18 April 2015

Tricks and Tips 3: Quadratics

Last month I presented a workshop at the National Mathematics Teacher Conference (#mathsconf2015) entitled 'Tricks and Tips: Clever Methods for Explaining Mathematical Concepts'. This is the third in a series of posts summarising the content of that workshop for those who were unable to attend. The aim of my workshop was to encourage people to reflect on their subject knowledge and the effectiveness of their explanations. I also hoped that delegates would learn new methods that they might consider using at school. In today's post I'm covering quadratics, specifically methods for finding a vertex. My previous posts were on methods for finding a highest common factor and methods for sequences, linear graphs and surds.

The vertex of a quadratic graph
This comes up in the new maths GCSE, in questions like this which is taken from the Pearson Edexcel GCSE (9-1) Mathematics Sample Assessment Materials:
Pause for a minute and look at this question, because it's a great example of the change in difficulty in the new GCSE. Note the use of function notation and the term turning point. The equation doesn't factorise. The graph doesn't cross the x-axis. The coordinates of the turning point aren't integers. This is a notable step up from the type of questions asked in current GCSE exams.

For the purpose of this post, let's consider the function y = x2 - 6x + 10. There are a number of ways to find the coordinates of the turning point - how would you do it?
1. Vertex Form
The term 'vertex form' is not commonly used in the UK. Vertex form is what you get when you complete the square. So we'd write the function as follows:
y = (x - 3)2 + 1
Now we can identify the turning point straight away. I've always explained it a bit like this:

"The (x - 3)2  is squared so it can never be negative. The lowest it can be is zero. It's zero when x is 3. The lowest possible y value is 0 + 1. So we know that the minimum is at (3,1)..."

This explanation is in line with my thought process - it's the way I identify the turning point - but my students really struggle with it. Of all the things I teach, this is the explanation that gets the most blank looks! So this year I tried a different approach when revising this topic with my Year 11s. This time I relied on my students' knowledge of graph transformations. I told them to think of the graph y = (x - 3)2 + 1 as a transformation of the graph y = x2. It's been translated 3 units right and 1 unit up. The vertex moves from (0,0) to (3,1). They found this approach really easy - it made a lot more sense to them. Suddenly all my students were able to find the vertex of a quadratic function.

As long as students have studied graph transformations then this approach seems to work. This teaching order is worth bearing in mind when designing a Scheme of Work. 

From now on, I'm going to use the transformation method. But there are alternatives...

2. A formula
In some countries, students simply memorise a formula. They learn that the x coordinate of the vertex is -b/2a. They then find the y coordinate by substituting that value into the equation.

By memorising this formula, you can find the coordinates of the turning point of any quadratic function without completing the square. At my conference session I showed the video below - watch it to see how the method is explained. I'm not a fan of this approach. I don't want my students to simply memorise a formula - there's no conceptual understanding here.

3. Differentiation
Differentiation is always a pleasure. We don't do calculus at GCSE, but I thought it worth mentioning here that another method to find a turning point of a function is to set the derivative equal to zero. As you can see below, for a quadratic that will always give us x = -b/2a.
4. Symmetry
For a quadratic that intercepts the x axis, the vertex is the midpoint of the two roots. This works because parabolas are symmetrical. Up until recently I thought this approach wasn't possible for quadratics that don't intercept the x axis, but then I discovered James Tanton's method. It's described below for the function  y = x2 + 4x + 5  - for more detail and examples, see this curriculum essay or this video.
If we apply this method to our original example, we rewrite y = x2 - 6x + 10 as y = x(x - 6) + 10. We can see that two points on this curve are (0,10) and (6,10), so the vertex has x coordinate 3 by symmetry. Simple!

James Tanton has produced a brilliant pamphlet 'Guide to Everything Quadratic' which is helpful for any maths teacher preparing to teach quadratics for the first time. My Algebra and Core AS resource libraries are packed full of recommended resources for teaching this fantastic topic, such as this activity from Susan Wall.
Preparing for the new GCSE
As I was writing this post it occurred to me that there's a lot of really important things that maths departments need to do this term to prepare for the new GCSE. Writing new Schemes of Work is a huge job, as is finding suitable resources for teaching the new GCSE topics.

CPD for maths teachers is also really important. All maths teachers need to be familiar with the new GCSE content - they need to know what's been added and what's been removed. They need to look at lots of example questions.

The other thing that all teachers need to do now is a subject knowledge check - are there any topics on the new GCSE syllabus that you're not familiar with? This is particularly relevant for teachers who've never taught A level maths. Has everyone in your department thought about how to teach the new GCSE topics? It's time for some vital maths department CPD.


  1. Great post Jo, I normally teach CTS and Symmetry. I don't like the formula method. Would like,to see calculus on GCSE.

  2. But one you've done the symmetry method a few times and 'get' the trick that the formula gives- it's a great time saver? ( never taught quadratics- doesn't really come up in primary!) I did however (re) teach myself how tomfind the vertex today. I remember finding it really hard when I was taught it many many years ago- but am not clear precisely which method. I think maybe something to do with from finding y when x was zero or the gradient was zero or....dunno like the symmetry way better.