4
$\begingroup$

I always say that the most difficult part of graphing or plotting points is labelling your axis/es. In the case of plotting the graph of a linear equation with integer coefficients in 2 variables it seems the most difficult part might be finding an integer point to start with.

Take the equation $6x+21y=20$ or similar.

One could use Euclid's algorithm (or similar) to find an integer point and step off another point, but this is not feasible to teach at high school.

My question is what procedure to teach students how to graph equations which will handle such equations?

Possible techniques include finding intercepts or finding y intercept and gradient, approximating as necessary. Judicious choice of examples to avoid annoying fractions could also be a strategy.

$\endgroup$
1
  • 1
    $\begingroup$ I agree with guest troll's answer that integer solutions aren't necessarily important for your students. I'd rather they understand why in your example there are no integer solutions (left side divisible by 3, right side not) than find them when they exist. $\endgroup$
    – Thierry
    Sep 13, 2023 at 13:00

3 Answers 3

8
$\begingroup$

If you're just trying to get students to graph lines, then while it's a good idea to start students off graphing lines whose equations permit obvious integer points, I think it's a mistake to necessitate integer points when the equations get more complicated.

For instance, the line $y= x + \sqrt{2}$ could easily show up in a high school math course, but there aren't any integer points that lie on it.

Personally, I think the easiest procedure for graphing lines is to just plot the $x$ and $y$-intercepts and draw a line through them

In your case, for the line $6x+21y=20,$ it's easy to see that the $x$ and $y$ intercepts are as follows:

\begin{align*} 6x + 21 \cdot 0 = 20 \quad &\Rightarrow \quad x = \frac{20}{6} = 3 \, \frac{1}{3} \\[5pt] 6 \cdot 0 + 21 y = 20 \quad &\Rightarrow \quad y = \frac{20}{21} \end{align*}

And when you write the intercepts in mixed number form (or decimal form) it's easy to see where they go on a graph.

That said, if both of the intercepts come out to $0,$ then you need to either plug in a nonzero number for one of the variables or use the slope (gradient) to find another point.

$\endgroup$
2
  • 1
    $\begingroup$ I would rephrase the instruction 'find integer points' to 'use easy to find points'. Integer points are often but not always easy to find points. Intercepts are generally easy to find points. Depending on context some other points on the line might already be known and therefore easyto find. $\endgroup$
    – quarague
    Sep 14, 2023 at 7:30
  • $\begingroup$ Thanks, @Stef -- fixed. $\endgroup$ Sep 28, 2023 at 19:03
2
$\begingroup$

This is a little bit of a frame challenge, but I don't think that it matters all that much. What is the graph of a line trying to convey? What data are really necessary to get right? In my opinion,

  1. an approximation of the slope (very negative, slightly negative, flat, slightly positive, very positive),
  2. a general indication of the location of the intercepts, and
  3. there is no (3).

My general advice to students when graphing these kinds of things is to either (a) sketch two easy-to-find points and connect-the-dots, or (b) sketch one easy-to-find point and use the slope to finish the graph. "Easy-to-find" is a bit of a dodge—it doesn't really explain what is going on—which is why I like to give a few random examples of different kinds of lines, generally by using a set of 20-sided dice to determine coefficients (plus a coin to determine the signs of coefficients). What is "easy-to-find" is going to be down to taste and experience.

In the example given, I think that the intercepts are relatively easy-to-find, so I would expect an answer from a student which looks something like the following:

enter image description here

My students will often work very hard to draw everything to scale, but you'll notice that I haven't bothered. The important information is there: the exact location of the intercepts (more importantly, the $x$-intercept is on the positive $x$-axis, the $y$-intercept is on the positive $y$-axis, and the $y$-intercept is closer to the origin than the $x$-intercept), and an indication that the slope is slightly negative (i.e. negative, but not super-steep, i.e. somewhere between $0$ and $-1/2$, perhaps)

I think that the philosophy that students should be taught early is that a graph should show key information, and that everything else can and should be left a little imprecise. This is true for lines, for parabolas, and for all the curves graphed in calculus and beyond. If one needs a really pretty picture, computers are very good at making super accurate graphs—the point of doing things by hand is to identify important features, not to get a publication-ready image.

$\endgroup$
1
  • $\begingroup$ I completely agree and I'd add: there is no (3) but there is a (0). The graph of the solution set of a linear equation conveys the very visual information that the solution set is a line, which as obvious as it might seem to the teacher when they look at the linear equation, is easily forgotten by some students when they manipulate abstract algebra symbols $\endgroup$
    – Stef
    Sep 28, 2023 at 17:56
1
$\begingroup$

You are better off just having them do the x and y intercepts (yes approximate if it's 3 and 10/21 or whatever). They are drawing by eye anyhow.

And the point is to start getting intuitive feeling for (first, simple) curves. Later in analytic geometry, they will learn to graph more complex functions and will do so by some process of looking at intercepts, asymptotes, symmetry, blabla, etc. There will really never be a point where you want to push them towards an "integer point". The whole idea is to be able have an intuitive feel for graphs by honing in on key features, versus too fussy precision, at the cost of intuitive understanding. I.e. intercepts uber integer points.

$\endgroup$
1
  • 2
    $\begingroup$ Although you call yourself a troll, your answer is not that bad :-) $\endgroup$
    – Dominique
    Sep 13, 2023 at 12:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.