### Video Transcript

In this video weβre gonna explore the general form of the equation of a
straight line function: π¦ equals ππ₯ plus π. Now depending on where you live, you may have seen this written as π¦
equals ππ₯ plus π. Well, it means just the same thing. But some bright spark
thought itβd be a good idea to use the letter βπβ instead of the letter
βπβ to represent the value of the coordinate of the point that cuts the
π¦-axis. Now you think thatβll it make it easier to remember which value tells
you where the line cuts the π¦-axis β π stands for cut. But after many
years of teaching in places where they use π¦ equals ππ₯ plus π rather than π¦
equals ππ₯ plus π, believe me it doesnβt work. You might as well just use π¦ equals
ππ₯ plus π.
Anyway whichever version of the formula you use, this is the big idea weβre
gonna be looking at in this video.

So letβs look at some graphs first. Letβs plot π¦ equals π₯. Well
π¦ equals π₯ could also be written as π¦ equals one π₯ or π¦ equals one times π₯. So whatever
π₯-coordinate we give, we just multiply it by one and that will
also be the π¦-coordinate. So weβve got a series of points that have exactly the
same π₯- and π¦-coordinates. So for example, if my π₯-coordinate is zero, my π¦-coordinate
would also be zero; if my π₯-coordinate is five, my π¦-coordinate is also
five; or if my π₯-coordinate is negative seven, the π¦-coordinate
would also be negative seven. And that leads us to a whole series of points and loads of
points in between them that look like this on the graph. And when I join them up, they look like this, giving us the graph of the line π¦ equals π₯.

Now letβs plot the line π¦ equals two π₯. And for this one, we had to take the π₯-coordinate and times it
by two to get the π¦-coordinate. So if our π₯-coordinate was zero, weβd multiply that by two and weβd
also get zero. And if the π₯-coordinate was three, weβd multiply that by two and get six
and so on. So our graph would look like this.

And in the same way, we could plot π¦ equals three π₯. Now already I think weβll plot two- weβll plot π¦ equals a half
π₯ and π¦ equals zero π₯. Well clearly π¦ equals zero π₯ is just π¦ equals
zero. So the π¦-coordinate is always gonna be zero in that case. So thereβs π¦ equals a half π₯ and we can see that the
π₯-coordinate is ten at this point here and our π¦-coordinate is five.
So the π¦-coordinate is half of the π₯-coordinate and thatβs true for all of
the coordinates on that line. And thereβs π¦ equals zero π₯, where the π¦-coordinate is
always zero.

So now weβve plotted all those lines together. What you should notice is that
the multiplier of the π₯ tells us how steep that line is gonna be. We got a series
of lines which start off horizontal and get steeper and steeper and steeper as the multiplier
of π₯ gets bigger. Letβs pick a point in one of those lines and increase our
π₯-coordinate by one. Now when we do that, the corresponding π¦-coordinate back on the
line has gone up by a half and look thatβs the multiplier of
π₯. When I look at the line π¦ equals one π₯ if I take a point on the
line, increase the π₯-coordinate by one, the corresponding π¦-coordinate
if I move back to the line has also gone up by one. If I do the same on the line π¦ equals two π₯, increase the
π₯-coordinate by one, the corresponding π¦-coordinate to get back to
the line goes up by two and thatβs the multiplier of π₯.

So for any straight line in this π¦ equals ππ₯ plus π format, so
π¦ equals a number times π₯ plus or minus another number; I mean in our case the
numbers are plus zero, so that π or π number on the end is zero, any line in
that format the multiplier of π₯ tells you about the slope of the line β how steep
it is or how shallow it is: is it horizontal? or is it getting more and more vertical? But the number itself specifically tells you how much will the π¦-coordinate
change if I increase my π₯-coordinate by one at any point on that line.

Right, letβs go through that exercise again. But this time our multipliers of
π₯ are going to be negative numbers. So weβre gonna do zero π₯, negative a half π₯, negative one π₯,
negative two π₯, and negative three π₯. So π¦ equals zero π₯ is still this horizontal line. And for π¦
equals negative a half π₯, the π¦-coordinate is negative a half times the π₯-coordinate.
So for example, when the π₯-coordinate is five, the π¦-coordinate is
negative a half times that. So itβs negative two point five. So you can see this point on
the line here.

Now we can also plot π¦ equals negative one π₯ or just negative
π₯, and π¦ equals negative two π₯, and π¦ equals negative three π₯. Now again π¦ equals zero π₯. We said it was a horizontal line and
youβll see that as this number gets bigger and more negative the line gets steeper. But instead
of going from the bottom left to the top right, going uphill, itβs going in the other direction.
Itβs going from the top left to the bottom right; itβs going where we would call downhill. So the number on the multiplier of π₯ tells us about the slope of
that line, but the sign of that number also gives us some information about the slope. If itβs
negative, itβs going in this downhill direction top left to bottom right. If it was positive,
itβs going in this uphill direction as we increase π₯ from the bottom left up to
the top right.

Now because every time I increase my π₯-coordinate by one on a
given line, the π¦-coordinate always changes by the same amount; thatβs what makes
it a straight line. So on the line π¦ equals negative three π₯ if I start up here and I
increase the π₯-coordinate by one, to get back into that line Iβve got to decrease
the π¦-coordinate by three. And that same thing is true no matter whereabouts on the straight line I
start. And looking at π¦ equals negative a half π₯, the same thing is true.
Every time no matter wherever I start on that line if I increase my π₯-coordinate
by one, the π¦-coordinate decreases by the same amount β negative a half
in this case. So whether itβs from here or itβs here or here or even over here, the slope of
that line is always the same, everywhere on the line.

Well now weβre gonna plot some different lines: π¦ equals π₯ plus zero, π¦ equals π₯ plus one, π₯ plus two,
π₯ plus three, π₯ plus four, and π₯ plus five. And you can do this using some software or you
could do it using tables of values. But Iβm going to do it for you here; what we do is we take
our π₯-coordinate and we add something to it. Now depending on which equation
weβre using, weβll either add one or nothing and two or three or four or five. So if I take my π₯-coordinate and add nothing to it to get my
π¦-coordinate, it means all my π₯- and π¦-coordinates are
the same. So this is the line that we get. If I take my π₯-coordinate and add one
to it all the time, then this is the line I get. Look when my π₯-coordinate is zero,
zero plus one is one, so my π¦-coordinate will be one; so this is the point
Iβm gonna get. If my π₯-coordinate is four, four plus one is five, so five
will be the π¦-coordinate I get. So thatβs the line π¦ equals π₯ plus one.
Now Iβm gonna plot π₯ plus two, π₯ plus three, π₯ plus four, and π₯ plus five. And here are the lines that I get. Now this isnβt massively surprising because look theyβve all got the same
multiplier of π₯. We didnβt have anything in front of it; so itβs all one
times π₯. So the slope of those lines should all be the same. Every time I increase my
π₯-coordinate by one, the π¦-coordinate goes up by one. Theyβre
parallel lines because theyβve all got the same slope of one.

But now letβs look at this other thing: the plus π. Weβre adding
zero, one, two, three, four, and five. π¦ equals π₯ plus zero cuts the π¦-axis here at zero, π¦
equals π₯ plus one cuts the π¦-axis here at one, π¦ equals π₯ plus two
cuts the π¦-axis here at two, and no surprises π¦ equals π₯ plus three, four, and five
cuts the π¦-axis here at three, four, and five. So this second term here, the plus
number on its own or the minus number on its own, will tell us where that line is gonna cut the
π¦-axis.

Okay letβs take a look at some lines then when we are subtracting a number
rather than adding a number on the end. So π¦ equals π₯ plus zero, π₯ minus one,
π₯ minus two, and so on down to π₯ minus five. So this is what those lines look like. Again remember that they all mean one times π₯. So the slope of
the line is one. Theyβve all got the same slope; therefore, theyβre parallel. And the thing thatβs different about all of those lines is the number weβre
adding on the end and that tells us whereabouts it cuts the π¦-axis.

So now we know how equations of straight lines work; we know the rules. We can
easily plot linear graphs from the equations without having to make a table of values first. So we just need first of all to plot a π¦-intercept; itβs minus three, so
thatβs here. And then as I increase my π₯-coordinate, my π¦-coordinate
is increasing by two because Iβve got a positive two in front of the π₯. So as I go
increase my π₯-coordinate by one, my π¦-coordinate is gonna go up by two,
increase the π₯-coordinate by one, the π¦-coordinate goes up by
two. And if I start decreasing my π₯-coordinate by one, the
π¦-coordinate is gonna do the opposite thing; itβs gonna come down by two. So as I
decrease my π₯-coordinate by one, the π¦-coordinate comes down by two,
decrease by one, comes down by two, and so on. Then I just have to join all these up. And hereβs my line. So remember the number in front of the π₯
tells us the slope of the line. If itβs positive, itβs an uphill line. If it was negative, it will
be a downhill line. So thatβs a little check that you need to sort of learn and remember and
the number on the end tells you where it cuts the π¦-axis. So negative three; itβs
cutting the π¦-axis at negative three.

Now letβs plot π¦ equals one and a half π₯ plus two. So here is where
it cuts the π¦-axis. And this multiplier of π₯ tells us the slope of the
graph. Every time I increase the π₯-coordinate by one, the
π¦-coordinate goes up by one and a half. Or to make it easier with whole numbers doubling those, if I increase the
π₯-coordinate by twos β thatβs twice as many, then the π¦-coordinate
goes up by three β thatβs twice as many. So increasing the π₯-coordinate by two, the π¦-coordinate
goes up by three. Increasing the π₯-coordinate by two, the π¦-coordinate
goes up by three, and so on. And likewise coming back, if I decrease the π₯-coordinate by two,
the π¦-coordinate is gonna go down by three, decrease the π₯-coordinate by
two, the π¦-coordinate goes in the opposite direction down by three, and so on. And joining up the points, thereβs our line. The slope was a positive number; the multiplier of π₯ was a
positive number. So we know it should be an uphill line, which it is; so thatβs good. And it cuts
the π¦-axis at positive two, which it does; so thatβs good.

Now letβs plot π¦ equals negative π₯ plus five. So looking at negative π₯, remember that means a negative
one π₯. So our slope is negative one. Every time I increase my π₯-coordinate
by one, my π¦-coordinate goes down by one. Now the number on its own is positive five. So this cuts the π¦-axis
at positive five; the intercept is five. So letβs plot that intercept there. And then every time I increase the
π₯-coordinate by one, the π¦-coordinate goes down by one, increase the
π₯ by one, the π¦-coordinate goes down by one. And letβs carry on with
that pattern to get these points and then go backwards. Every time I decrease my
π₯-coordinate by one, the π¦-coordinate is gonna go up by one; the
opposite of the negative side, itβs gonna go up by one. Decrease my π₯-coordinate by
one, π¦-coordinate goes up by one, and so on. And joining up those points, thereβs my line. Remember cuts the π¦-axis at five yup. And itβs a negative slope, so
thatβs a downhill slope. Every time I increase my π₯-coordinate by one, the
π¦-coordinate decreases by one; thatβs negative one π₯, which matches my
graph.

Now letβs plot π¦ plus π₯ equals two. Well weβve a slight problem
here because this isnβt quite in this right format. Itβs not in our π¦ equals something times π₯ plus another number
format. So what Iβm gonna do is Iβm going to subtract π₯ from both sides of my
equation, which gives me π¦ plus π₯ take away π₯ on the left-hand side and
two take away π₯ on the right-hand side. So if I start off with positive
π₯ and then I take away π₯, those two things are gonna cancel each
other out. So Iβve just got π¦ on the left-hand side of my equation. Now on
the right-hand side, Iβve got two take away π₯; now it doesnβt matter whether I say
two take away π₯ or if I start off with negative π₯ and then I add two
onto it. Remember this two on its own here is really a positive two. And in this format, itβs really easy to recognize that the negative
π₯ means the negative one π₯. So the slope is negative one. And that plus
two on its own here tells us that weβre cutting the π¦-axis at positive two. So I
can put that on the graph. Here we go, cut the π¦-axis at positive two and then every time I
increase my π₯-coordinate by one, the π¦ coordinate decreases by
one. So these are the points youβre gonna generate. And when you join them up thatβs what theyβre gonna look like.

Now itβs also worth noting at this point Iβve been drawing all these little
orange lines and showing you know increase π₯ by one the π¦-coordinate
goes down by one, you donβt actually need to be doing those. So you only need to be plotting the
points. So Iβm just putting them on the page so you can see them. It makes it nice and clear
what weβre doing β how weβre counting this up, but you wouldnβt normally draw those orange lines
in when youβre plotting these graphs. Now the key learning point for this example was the fact that sometimes we
have to rearrange our equation in order to get that π¦ equals ππ₯ plus π format,
which makes it much easier to plot.

Letβs do another couple of examples then. So for this one, Iβve got π₯
plus π¦ plus three equals zero. Iβm gonna have to do a bit more rearranging. So first of all,
Iβm gonna subtract π₯ from both sides. And when I do that, Iβve got π₯ and Iβm taking away π₯
on the left-hand side, so I can cancel those two out. And on the right-hand side, I got
zero take away π₯. Well that is just negative π₯. So Iβve got π¦
plus three is equal to negative π₯, still not quite the right format. So I need to take away
three from both sides. And when I do that over on the left-hand side, Iβve got π¦ add three take away
three. So if I have to do three take away three, thatβs nothing; so these two terms
over here cancel out. So Iβve got π¦ is equal to negative π₯ take away three. Great, thatβs
now in my π¦ equals ππ₯ plus π format. And negative π₯ is the same as negative one π₯, so our
slope is negative one. So cutting the π¦-axis at negative three looks like that. And without doing all the ziggity zaggity lines all over the place, we can see
that increasing our π₯-coordinate decreases our π¦-coordinate by one,
leaves with these points which look like this when I join them up. So thatβs the equation π₯ plus
π¦ plus three equals zero, which I rearranged to π¦ equals negative π₯ take away three in order
to quickly be able to see what the slope was, negative one, and where the π¦-intercept
was, negative three.

So the big lesson of this example is that sometimes we need to rearrange our
equation in order to get it into that π¦ equals ππ₯ plus π format so that we can
easily work out the slope and the intercept. So to summarize then, π¦ equals ππ₯ plus π or ππ₯ plus
π is the general form of the straight line equation. The multiplier π₯ is the slope, which means that when I increase my π₯-coordinate by one, the
π¦-coordinate will increase by whatever that π value is β whether
itβs positive or whether itβs negative. And the π value tells us the π¦-coordinate of the
point on the line that cuts the π¦-axis. If the π value was positive, weβll have an uphill line that looks
like this. And if it was negative, weβll have a downhill line like that. Sometimes we need to rearrange the equation to get the π¦ equals ππ₯ plus
π format and hence the slope and where it cuts the π¦-axis. And lastly remember not all equations rearrange into the π¦ equals ππ₯
plus π format. Not all functions represent straight line graphs.