### Video Transcript

Plot Linear Equations

We know that every single linear equation is in the form π¦ equals ππ₯ plus π, where π is the gradient and π is the π¦-intercept or where it cuts the π¦-axis β cept meaning to cut. Now in this video weβre just going to be able to plot graphs and get them into the form π¦ equals ππ₯ plus π to be able to plot them. So if we want to plot the graph π¦ equals three π₯ minus one or any graph for that matter, the first thing we need is a table of values. Well you have to choose at least three π₯-values. So we are choosing at least three π₯-coordinates. And the reason you choose three is basically just in case we get something wrong. So if you choose two and they give you those two points, we could say okay Iβll draw a line in between them. And here we are, hereβs my graph. But actually what happens if we needed the third one. It was here and that tells you either the first, the second, or the third point is wrong. So by doing at least three, it gives us a way of checking what points actually correct into our calculations. Now I like to choose negative one, zero, and one because it involves the least calculation, but you can choose whichever points you like.

So if we do choose negative one, zero, and one, what we need to do is substitute each of those π₯-coordinates into our function to find the π¦. So in the first case, weβre going to do π¦ equals three multiplied by negative one minus one. Well, three multiplied by negative one is negative three. And then subtracting one, we get negative four. So our first π¦-coordinate is negative four. And substituting our next π₯-coordinate of zero, weβve got three multiplied by zero, which we know is zero. And subtracting one, so zero minus one is minus one. So our next π¦-coordinate is negative one. And then finally three multiplied by one minus one, so three multiplied by one is three and minus one is equal to two.

So now we need a set of axes to be able to plot our points. Now we can see that our table of value leads us to coordinates to be able to plot. For example, the first one, weβve got negative one as the π₯-coordinate and then negative four as the π¦. And the next one is zero as the π₯ and negative one as the π¦. And then finally, weβve got one in the π₯-coordinate and two in the π¦. Now plotting each of these, negative one, negative four, then zero, negative one, and then one, two. And then joining them up, we should get something like this that I prepared earlier.

Now in our next example, we havenβt got it in the form π¦ equals ππ₯ plus π. So weβre going to have to do that first. Plot the graph four π₯ plus two π¦ equals ten. So as we can see that as I just said itβs not in the form π¦ equals ππ₯ plus π, and it must do- first do a table of values. So first we want it to be in that form. So basically what we need to do is rearrange this to get π¦ as the subject. So what weβre gonna do first is subtract four π₯ from both sides. And that will give us just two π¦ on the left and then ten minus four π₯ on the right. And then this is two multiplied by π¦ or the opposite of multiplied or times by is divide by, so you must now divide both sides by two. Be careful as we need to divide every single term by two. On the left-hand side, we will just have π¦ now. Now weβve got five minus two π₯.

Now although this isnβt in the form exactly βπ¦ equals ππ₯ plus π,β weβve got the π¦ by itself and weβve got the ππ₯ and the π on the same side. So as long as itβs roughly in the same form like that, then youβre okay. So now weβve got this; we can do a table of values. And again Iβm gonna choose the same three coordinates as we did last time, so negative one, zero, and one. And substituting them each individually into our function, five minus two multiplied by negative one. So the two negatives will cancel out giving us a positive two. So the five plus two, and we know the answer to five plus two is seven.

And then for the next case where π₯ is equal to zero, so five minus two times zero, well two times zero is zero. So itβs five minus zero, which we all know is just five. And then for the last π₯-coordinate of one, weβll substitute that into the function. So weβve got five minus two times one. Two times one is two; so itβs five minus two and five minus two is three. So there we have our table of values and we know that the table of values leads us to coordinates: the first one being negative one, seven, then zero, five, and one, three.

Finally, we just need to plot the points and join them up with a line. So looking for negative one, seven will be minus [minus one] in the π₯ and seven in the π¦. Then zero, five, zero in the π₯, five in the π¦, and then one, three, so across one and up three. And then joining them up, it gives us our straight line here. So we can see it goes through five on the π¦-axis and it goes down. And this down is related to whatβs in front of the two, being a negative. So this goes on to gradients. So there we have it. To plot linear graphs, what we need is a table of values. But before itβs in a table of values, we must rearrange it to the form π¦ equals ππ₯ plus π. Now once we have our table of values, we have some coordinates. Obviously weβve chosen the π₯-coordinates, but the π¦-coordinates weβve got from substituting the π₯-coordinates into the functions individually. And then these give us the coordinates to plot the points on the graph. And once weβve plotted the points, we get a straight edge or a ruler which I havenβt done here and draw a nice line through all these points, giving us the function.