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Differentiate the function, set the derivative equal to zero and solve for x. Free IGCSE Edexcel Maths notes with step-by-step methods and 10 FAQs.
Gradients and Stationary Points by Differentiation is part of Quadratic and Other Graphs in IGCSE Edexcel Higher Maths.
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IGCSE Edexcel Higher
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Differentiate the function, set the derivative equal to zero and solve for x. Free IGCSE Edexcel Maths notes with step-by-step methods and 10 FAQs.
Use this lesson to stay inside Quadratic and Other Graphs while connecting the detail back to the wider Topic 3: Sequences, Functions and Graphs topic.
Read the core explanations from Gradients and Stationary Points by Differentiation before testing yourself from memory.
A stationary point happens when the curve flattens, so the tangent there has gradient zero. Differentiation gives a clean algebraic way to find those points.
Once you know the x-value, substitute back into the original equation to get the full coordinate.
Worked example: Finding a minimum point
Find the stationary point of y = x^2 - 4x + 1.
Answer: Stationary point at (2, -3), and it is a minimum.
Worked example: Finding more than one stationary point
Find the stationary points of y = x^3 - 3x.
Answer: Stationary points at (-1, 2) and (1, -2).
Worked example: Finding and Classifying a Stationary Point
Find the stationary point of y = x^2 - 6x + 5 and decide whether it is a maximum or minimum.
Answer: The stationary point is (3, -4) and it is a minimum.
Setting the derivative to zero only finds the x-coordinate. You still need to substitute back into the original function to get the point.
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Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.
Differentiate the function, set the derivative equal to zero and solve for x. Then substitute each x-value back into the original function to get the coordinates. For y = x² - 4x + 1, 2x - 4 = 0 gives x = 2 and y = -3, so the stationary point is (2, -3).
A stationary point is a point on a graph where the gradient is zero, so the curve flattens and the tangent there has gradient zero. It may be a maximum, a minimum or neither.
Use the shape of the graph or the sign of the gradient around the point. For a parabola that opens upwards, such as y = x² - 6x + 5, the graph falls to the point and then rises, so the stationary point (3, -4) is a minimum.
Differentiate to get dy/dx = 3x² - 3, then set 3x² - 3 = 0, so x = ±1. Substituting gives (-1, 2), a local maximum, and (1, -2), a local minimum, as the cubic rises, falls, then rises again.
Setting the derivative to zero only finds the x-coordinate of a stationary point. Substituting that x-value back into the original function, not the derivative, gives the y-coordinate and the full point.
The stationary point of y = x² - 4x + 1 is (2, -3), and it is a minimum. dy/dx = 2x - 4 = 0 gives x = 2, then y = 2² - 4(2) + 1 = -3, and the parabola opens upward.
A maximum is a highest turning point in the local shape of a graph, where the gradient is zero. For y = x³ - 3x, the point (-1, 2) is a local maximum, because the cubic rises, then falls.
A minimum is a lowest turning point in the local shape of a graph, where the gradient is zero. For y = x² - 6x + 5, the minimum is (3, -4), because the graph falls to the point and then rises away from it.
The gradient is zero at a stationary point because the curve flattens there, so the tangent is flat. This is why stationary points are found by setting the derivative equal to zero.
To find stationary points, set the derivative equal to zero, not the original function. Setting the original function to zero instead of the derivative is a common pitfall.