Families Of Cubic Polynomial Functions
Explore families of cubic functions for Queensland Year 11 Mathematical Methods (QCAA). A family shares the same basic cubic shape, with parameters that shift, stretch and reflect the basic cubic.
You will learn to describe these transformations, locate the point of inflection, and find a cubic's rule from a centre of symmetry and a point, or from its x-intercepts and a point — building fluency for QCAA modelling and graphing.
Every question with a fully worked solution.
- Families Of Cubic Polynomial Functions - Video - Families of cubic functions and determining the rule Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), the family of cubics \(y=a(x-h)^3+k\) is built from \(y=x^3\) by a translation to the point of inflection \((h,k)\), a dilation by \(a\) and, when \(a<0\), a reflection. This page shows how the parameters \(a,h,k\) move the graph and how to find the rule of a cubic from a centre of symmetry and a point, from its intercepts and a point, or from a graph.
The cubic family \(y=a(x-h)^3+k\) has its point of inflection (the centre of symmetry) at \((h,k)\): the graph of \(y=x^3\) shifted right \(h\) and up \(k\). The number \(h\) is a horizontal shift and \(k\) a vertical shift.
The dilation factor \(a\) stretches the graph vertically. A larger \(|a|\) makes it steeper; a negative \(a\) also reflects it in the horizontal through \((h,k)\), so it falls instead of rising.
A cubic can also be given in factor form \(y=a(x-x_1)(x-x_2)(x-x_3)\). To pin down a particular member of a family you need the shape information (inflection or intercepts) plus one extra point to solve for \(a\).
The transformed cubic family, with inflection at \((h,k)\):
The factor form, with zeros \(x_1,x_2,x_3\):
How to find the rule of a cubic
- Choose the form: a centre of symmetry gives \(y=a(x-h)^3+k\); a set of intercepts gives \(y=a(x-x_1)(x-x_2)(x-x_3)\).
- Insert the features: write in the values of \(h,k\) (or the zeros), marking a squared factor for any touch.
- Solve for \(a\): substitute the extra point and solve for \(a\).
- State the rule: write the final equation with \(a\) filled in.
Compare with \(y=a(x-h)^3+k\):
| \(a\) | \(=\) | \(2\) |
| \(h\) | \(=\) | \(1\) |
| \(k\) | \(=\) | \(3\) |
Dilation: a vertical stretch by factor \(2\) (steeper, \(a>0\) so no reflection).
Translation: right \(1\) unit (\(h=1\)) and up \(3\) units (\(k=3\)).
Point of inflection — the centre \((h,k)\):
| \((h,k)\) | \(=\) | \((1,3)\) |
Check with the \(y\)-intercept, \(x=0\):
| \(y\) | \(=\) | \(2(0-1)^3+3\) |
| \(=\) | \(2(-1)+3\) | |
| \(=\) | \(1\) |
Stretch by \(2\), right \(1\), up \(3\); inflection \((1,3)\), through \((0,1)\).
Form — use the inflection \((h,k)=(2,1)\):
| \(y\) | \(=\) | \(a(x-2)^3+1\) |
Solve for \(a\) — substitute \((4,17)\):
| \(17\) | \(=\) | \(a(4-2)^3+1\) |
| \(17\) | \(=\) | \(a(2)^3+1\) |
| \(17\) | \(=\) | \(8a+1\) |
| \(16\) | \(=\) | \(8a\) |
| \(a\) | \(=\) | \(2\) |
Write the rule with \(a=2\):
| \(y\) | \(=\) | \(2(x-2)^3+1\) |
Rule: \(y=2(x-2)^3+1\).
Form — use the three zeros:
| \(y\) | \(=\) | \(a(x+1)(x-2)(x-3)\) |
Solve for \(a\) — substitute \((0,-12)\):
| \(-12\) | \(=\) | \(a(0+1)(0-2)(0-3)\) |
| \(-12\) | \(=\) | \(a(1)(-2)(-3)\) |
| \(-12\) | \(=\) | \(6a\) |
| \(a\) | \(=\) | \(-2\) |
Write the rule with \(a=-2\):
| \(y\) | \(=\) | \(-2(x+1)(x-2)(x-3)\) |
Rule: \(y=-2(x+1)(x-2)(x-3)\).
Form — a touch is a squared factor, a cross is a single factor:
| \(y\) | \(=\) | \(a(x+2)^2(x-3)\) |
Solve for \(a\) — substitute \((0,6)\):
| \(6\) | \(=\) | \(a(0+2)^2(0-3)\) |
| \(6\) | \(=\) | \(a(4)(-3)\) |
| \(6\) | \(=\) | \(-12a\) |
| \(a\) | \(=\) | \(-\dfrac{1}{2}\) |
Write the rule with \(a=-\dfrac{1}{2}\):
| \(y\) | \(=\) | \(-\dfrac{1}{2}(x+2)^2(x-3)\) |
Rule: \(y=-\dfrac{1}{2}(x+2)^2(x-3)\).
Common pitfalls
Frequently asked questions
What do a, h and k do in y = a(x-h)^3 + k?
\(h\) and \(k\) translate the inflection to \((h,k)\); \(a\) dilates the curve vertically, and if \(a<0\) it also reflects it.
How do you find the equation of a cubic from its graph?
Read the shape feature (inflection or the intercepts), write the matching form, then substitute one extra point and solve for \(a\).
Why do you need an extra point to find the rule?
The intercepts or inflection fix the shape but not the vertical stretch. One more point gives a linear equation you solve for \(a\).
What is the point of inflection of y = a(x-h)^3 + k?
It is the centre of symmetry \((h,k)\) — the point about which the cubic has half-turn symmetry.
How do you show a graph touches rather than crosses the axis?
Use a squared factor, e.g. \((x-h)^2\), which makes the curve touch the \(x\)-axis at \(x=h\) and turn back.
Do you need calculus to work with cubic families?
No. In Year 11 you use translations, dilations and reflections and fit the rule from points; derivatives are not required.