PAULA KELLY: OK. So we’re going to have a look at how we could use our function machines to give us some coordinates. So we’ll start off with our function machine. We’ll have our multiplied by 3.
And our input– any number could come in. Whatever number we have, we multiply it by 3. So we have 3x.
MICHAEL ANDERSON: So if we started with, say, an input of negative 2, multiply that by 3. That would be negative 6.
PAULA KELLY: Lovely. Very good. Yeah. What would be a sensible choice to have next, do you think?
MICHAEL ANDERSON: Well, let’s go to negative 1.
PAULA KELLY: OK. We’ll keep our pattern.
MICHAEL ANDERSON: And that output would be negative 3.
PAULA KELLY: Yeah. A good one– if we have 0.
MICHAEL ANDERSON: Well, 0 times 3, three lots of nothing, is also 0.
PAULA KELLY: Very good. And then we can spot we’re getting a pattern here.
MICHAEL ANDERSON: Oh, we’re increasing by 3 each time, just like the 3 times table. So if we went 1 as an input, the output would be 3. 2 goes in, and 6 comes out. And then maybe 3 and 9.
PAULA KELLY: And then we could continue like this forever and ever.
MICHAEL ANDERSON: Could we put any kind of decimals in? So if we put 0.5 in as the input.
PAULA KELLY: Absolutely. That would work.
MICHAEL ANDERSON: And three lots of 0.5 is 1 and 1/2, 1.5.
MICHAEL ANDERSON: And I suppose we could put 100 in again.
PAULA KELLY: Any number at all. Anything at all.
MICHAEL ANDERSON: And 100 times 3 would give us 300.
PAULA KELLY: So we’ve got now lots of our inputs and outputs. These are really useful now because we can use them– rather than inputs and outputs, we’re going to use them as some coordinates.
MICHAEL ANDERSON: So we’ve got some pre-drawn axes here. This is the horizontal one, which we’ve labelled x, which is going to be our inputs. And we’ve labelled the vertical axis y, which would be our outputs, which are those 3x’s, the things that we’ve times by 3.
PAULA KELLY: Lovely. So if we’re going to draw this function on this graph, we need to know where to put these coordinates. So we got 10 squares, and we could table these as 1, 2, 3, 4.
MICHAEL ANDERSON: Yeah. So if this is the origin here, that’s 0, 0.
MICHAEL ANDERSON: So for every one square we go across, that’s one unit. So I could say, well, here, that’s 1, 2, 3, 4, 5, and keep going. That would be 6, 7, 8, 9, and then we’d get up to 10. I’m going to save myself some writing and maybe not write them all in. And then we’re going to go the same unit up, so that would be 1, 2, 3, 4, 5 here. And then 6, 7, 8, 9, 10 here. What about down at the bottom or that way as well?
PAULA KELLY: So same again. Like a thermometer, it would go into our negative values. We could just label our negative 5.
MICHAEL ANDERSON: So we keep on counting downwards, so that would be 0, negative 1, negative 2, negative 3, negative 4, negative 5. And negative 6, 7, 8, 9, negative 10. And same for this way, so 1, 2, 3, 4, 5, but that is going to be negative. And then I presume that is negative 10 as well. Let’s just check. 6, 7, 8, 9, 10.
PAULA KELLY: Lovely. So we could have labelled these. We could have had each of these squares being a value of 2 and go up to 20. In this case, we’ve got the same scale for our horizontal axis as we have for our vertical. We wouldn’t necessarily need to have that.
MICHAEL ANDERSON: So I counted one every time, but we could have quite easily gone 2, 4, 6, 8, with each square being worth two units, as long as that’s kind of consistent. And we could have done any similar scale on the y-axis as well.
PAULA KELLY: Yeah. It does need to be consistent with your horizontal. Your vertical– as long as it’s consistent within the axis, these could be worth 10 units each, if you wanted to.
MICHAEL ANDERSON: Yeah, but so long as the spaces are even because I know a lot of my students get slightly confused when drawing and labelling these axes.
PAULA KELLY: Yeah. And also temptations to label the middle of the square, but quite rightly you’ve labelled your axes here. So let’s put some of these coordinates onto our graph. So we’ve got negative 2 and negative 6. So we go along the corridor to negative 2 add up to– down rather to negative 6. Very good. Again, we’re going to put a cross rather than a spot because the most accurate point is right in the middle of that cross.
MICHAEL ANDERSON: So these values are telling us where we’re moving from the origin, either left or right. And these values are going to tell us whether we go up or down. So we go negative 2 across and then down to this negative 6.
PAULA KELLY: Perfect. So similarly, along to negative 1, down to negative 3. I’m going to put a cross. 0, 0 is our origin. And we can see we’re getting a start of a pattern here.
MICHAEL ANDERSON: Yeah. They kind of look to be in a straight line. I don’t know whether that’s a coincidence or not. So maybe plot a few more points and see.
PAULA KELLY: Absolutely. So we’ve got our 1 and 3, so along to 1, up to 3. 2 and 6. Along to 2, up to 6. 3 and 9. And go up to 9.
MICHAEL ANDERSON: Yeah, they definitely look to be on that straight line. So for our more trickier examples, 0.5 went into the function machine, and 1.5 came out. I suppose that means we have to go half a square across, which would be just about there, and then 1 and 1/2 squares up. So I’ll try and label that one. If we go 1/2 square up and 1 and 1/2 squares there, somewhere there. And even though it’s a decimal, it looks to be lying on that line that we’ve kind of mapped out there.
PAULA KELLY: Yeah, so it should be obviously on our line. And we’ll see what the rule of it is. As you were saying, it’s difficult to be absolutely accurate.
MICHAEL ANDERSON: As long as it’s estimated accurately.
PAULA KELLY: So yes, it’s a rough estimate. When we come to our 100 and 300–
PAULA KELLY: –we might struggle.
MICHAEL ANDERSON: Should we leave that one because it’s off the scale?
PAULA KELLY: Off the scale. But if we continued that line–
MICHAEL ANDERSON: So it’s somewhere over there.
PAULA KELLY: –it would go on. So we’re going to draw a line through all of these coordinates. If we’ve put them perfectly on our grid, it should go through the centre of all those crosses. I’m going to extend the line because really this line will continue infinitely. And then just so we know what line we’ve actually drawn, we need to label it with our function. So we’re going to call our function to find our output, our y value– we’ve got three lots of whatever our x value is.
MICHAEL ANDERSON: So that is the line y equals 3x.
MICHAEL ANDERSON: And interestingly, I’ve noticed that it goes through this point here. And when I read off the values, that is when our x value is negative 3, the y value is negative 9, which kind of makes sense because if negative 3 goes in, we times it by 3, and we get negative 9 out as a y value. So for all of these coordinate points, wherever we picked a point on the line, the y value is going to be 3 times larger than the x value.
PAULA KELLY: Absolutely. We could think about it inversely. Whatever our x value is, it’s going to be 1/3 of our y value.
MICHAEL ANDERSON: Yeah, that makes sense.