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Year 11 Methods (Unit 1 & 2) Polynomials

Cubic Functions Of The Formf(X) =A(X−H)3+K

20 practice questions 1 video lesson Theory + worked examples

Understand transformed cubics of the form a times x minus h cubed plus k for Queensland Year 11 Mathematical Methods (QCAA). This rule shifts, stretches and reflects the basic cubic curve.

You will learn to read the stationary point of inflection, describe the effect of the parameters a, h and k, find the single x-intercept, and note the behaviour for large positive and negative x — core QCAA graphing.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), the cubic \(y=a(x-h)^3+k\) is the graph of \(y=x^3\) after a dilation, a possible reflection, and translations. Its stationary point of inflection is \((h,k)\). This page covers reading \((h,k)\), the effect of \(a\), the transformations from \(y=x^3\), the intercepts, and finding the rule from features — all without calculus.

The graph of \(y=x^3\) is the basic cubic: it increases everywhere and has a stationary point of inflection at the origin, where it briefly flattens before continuing. It has no turning points.

The transformed cubic \(y=a(x-h)^3+k\) keeps this shape. The inflection point moves to \((h,k)\): \(h\) shifts it horizontally and \(k\) vertically. The number \(a\) is a dilation that changes steepness, and if \(a\lt 0\) it reflects the curve in the \(x\)-axis so it decreases everywhere.

Because the cube is a one-to-one function, \(y=a(x-h)^3+k\) has exactly one \(x\)-intercept, found by taking a cube root (which always gives one real value).

Read \((h,k)\) straight off the rule. In \(a(x-h)^3+k\) the inflection is at \((h,k)\) — note the sign: \((x-2)^3\) gives \(h=2\), while \((x+1)^3\) gives \(h=-1\).
The basic cubic y=x cubedThe graph of y=x cubed with its stationary point of inflection at the origin. x y (0, 0)
\(y=x^3\): increases everywhere, with a stationary point of inflection at the origin.
Translated inflection pointThe graph of y=(x minus one) cubed plus two, with its inflection point shifted to one comma two. x y (1, 2)
\(y=(x-1)^3+2\): the same shape with its inflection point moved to \((1,2)\).

The transformed cubic and its inflection point:

\[y=a(x-h)^3+k\quad\text{with inflection at }(h,\,k)\]
y=a(x-h)3+k

The \(y\)-intercept (set \(x=0\)):

\[y=a(0-h)^3+k=-ah^3+k\]
y=-ah3+k

The single \(x\)-intercept (set \(y=0\), then cube-root):

\[x=h+\sqrt[3]{-\dfrac{k}{a}}\]
x=h+-ka3
End behaviour follows \(a\). If \(a\gt 0\) the curve rises to the right (falls left); if \(a\lt 0\) it falls to the right (rises left). Either way there is one \(x\)-intercept and no turning point.

How to work with \(y=a(x-h)^3+k\)

  1. Inflection: read \((h,k)\) directly from the rule, watching the sign of \(h\).
  2. Shape: use \(a\) for steepness and direction — \(a\gt 0\) increasing, \(a\lt 0\) decreasing (reflected) — and state the end behaviour.
  3. Intercepts / rule: put \(x=0\) for the \(y\)-intercept and \(y=0\) then cube-root for the \(x\)-intercept; or, given the inflection and one point, substitute to solve for \(a\).
Example 1 — Inflection and intercepts
For \(y=(x-2)^3+1\), state the stationary point of inflection and find both intercepts.
Solution

Comparing with \(y=a(x-h)^3+k\) gives \(a=1,\ h=2,\ k=1\), so the inflection is \((2,1)\).

\(y\)-intercept — substitute \(x=0\):

\(y\)\(=\)\((0-2)^3+1\)
\(=\)\(-8+1\)
\(=\)\(-7\)

\(x\)-intercept — set \(y=0\) and cube-root:

\((x-2)^3+1\)\(=\)\(0\)
\((x-2)^3\)\(=\)\(-1\)
\(x-2\)\(=\)\(-1\)
\(x\)\(=\)\(1\)

Inflection \((2,1)\); \(y\)-intercept \((0,-7)\); \(x\)-intercept \((1,0)\).

Graph of y=(x-2)^3+1Increasing cubic with inflection at two comma one, x-intercept at one and y-intercept at minus seven. x y (2, 1)
(2,1)
Example 2 — Transformations and end behaviour
Describe the transformations mapping \(y=x^3\) onto \(y=-2(x-3)^3+4\), state the inflection point and the end behaviour.
Solution

Reading the constants: \(a=-2,\ h=3,\ k=4\).

Interpret each constant:

\(a=-2\)\(=\)\(\text{dilate by factor }2,\ \text{reflect in the }x\text{-axis}\)
\(h=3\)\(=\)\(\text{translate right }3\)
\(k=4\)\(=\)\(\text{translate up }4\)

The inflection point is \((3,4)\). Because \(a\lt 0\) the curve decreases, so as \(x\to\infty,\ y\to-\infty\) and as \(x\to-\infty,\ y\to\infty\).

Dilation factor \(2\), reflection in the \(x\)-axis, then right \(3\) and up \(4\); inflection \((3,4)\), decreasing.

Graph of y=-2(x-3)^3+4Decreasing cubic reflected by the negative leading factor, with inflection at three comma four. x y (3, 4)
(3,4)
Example 3 — Exact cube-root intercept
Find the intercepts of \(y=(x+1)^3-4\), giving the \(x\)-intercept exactly.
Solution

Here \(h=-1,\ k=-4\), so the inflection is \((-1,-4)\).

\(y\)-intercept — substitute \(x=0\):

\(y\)\(=\)\((0+1)^3-4\)
\(=\)\(1-4\)
\(=\)\(-3\)

\(x\)-intercept — set \(y=0\) and cube-root:

\((x+1)^3-4\)\(=\)\(0\)
\((x+1)^3\)\(=\)\(4\)
\(x+1\)\(=\)\(\sqrt[3]{4}\)
\(x\)\(=\)\(\sqrt[3]{4}-1\)

\(y\)-intercept \((0,-3)\); \(x\)-intercept \(\left(\sqrt[3]{4}-1,\,0\right)\).

Graph of y=(x+1)^3-4Increasing cubic with inflection at minus one comma minus four and one real x-intercept near zero point five nine. x y (-1, -4)
x=43-1
Example 4 — Find the rule from features
A cubic \(y=a(x-h)^3+k\) has a stationary point of inflection at \((2,3)\) and passes through \((0,-1)\). Find its rule and \(x\)-intercept.
Solution

The inflection gives \(h=2,\ k=3\), so \(y=a(x-2)^3+3\).

Substitute the point \((0,-1)\) to find \(a\):

\(-1\)\(=\)\(a(0-2)^3+3\)
\(-1\)\(=\)\(-8a+3\)
\(-8a\)\(=\)\(-4\)
\(a\)\(=\)\(\dfrac{1}{2}\)

\(x\)-intercept of \(y=\dfrac{1}{2}(x-2)^3+3\) — set \(y=0\):

\(\dfrac{1}{2}(x-2)^3+3\)\(=\)\(0\)
\((x-2)^3\)\(=\)\(-6\)
\(x-2\)\(=\)\(-\sqrt[3]{6}\)
\(x\)\(=\)\(2-\sqrt[3]{6}\)

Rule \(y=\dfrac{1}{2}(x-2)^3+3\); \(x\)-intercept \(\left(2-\sqrt[3]{6},\,0\right)\).

Graph of y=one half (x-2)^3+3Increasing cubic with a half dilation, inflection at two comma three, passing through the point zero comma minus one. x y (2, 3)
y=12(x-2)3+3

Common pitfalls

Sign of \(h\). The inflection is at \(x=h\) where the bracket is zero. \((x-2)^3\) gives \(h=2\), but \((x+1)^3\) gives \(h=-1\), not \(+1\).
Looking for turning points. A cubic of this form has none — it has a single stationary point of inflection at \((h,k)\). Do not try to find a maximum or minimum.
Expecting three \(x\)-intercepts. \(y=a(x-h)^3+k\) has exactly one \(x\)-intercept, since a cube root gives a single real value.
Cubing away the coefficient. When solving \(a(x-h)^3+k=0\), first isolate \((x-h)^3=-\dfrac{k}{a}\), then cube-root; do not cube-root term by term.

Frequently asked questions

What is the point of inflection of \(y=a(x-h)^3+k\)?

It is the stationary point of inflection at \((h,k)\), read straight from the rule. Watch the sign: \((x+1)^3\) gives \(h=-1\).

How many x-intercepts does \(y=a(x-h)^3+k\) have?

Exactly one, because solving it needs a cube root, which always gives a single real value.

What does the value of \(a\) do to the graph?

It is a dilation that changes the steepness, and if \(a\) is negative it reflects the curve in the \(x\)-axis so the graph decreases instead of increases.

How do you find the x-intercept exactly?

Set \(y=0\), isolate \((x-h)^3=-\dfrac{k}{a}\), then cube-root: \(x=h+\sqrt[3]{-\dfrac{k}{a}}\), leaving the cube root exact.

How do you find the rule from the inflection and a point?

Read \(h\) and \(k\) from the inflection point, substitute the given point into \(y=a(x-h)^3+k\), and solve for \(a\).

Does \(y=a(x-h)^3+k\) have turning points?

No. It increases everywhere (or decreases if \(a\lt 0\)) with just a stationary point of inflection at \((h,k)\); there is no maximum or minimum.