Stationary Points & Turning Points (AQA GCSE Further Maths)

Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Stationary Points & Turning Points

What are stationary points?

  • A stationary point is any point on a curve where the gradient is zero

  • To find stationary points of a curve

Step 1
Find the first derivative fraction numerator d y over denominator d x end fraction

Step 2
Solve fraction numerator d y over denominator d x end fraction equals 0 to find the x-coordinates of any stationary points

Step 3
Substitute those x-coordinates into the equation of the curve to find the corresponding y-coordinates

  • A stationary point may be either a local minimum, a local maximum, or a point of inflection 

Stat Point Illustr 1, A Level & AS Maths: Pure revision notes

Stationary points on quadratics

  • The graph of a quadratic function only has a single stationary point

  • For a positive quadratic this is the minimum; for a negative quadratic it is the maximum

    • No need to talk about 'local' here, as it is the overall minimum/maximum for the whole curve

Stationary Points min max for parabola illustr, A Level & AS Maths: Pure revision notes
  • The y value/coordinate of the stationary point is therefore the minimum or maximum value of the quadratic function

  • For quadratics especially, minimum and maximum points are often referred to as turning points

How do I determine the nature of stationary points on other curves?

  • For a graph there are two ways to determine the nature of its stationary points 

  • Method A

    • Compare the signs of the first derivative, fraction numerator straight d y over denominator straight d x end fraction, (positive or negative) a little bit to either side of the stationary point

    • (After completing Steps 1 - 3 above to find the stationary points)

Step 4
For each stationary point find the values of the first derivative a little bit 'to the left' (i.e. for a slightly smaller x value) and a little bit 'to the right' (i.e. for a slightly larger x value) of the stationary point

Stat Points left right proviso, A Level & AS Maths: Pure revision notes

Step 5
Compare the signs (positive or negative) of the derivatives on the left and right of the stationary point

  • If the derivatives are negative on the left and positive on the right, the point is a local minimum (a u-shape)

  • If the derivatives are positive on the left and negative on the right, the point is a local maximum (an n-shape)

  • If the signs of the derivatives are the same on both sides (both positive or both negative) then the point is a point of inflection (ablank subscript bold divided by bold minus to the power of bold divided by shape)   

incr decr min max, A Level & AS Maths: Pure revision notes
Stationary Points point of inflection, A Level & AS Maths: Pure revision notes
  • Method B

    • Look at the sign of the second derivative (positive or negative) at the stationary point

    • (After completing Steps 1 - 3 above to find the stationary points)

Step 4
Find the second derivative fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction

Step 5
For each stationary point find the value of fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction at the stationary point
i.e. substitute the x-coordinate of the stationary point into fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction and evaluate

  • If fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction is positive then the point is a local minimum

  • If fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction is negative then the point is a local maximum

  • If fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction is zero then the point could be a local minimum, a local maximum OR a point of inflection

    • use Method A to determine which

Examiner Tips and Tricks

  • Usually, using the second derivative (Method B above) is a much quicker way of determining the nature of a stationary point.

    • However, if the second derivative is zero it tells you nothing about the point

      • In this case you will have to use Method A

      • This method always works – see the Worked Example

Worked Example

Find the stationary points of

y equals x cubed open parentheses 3 x squared minus 20 close parentheses

and determine the nature of each.

Start by expanding the brackets in the expression for y

y equals 3 x to the power of 5 minus 20 x cubed

Step 1
Find the first derivative

fraction numerator straight d y over denominator straight d x end fraction equals 15 x to the power of 4 minus 60 x squared

Step 2
Solve fraction numerator straight d y over denominator straight d x end fraction equals 0

table attributes columnalign right center left columnspacing 0px end attributes row cell 15 x to the power of 4 minus 60 x squared end cell equals 0 end table

Divide through by 15

table attributes columnalign right center left columnspacing 0px end attributes row cell x to the power of 4 minus 4 x squared end cell equals 0 end table

Fully factorise, spotting the difference of two squares

table row cell x to the power of 4 minus 4 x squared end cell equals 0 row cell x squared open parentheses x squared minus 4 close parentheses end cell equals 0 end table

Solve to find the x-coordinates of the stationary (turning) points

x equals 0 comma space minus 2 comma space 2

Step 3
Find the corresponding y-coordinates

table row x equals cell 0 comma space space y equals open parentheses 0 close parentheses cubed open parentheses 3 open parentheses 0 close parentheses squared minus 20 close parentheses equals 0 end cell row x equals cell negative 2 comma space y equals open parentheses negative 2 close parentheses cubed open parentheses 3 open parentheses negative 2 close parentheses squared minus 20 close parentheses equals 64 end cell row x equals cell 2 comma space y equals open parentheses 2 close parentheses cubed open parentheses 3 open parentheses 2 close parentheses squared minus 20 close parentheses equals negative 64 end cell end table

Write the answers as coordinates, being careful to correctly match x's and y's

The stationary points are (0, 0), (-2, 64) and (2, -64)

To find the nature of each stationary point, start with Method B
Step 4
Find the second derivative

fraction numerator straight d squared y over denominator straight d x squared end fraction equals 60 x cubed minus 120 x

Step 5
Evaluate the second derivative at each stationary point

table row x equals cell 0 comma fraction numerator space straight d squared y over denominator straight d x squared end fraction equals 60 open parentheses 0 close parentheses cubed minus 120 open parentheses 0 close parentheses end cell end table = 0

table row x equals cell negative 2 comma space fraction numerator space straight d squared y over denominator straight d x squared end fraction equals 60 open parentheses negative 2 close parentheses cubed minus 120 open parentheses negative 2 close parentheses equals negative 240 end cell end table < 0

table row x equals cell 2 comma space fraction numerator space straight d squared y over denominator straight d x squared end fraction equals 60 open parentheses 2 close parentheses cubed minus 120 open parentheses 2 close parentheses equals 240 end cell end table > 0

The nature of two of the stationary points is now determined

(-2, 64) is a local maximum point
(2, -64) is a local minimum point

For the third stationary point, (0, 0) switch to Method A
Step 4
Compare first derivatives a little to the left and right of x equals 0

Choose x equals negative 1 and x equals 1

x equals negative 1 comma fraction numerator space straight d y over denominator straight d x end fraction equals 15 open parentheses negative 1 close parentheses to the power of 4 minus 60 open parentheses negative 1 close parentheses squared equals negative 45 < 0

x equals 1 comma space fraction numerator space straight d y over denominator straight d x end fraction equals 15 open parentheses 1 close parentheses to the power of 4 minus 60 open parentheses 1 close parentheses squared equals negative 45 < 0

Step 5
Interpret the result and state the answer

fraction numerator straight d y over denominator straight d x end fraction has the same sign on both sides of (0, 0)

(0, 0) is a point of inflection

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Roger B

Author: Roger B

Expertise: Maths

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.