### 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.