Finding Stationary Points & Turning Points (Edexcel IGCSE Maths B): Revision Note

Exam code: 4MB1

Finding stationary points & turning points

What is a stationary point?

  • A stationary point is a point on the graph at which the gradient is zero (a tangent drawn at this point will be horizontal)

    • These include peaks (maximum points) and troughs (minimum points)

  • Maximum points and minimum points are collectively know as turning points

    • At a turning point a curve changes from moving upwards to moving downwards, or vice versa

A curve on a graph with the direction of the graph at different points highlighted and the stationary points labelled where the direction changes.
  • At a turning point the gradient of the curve is zero

    • If a tangent is drawn at a turning point it will be a horizontal line

    • Horizontal lines have a gradient of zero

  • Because the gradient at a turning point is zero

    • Substituting the x-coordinate of a turning point into the gradient function dydx (the derivative) will give an output of zero 

Graph of a curve with the stationary points highlighted. Tangents are drawn at the stationary point, which are horizontal lines.

How do I find the coordinates of a turning point?

  • STEP 1
    Find the gradient function (derivative) dydx of the original equation

    • E.g. Find the coordinates of the turning point of the equation y=4x28x+9 

    • dydx=8x8

  • STEP 2
    Set the gradient function (derivative) equal to zero and solve for x

    • This will find the x-coordinate of the turning point

    • E.g. 0=8x8, so x=1

  • STEP 3
    Substitute the x-coordinate into the original equation of the graph

    • This will find the y-coordinate of the turning point

    • E.g. y=4(1)28(1)+9=5

  • STEP 4

    Write out the coordinates of the turning point

    • E.g. The turning point is at (1, 5)

Examiner Tips and Tricks

  • Remember to read the questions carefully

    • Sometimes only the x-coordinate of a turning point is required

Worked Example

Find the coordinates of the turning point on the curve with equation y=2x2+8x9.

Answer:

Use differentiation to find the gradient function (derivative) of the equation

dydx=4x+8

Set the derivative equal to 0 and solve for x

0=4x+88=4x2=x

Substitute the x-coordinate into the origination equation of the curve and solve for y

y=2(2)2+8(2)9y=8169y=17

Form a set of coordinates

The turning point has coordinates (2, 17)

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