Stationary Points
Understand stationary points for Queensland Year 11 Mathematical Methods (QCAA). A stationary point is where a curve momentarily levels off — the gradient is zero and the tangent is horizontal.
You will learn to find stationary points by solving where the derivative is zero, substitute back for their coordinates, and count how many a cubic or quartic can have — the first step in sketching and optimising curves.
Every question with a fully worked solution.
- Stationary Points - Video - Stationary points Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), a stationary point of a curve \(y=f(x)\) is a point where the gradient is zero, so the tangent is horizontal. This page shows how to locate stationary points of a polynomial by solving \(f'(x)=0\) for the \(x\)-value(s), substituting back to get the coordinates, and how many stationary points a cubic or quartic can have.
A stationary point of \(y=f(x)\) is a point where the curve is momentarily flat: the gradient is zero, so the tangent there is horizontal. Because the gradient is the derivative, the defining condition is \(f'(x)=0\).
To find a stationary point you solve \(f'(x)=0\) to get the \(x\)-coordinate(s), then substitute each back into the original \(f(x)\) to get the matching \(y\)-coordinate(s). A quadratic has one stationary point, a cubic has up to two, and a quartic up to three — one for each distinct real solution of \(f'(x)=0\).
A point \((a,\,f(a))\) is stationary when the derivative is zero there:
The coordinates come from solving that equation, then substituting back:
How to find the stationary points of \(y=f(x)\)
- Differentiate: find \(f'(x)\) using the power rule.
- Solve: set \(f'(x)=0\) and solve for every \(x\)-value (factorise where you can).
- Substitute back: put each \(x\)-value into the original \(f(x)\) to get the \(y\)-coordinate, and state each stationary point as \((x,\,y)\).
Differentiate and set \(f'(x)=0\):
| \(f'(x)\) | \(=\) | \(2x+4\) |
| \(2x+4\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-2\) |
\(y\)-coordinate — substitute into the curve:
| \(f(-2)\) | \(=\) | \((-2)^2+4(-2)+1\) |
| \(=\) | \(4-8+1\) | |
| \(=\) | \(-3\) |
Stationary point at \((-2,\,-3)\).
Differentiate and set \(f'(x)=0\):
| \(f'(x)\) | \(=\) | \(3x^2-6x-9\) |
| \(3x^2-6x-9\) | \(=\) | \(0\) |
| \(3(x^2-2x-3)\) | \(=\) | \(0\) |
| \(3(x-3)(x+1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(3 \ \text{or}\ -1\) |
\(y\)-coordinates — substitute each back:
| \(f(3)\) | \(=\) | \((3)^3-3(3)^2-9(3)+5=-22\) |
| \(f(-1)\) | \(=\) | \((-1)^3-3(-1)^2-9(-1)+5=10\) |
Stationary points at \((3,\,-22)\) and \((-1,\,10)\).
Differentiate and set \(f'(x)=0\):
| \(f'(x)\) | \(=\) | \(3x^2+6x+3\) |
| \(3x^2+6x+3\) | \(=\) | \(0\) |
| \(3(x+1)^2\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-1\) |
The factor \((x+1)^2\) gives a single (repeated) root, so there is one stationary point.
\(y\)-coordinate:
| \(f(-1)\) | \(=\) | \((-1)^3+3(-1)^2+3(-1)+2\) |
| \(=\) | \(-1+3-3+2\) | |
| \(=\) | \(1\) |
One stationary point, at \((-1,\,1)\).
Use \(f'(2)=0\) to find \(a\):
| \(f'(x)\) | \(=\) | \(3x^2+2ax-24\) |
| \(f'(2)\) | \(=\) | \(3(2)^2+2a(2)-24\) |
| \(0\) | \(=\) | \(12+4a-24\) |
| \(4a\) | \(=\) | \(12\) |
| \(a\) | \(=\) | \(3\) |
With \(a=3\), solve \(f'(x)=0\):
| \(f'(x)\) | \(=\) | \(3x^2+6x-24\) |
| \(3(x^2+2x-8)\) | \(=\) | \(0\) |
| \(3(x+4)(x-2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-4 \ \text{or}\ 2\) |
Other point — substitute \(x=-4\):
| \(f(-4)\) | \(=\) | \((-4)^3+3(-4)^2-24(-4)+7\) |
| \(=\) | \(-64+48+96+7\) | |
| \(=\) | \(87\) |
\(a=3\); the other stationary point is \((-4,\,87)\).
Common pitfalls
Frequently asked questions
What is a stationary point?
A point on a curve where the gradient is zero, so the tangent is horizontal. The condition is \(f'(x)=0\).
How do you find the coordinates of a stationary point?
Solve \(f'(x)=0\) for the \(x\)-value(s), then substitute each back into the original \(f(x)\) for the matching \(y\)-value.
How many stationary points can a cubic have?
Up to two — one for each distinct real solution of \(f'(x)=0\). A repeated root gives just one.
Do I use the derivative to find the y-coordinate?
No. The derivative gives only the \(x\)-value where the gradient is zero; the \(y\)-value comes from the original curve \(f(x)\).
Is a stationary point always a maximum or minimum?
Not always — it can also be a stationary point of inflection. Finding the coordinates comes first; classifying the nature is a separate step (the first-derivative sign test).
What is the difference between locating and classifying a stationary point?
Locating means finding its coordinates by solving \(f'(x)=0\); classifying means deciding whether it is a maximum, minimum or inflection using a sign test.