unknown \(a\), then substitute the coordinates of one more point on the curve, often the \(y\)-intercept. Solve the resulting equation for \(a\); for example \(12=6a\) gives \(a=2\).
How do you tell the sign of the leading coefficient from a graph?
Look at the end behaviour. If a cubic runs from bottom-left to top-right, or a quartic opens upward, \(a\) is positive; if the graph is reflected so it falls to the right or opens downward, \(a\) is negative. The sign of \(a\) must match the direction of the tails.
What is a stationary point of inflection on a polynomial graph?
It is where the curve flattens against a horizontal tangent and then continues in the same overall direction, crossing the axis if it sits on it. It corresponds to a cubed factor \((x-p)^3\), an odd power greater than one.
Do you always have to expand the factor form?
No. The factor form \(y=a(x-p)(x-q)\cdots\) is a complete and valid rule. Expand only when the question asks for the polynomial in the form \(ax^3+bx^2+cx+d\), or when you need the individual coefficients.