The linear relationship is probably one of the most fundamental relationships in all of mathematics. Functions that have a constant rate of change are the basis of our most rudimentary geometrical transformations, conversions and correlations. It should be fair to say that ensuring pupils have a proper grasp of linear relationships should be an important part of any mathematics curriculum; and yet many pupils are only given a very narrow view of these key mathematical constructs.
Most pupils first view of the graphs of linear relationships between two variables are through algebra in the form y = mx + c. Pupils will be given equations of this form, and asked to substitute to find coordinates and then plot coordinates to draw lines. Some pupils may be given the opportunity to draw parallels between the equation and the relationship between the variables x and y but not all. Eventually concepts like gradients and intercepts will be taught, and here is where the narrowing will begin. Most pupils will be given an algebraic definition of gradient, such as "change in y over change in x" or similar. Can we first be very clear from the start please that this is not what gradient is, this is just one way to find the gradient if you happen to know the horizontal and vertical distance travelled (for those people who think I am being picky, another way to find the gradient is to take the tangent of the angle the line makes with the horizontal, which is seldom taught in this way).
What gradient actually is is the vertical distance travelled for a unit increase in horizontal distance. Dividing a given vertical by a given horizontal will calculate the the value, as will applying the tangent function to the angle made with the horizontal, but neither tell you what it actually is. Pupils should have a proper understanding of what gradient is, before they begin calculating it (in my opinion). But this is not actually the point of this blog post so I will get back on track...
Once gradient is 'taught' the link between its value and the value of m in the formula given above is very quickly highlighted, often either explicitly or through some form of 'discovery'. Here comes the second narrowing - from this point onward virtually every attempt to ascertain the value of the gradient of a particular line when given any form of linear algebraic relationship invariably leads back to writing the equation in the form y = mx + c. Remember lines are very often defined in a different form; x + y = 5, 3x + 2y + 4 = 0 etc. Ask any competent school age pupil to find the value of the gradient of these lines, and I will guarantee that the vast majority of the time a rearrangement into the form y = mx + c is attempted if the pupil is even able to attempt the problem at all. And while this approach is perfectly correct and if done well will reveal the value of the gradient, it isn't the only approach; many pupils labour in ignorance when better methods may be applied.
Take the line x + y = 5 for example. Now for most mathematicians it would be straightforward to rearrange this to give y = -x + 5, and hence find the value of the gradient of -1, and the y-intercept of (0,5). However I would argue at least equally straightforward would be to say "the points (0,5) and (5,0) are on the line, and so the value of the gradient = -5/5 = -1 and the y-intercept is (0,5) [and, by the way, the x intercept is (5,0) - which is not nearly so often asked about]. To be fair, there is probably not a huge difference in the mechanics, but as Anne Watson highlights in her blog (see postscript below) there is perhaps a difference in pupils understanding of what this line actually looks like, as well as providing more of an opportunity to reinforce the idea of vertical distance travelled for unit horizontal distance.
If we then take the line 3x + 2y + 4 = 0, the rearrangement is a bit messier - I know plenty of pupils that wouldn't be able to rearrange successfully. However it is still a rearrangement that you would want pupils to be able to do and expect that they could if they had the proper grounding in inverse operations etc. The other side of this though is that I can quite quickly see that the point (0,-2) is on this line, and that the point (-1⅓, 0) is on this line. So I can also calculate the gradient as -2/1⅓ = 1½, as well as tell you about the x-intercept and y-intercept. Perhaps even more straightforwardly I could have told you that the point (1, -3½) is on the line, and so arrived at the value of the gradient immediately, I have gone 1½ units down when x increased by 1 (from 0 to 1).
Whether you want to consider rearrangement to the form y = mx + c as a 'method of last resort' or not is up to you; clearly it is an important mathematical idea that relationships can be expressed in different forms. However I would suggest that it is not the only idea that pupils should be able to draw upon when talking and thinking about finding gradient values, and that we should be aiming to give pupils a range of strategies linked to a deeper understanding of what gradients, and lines of constant gradient, are.
Postscript: Emeritus Professor of Education at Oxford University Anne Watson recently released a blog about a similar topic (and actually using one of the same equations!) here. I have actually been writing this blog post since late January and was just trying to find time to finish it off, so wanted to go ahead and publish it anyway!
Most pupils first view of the graphs of linear relationships between two variables are through algebra in the form y = mx + c. Pupils will be given equations of this form, and asked to substitute to find coordinates and then plot coordinates to draw lines. Some pupils may be given the opportunity to draw parallels between the equation and the relationship between the variables x and y but not all. Eventually concepts like gradients and intercepts will be taught, and here is where the narrowing will begin. Most pupils will be given an algebraic definition of gradient, such as "change in y over change in x" or similar. Can we first be very clear from the start please that this is not what gradient is, this is just one way to find the gradient if you happen to know the horizontal and vertical distance travelled (for those people who think I am being picky, another way to find the gradient is to take the tangent of the angle the line makes with the horizontal, which is seldom taught in this way).
What gradient actually is is the vertical distance travelled for a unit increase in horizontal distance. Dividing a given vertical by a given horizontal will calculate the the value, as will applying the tangent function to the angle made with the horizontal, but neither tell you what it actually is. Pupils should have a proper understanding of what gradient is, before they begin calculating it (in my opinion). But this is not actually the point of this blog post so I will get back on track...
Once gradient is 'taught' the link between its value and the value of m in the formula given above is very quickly highlighted, often either explicitly or through some form of 'discovery'. Here comes the second narrowing - from this point onward virtually every attempt to ascertain the value of the gradient of a particular line when given any form of linear algebraic relationship invariably leads back to writing the equation in the form y = mx + c. Remember lines are very often defined in a different form; x + y = 5, 3x + 2y + 4 = 0 etc. Ask any competent school age pupil to find the value of the gradient of these lines, and I will guarantee that the vast majority of the time a rearrangement into the form y = mx + c is attempted if the pupil is even able to attempt the problem at all. And while this approach is perfectly correct and if done well will reveal the value of the gradient, it isn't the only approach; many pupils labour in ignorance when better methods may be applied.
Take the line x + y = 5 for example. Now for most mathematicians it would be straightforward to rearrange this to give y = -x + 5, and hence find the value of the gradient of -1, and the y-intercept of (0,5). However I would argue at least equally straightforward would be to say "the points (0,5) and (5,0) are on the line, and so the value of the gradient = -5/5 = -1 and the y-intercept is (0,5) [and, by the way, the x intercept is (5,0) - which is not nearly so often asked about]. To be fair, there is probably not a huge difference in the mechanics, but as Anne Watson highlights in her blog (see postscript below) there is perhaps a difference in pupils understanding of what this line actually looks like, as well as providing more of an opportunity to reinforce the idea of vertical distance travelled for unit horizontal distance.
If we then take the line 3x + 2y + 4 = 0, the rearrangement is a bit messier - I know plenty of pupils that wouldn't be able to rearrange successfully. However it is still a rearrangement that you would want pupils to be able to do and expect that they could if they had the proper grounding in inverse operations etc. The other side of this though is that I can quite quickly see that the point (0,-2) is on this line, and that the point (-1⅓, 0) is on this line. So I can also calculate the gradient as -2/1⅓ = 1½, as well as tell you about the x-intercept and y-intercept. Perhaps even more straightforwardly I could have told you that the point (1, -3½) is on the line, and so arrived at the value of the gradient immediately, I have gone 1½ units down when x increased by 1 (from 0 to 1).
Whether you want to consider rearrangement to the form y = mx + c as a 'method of last resort' or not is up to you; clearly it is an important mathematical idea that relationships can be expressed in different forms. However I would suggest that it is not the only idea that pupils should be able to draw upon when talking and thinking about finding gradient values, and that we should be aiming to give pupils a range of strategies linked to a deeper understanding of what gradients, and lines of constant gradient, are.
Postscript: Emeritus Professor of Education at Oxford University Anne Watson recently released a blog about a similar topic (and actually using one of the same equations!) here. I have actually been writing this blog post since late January and was just trying to find time to finish it off, so wanted to go ahead and publish it anyway!
