How to find stationary points
Learn how to find and classify stationary points for IGCSE Maths. This page covers maximum and minimum turning points on quadratic and cubic curves, how to locate them by solving dy/dx = 0, and two methods to classify them: using the shape of the curve and the local gradient check. Step-by-step worked examples and a randomised practice room with auto-marked questions help you build the skills and confidence you need for your exam.
You need to be confident differentiating and finding gradients at a point before starting. Revise Differentiation (Power Rule) and Equation of a Tangent to a Curve first.
What is a Stationary Point?
A stationary point (also called a turning point) is a point on a curve where the gradient is exactly zero. The curve is momentarily flat at these points; neither rising nor falling. They occur wherever \(\dfrac{dy}{dx} = 0\).
The curve rises to this point, then falls away. The gradient changes from positive to zero to negative.
The curve falls to this point, then rises. The gradient changes from negative to zero to positive.
At every stationary point: \(\dfrac{dy}{dx} = 0\). Solving this equation gives the \(x\)-values of the turning points.
Stationary Points on a Cubic Curve
At every stationary point the tangent is horizontal; \(\dfrac{dy}{dx} = 0\). The gradient changes sign on either side.
How to Find and Classify Stationary Points
⚡ Four Steps
- Differentiate to find \(\dfrac{dy}{dx}\).
- Solve \(\dfrac{dy}{dx} = 0\) to find the \(x\)-value(s) of the stationary points.
- Substitute each \(x\)-value back into the original equation to find the \(y\)-coordinates.
- Classify each point as a maximum or minimum using one of the two methods below.
Classifying Stationary Points
Method 1: Shape of Curve
Look at the sign of the leading coefficient of the original equation.
Quadratic \(y = ax^2 + \ldots\)
- \(a > 0\): U-shape; stationary point is a minimum
- \(a < 0\): n-shape; stationary point is a maximum
Cubic \(y = ax^3 + \ldots\)
- \(a > 0\): the point with the smaller \(x\)-value is the maximum
- \(a < 0\): the point with the smaller \(x\)-value is the minimum
Method 2: Local Gradient Check
Substitute values 0.1 either side of the stationary \(x\)-value into \(\dfrac{dy}{dx}\) and read the signs.
| \(\dfrac{dy}{dx}\) before | At point | \(\dfrac{dy}{dx}\) after | Type |
|---|---|---|---|
| +ve | 0 | −ve | Maximum |
| −ve | 0 | +ve | Minimum |
Recognising Curve Shapes
Knowing the shape from the equation is the quickest classification method.
U-shape: one minimum
n-shape: one maximum
Left turning point: max
Left turning point: min
Worked Examples
💡 Example 1: Quadratic (Method 1)
Find and classify the stationary point of \(y = x^2 - 6x + 11\).
Step 1: Differentiate.
\[\frac{dy}{dx} = 2x - 6\]Step 2: Solve \(\dfrac{dy}{dx} = 0\).
\[\begin{array}{rcl} 2x - 6 &=& 0 \\ x &=& 3 \end{array}\]Step 3: Find \(y\).
\[\begin{array}{rcl} y &=& (3)^2 - 6(3) + 11 \\ &=& 2 \end{array}\]Step 4: Classify. \(a = 1 > 0\): U-shape, so minimum at \((3,\; 2)\).
💡 Example 2: Cubic (Both Methods)
Find and classify the turning points of \(y = x^3 - 3x^2 - 9x + 5\).
Differentiate: \(\dfrac{dy}{dx} = 3x^2 - 6x - 9\)
Solve:
\[\begin{aligned} 3x^2 - 6x - 9 &= 0 \\ (x-3)(x+1) &= 0 \end{aligned}\]\(x = 3\) or \(x = -1\)
Coordinates:
At \(x = -1\):
\[\begin{array}{rcl} y &=& -1 - 3 + 9 + 5 \\ &=& 10 \end{array}\]At \(x = 3\):
\[\begin{array}{rcl} y &=& 27 - 27 - 27 + 5 \\ &=& -22 \end{array}\]Method 1: \(a = 1 > 0\), so smaller \(x\) is max: maximum at \((-1,\;10)\), minimum at \((3,\;-22)\).
Method 2 check at \(x = -1\): test \(x = -1.1\) gives +ve, \(x = -0.9\) gives −ve. Confirms maximum.
💡 Example 3: Full Working
Find and classify the turning points of \(y = 4x^3 + 6x^2 - 24x + 5\).
Step 1: Differentiate.
\[\frac{dy}{dx} = 12x^2 + 12x - 24\]Step 2: Solve.
\[\begin{aligned} 12x^2 + 12x - 24 &= 0 \\ x^2 + x - 2 &= 0 \\ (x+2)(x-1) &= 0 \end{aligned}\]\(x = -2\) or \(x = 1\)
Step 3: Coordinates.
At \(x = -2\):
\[\begin{aligned} y &= 4(-2)^3 + 6(-2)^2 - 24(-2) + 5 \\ &= -32 + 24 + 48 + 5 \\ &= 45 \end{aligned}\]At \(x = 1\):
\[\begin{aligned} y &= 4(1)^3 + 6(1)^2 - 24(1) + 5 \\ &= 4 + 6 - 24 + 5 \\ &= -9 \end{aligned}\]Method 1: \(a = 4 > 0\), so the point with the smaller \(x\)-value is the maximum:
Maximum at \((-2,\; 45)\) and minimum at \((1,\; -9)\).
Method 2: gradient check at \(x = -2\)
At \(x = -2.1\):
\[\begin{aligned} \frac{dy}{dx} &= 12(4.41) + 12(-2.1) - 24 \\ &= 52.92 - 25.2 - 24 \\ &= +3.72 \quad (+\text{ve}) \end{aligned}\]At \(x = -1.9\):
\[\begin{aligned} \frac{dy}{dx} &= 12(3.61) + 12(-1.9) - 24 \\ &= 43.32 - 22.8 - 24 \\ &= -3.48 \quad (-\text{ve}) \end{aligned}\]+ve then −ve: confirms maximum. Both methods agree.
🌍 Example 4: Applied Optimisation
A company's daily profit \(P\) (in thousands of dollars) when producing \(x\) hundred units is modelled by \[P = 8x - x^2 - 7 \quad (0 \le x \le 8).\] Find the level of production that maximises profit and state that maximum profit.
Step 1: Differentiate.
\[\frac{dP}{dx} = 8 - 2x\]Step 2: Solve.
\[\begin{array}{rcl} 8 - 2x &=& 0 \\ x &=& 4 \end{array}\]Step 3: Maximum profit.
\[\begin{array}{rcl} P &=& 8(4) - (4)^2 - 7 \\ &=& 32 - 16 - 7 \\ &=& 9 \end{array}\]Step 4: Classify. Coefficient of \(x^2\) is \(-1 < 0\): n-shaped, so this is a maximum.
Producing 400 units per day gives a maximum profit of \$9,000 per day.
🔑 Key Points
- A stationary point always satisfies \(\dfrac{dy}{dx} = 0\).
- Always find both the \(x\) and \(y\) coordinates.
- Quadratic with \(a > 0\): one minimum. With \(a < 0\): one maximum.
- Cubic with \(a > 0\): the left turning point is the maximum.
- Local gradient check: test \(\dfrac{dy}{dx}\) at \(x \pm 0.1\) and read the sign change.
⚠️ Common Mistakes
- Finding \(x\) but forgetting to substitute back to find \(y\).
- Misreading the sign of the leading coefficient when using the curve shape method.
- Testing only one side in the gradient check; always test both sides.
- Not simplifying \(\dfrac{dy}{dx} = 0\) first; dividing by the common factor before solving saves errors.
- Confusing the \(x\)-value of the point with the gradient value when setting up the check.