Lesson video

Well done for making the decision to learn with this video today.

My name is Ms. Davis and I'm gonna be helping you as you work through this lesson.

Let's get started then.

Welcome today's lesson, we're gonna be checking and securing your understanding of finding the equation of a line from a graph.

A couple of key words that we're gonna use today.

So we're gonna talk loads about the gradient.

So the gradient is a measure of how steep a line is and it's calculated by finding the rate of change in the Y direction with respect to the positive X direction.

And we're gonna review how to find that gradient today.

And intercept is a coordinate where the line or curve meets a given axis.

So we might say the Y intercept, that's a coordinate where the graph would meet the Y axis, or we might say the X intercept.

That would be the coordinate of the point where the graph meets the X axis.

So let's start then by reminding ourselves how to calculate the gradient.

Aisha has plotted four graphs with linear equations.

I'd like you to pause the video and have a think, what is the same and what is different about her four graphs? Let's see if you can use some mathematical language.

Think about how you would describe this to somebody else.

Off you go.

So your answers may have included that all her graphs are straight lines.

You might have said all her graphs are linear graphs.

You might have said they all passed through the origin or they all have the same Y intercept or they all have a Y intercept of zero, zero.

You might have said that as the X values increase, so do the Y values or they all have positive gradient.

Some of the things you might have said as the differences between the graphs, well they all have a different steepness or a different rate of change.

You may have even said that they have different gradients.

So it's useful to refer to the rate of change in all graphs in a similar way, and that's why we have this term gradient.

So gradient is a measure of rate of change and it's calculated by finding the change in the Y direction with respect to the positive X direction.

So if we look at this example here, for every increase of one in the positive X direction, Y increases by two.

We can draw that on our graph with a triangle.

So if we increase one in the X direction, the Y direction increases by two and that will carry on along this line.

So what Izzy has said there is she's described the gradient because she's referring to an increase of one in the positive X direction.

Therefore the gradient of the line is two.

Lines have a positive gradient if X increases, Y also increases.

So these graphs all have a positive gradient because Y increases as the graph moves to the right.

In contrast, lines have a negative gradient.

If as X increases, Y decreases, and this is what this will look like on a graph.

So a quick check as to whether we're happy with the difference between a positive and a negative gradient.

So which of these lines have a positive gradient and which have a negative gradient? Give that a go and then we'll check our answers.

B is gonna have a positive gradient.

That green line, E is also gonna have a positive gradient.

That purple line, it is steeper but it is still positive.

Both those are gonna be positive.

For negative C, the pink line will have a negative gradient.

You can see how the Y values are decreasing as the X values increase.

And D, that black line's also got a negative gradient.

It's a very steep line that one, but you can also see that as X increases, Y decreases.

Now the only one we haven't mentioned is A.

A, that blue line is a horizontal line, so it has a gradient of zero.

So we can calculate the gradient from any graph.

If we have a look at this graph here, in order to calculate the gradient, we need to pick a point on the line.

We need to make sure we are picking either an integer coordinate or definitely a coordinate where two grid lines intersect.

It's often easiest to start with the Y intercept, which is what I've done here.

So the Y intercept is zero, negative one.

So I've picked point.

Now if you draw an arrow to show a step of one in the positive X direction and then we need to draw an arrow vertically until we reach the line again.

And then if you have a look at the values on your Y axis and count up, that's one, two, three.

So our X value has changed by one and our Y value has changed by three.

The gradient of our line then is three.

Let's have a look at this one.

We can see already that it's gonna have a negative gradient.

We can calculate negative gradients in the same way.

So pick any coordinate on the line.

Again, I've chosen the yin set here 'cause that's zero, five and draw an arrow to show a step of one in the positive X direction.

Right, this time I cannot go up to reach my line again, that won't work.

So instead I need to draw an arrow vertically downwards.

I can then see that when X increases by one, Y decreases by one.

So the change in X is one and the change in Y is negative one.

The gradient therefore is negative one.

Laura drew the triangle like this to help her work out the gradient.

How can Laura use her triangle to calculate the gradient? What do you think? Well to start with, it does not matter where on your graph you draw your triangle.

Also doesn't matter what size you draw your triangle, as long as it joins two coordinates that sit perfectly where two grid lines meet.

So we've got positive five change in the X direction and then negative five change in the Y direction.

So we put that into our ratio table for a change in X of five, we have a change in Y of negative five.

So if I want to find out the gradients, I need to work out what the change in Y is for every one increase in X.

So what I need to do is divide my values by five.

So the change in X is one, then the change in Y is negative one.

Using ratio tables could be really useful when a gradient is a non integer.

So let's have a look at this example.

If we have an increase of one in the X direction, you can see we have a non intercept value for our gradient.

Without more grid lines, it's really hard to see the exact amount that Y has increased by.

We know it's a value between zero and one, but it's hard to be exactly sure how much without looking further along the graph.

So that's what we're gonna do.

Instead, let's look at an increase of two in the X direction.

That then allows me to draw an increase of one in the Y direction.

If I put that into my ratio table then, when the change in X is two, the change in Y is one.

Just like before I need to divide my change in X so that it becomes one.

So we need to do two divided by two to get one.

Therefore we need to do one divided by two.

And of course that's a half.

The gradient of the line then is a half.

You can use decimals for the gradient of lines, however, fractions are easier to manipulate later on.

And if you had fractions such as 5/3, it's gonna be easier to leave that as a fraction than to try and turn it into a decimal.

The other thing it's important to look for is the scales on the axes.

Pause the video, what do you notice about the scales on these axes? Hopefully you spotted that the X axis has a step of two, so we can't just count the squares to find out the gradient because horizontally one square is worth two, but vertically a square is worth one.

We're gonna calculate the gradient exactly the same way as before.

So we need to pick a point where two grid lines meet.

Now I need to look at my axes to calculate my change in X.

So we've got two, four, six, So we've got an increase of six and then a decrease of one, two, three, four.

Now use my ratio table as before as X increases by six, Y decreases by four.

If I divide them both by six for every one increase in the X direction, Y decreases by four over six or 2/3.

Always good practise to write our fractions in simplest form, time for you to have a try yourself.

Which of these is the correct gradient of the line? I'd like you to see if you can explain why.

Well done if you spotted that it was five, you need to look carefully at the scales this time.

If we look at the coordinate two, zero and then the coordinate four, 10, we can see that's an increase of two in the X direction and an increase of 10 in the Y direction.

Then if we do 10 divided by two, we can see that for every increase of one in the X direction, Y increases by five.

Have a go at this one.

Which of these is the correct gradient? And again, can you explain why? Well done if you've got negative 4/3, straight away we can see it's a negative gradient.

So you should have been able to rule out C and D.

Then if we pick a couple of points, which are where two grid lines meet, so I've got zero, two and three, negative two, that's an increase of three in the X direction and a decrease of four, I put that into a ratio table.

Divide both sides by three.

I get negative 4/3.

Right, time to have a practise then.

Five lines have been drawn on this set of axes.

We don't have a known scale.

I'd like you to match each line to its gradient.

So what you're gonna need to do is you're gonna need to think about which ones are positive and negative.

Then you're gonna have to think about which lines are gonna be the steepest and so on.

Give that one a go.

So Laura is volunteering at a homework club.

She's marking three pupils' work.

What mistake have each of them made and can you think of some advice that Laura could give each pupil to help them progress? Give this a go.

Well done, those are all really common mistakes.

So I'd like you to think about those as you're trying to find the gradient of these lines.

There's space to draw a ratio table if you need to.

Off you go, there's three more for you to try this time.

I do recommend drawing that ratio table and look carefully at the scales on the axes.

Give these three a go.

I have three more for you to try.

You're getting really good at these now.

So make sure you're stretching yourself and you're really trying to get these trickier ones correct.

Off you go.

Superb work, let's see how we did.

The blue line, C has a gradient of negative three.

The black line E had a gradient of four.

The green line D was the one with a gradient of five.

The purple line or A had a gradient of negative 1/3 and the pink line which was B, had a gradient of negative 3/4.

Check your answers off the screen if you missed any of those.

What happened for the first question, they didn't mark a movement of one in the positive X direction and the reason why that happened is that they didn't pick points where the grid lines clearly met, often they're points with integer coordinates.

So for B, this person has divided the values incorrectly.

What might help is using a ratio table and then we can clearly see that there's a change in X of two.

A change in Y of one, dividing both by two gives you a gradient of 1/2.

The correct gradients then should be three for that first one and 1/2 for that second one.

Okay, the final one, the mistake here is that the gradient should be negative.

We can see that straight away.

Where this pupil has gone wrong is they've drawn their triangle the wrong way.

It needs to show a movement of one in the positive X direction.

Equally, they haven't looked at the scale.

If you look at the Y axis, it has a step of two.

So what we should have, if we use a ratio table, we should have a change of X as one and a change in Y is negative four.

Looking at that Y-axis scale.

So now you've pulled apart all those common mistakes.

Let's hope that we haven't made any when we try to do it ourselves.

So you should have got a gradient of two, negative four and three.

I've used ratio tables here to help me.

If I look clearly at the axes for D, an increase of one in the X direction is an increase of 10 in the Y direction.

So gradient 10, for E, an increase of 10 in the X direction is a decrease of 10 in the Y direction.

So that's a gradient of negative one.

And then if we look at the two marked coordinates to help us, a change in X of 10 is a change in Y is 30.

So we have a gradient of three.

For G, H and I, we'll do them step by step.

So for G, a change in X of two is a change in Y is negative one.

So we've got a gradient of negative a half.

For H, if we increase X by 10, Y increases by four.

That gives us a grey agent of 4/10 or 2/5 or 0.

4.

And for I, if we look carefully at the scales, when we increase by 0.

8 in the X direction, that's a decrease of one in the wide direction.

So to get from 0.

8 to one, we can divide by 0.

8 or divide by 4/5.

Negative one divided by 4/5 is negative 5/4.

That last one was a little bit tricky, so well done if you've got that one.

Right, we're experts now at calculating the gradient.

So we're gonna use that to find the equation of a line.

Equations and straight lines can be written in a variety of ways.

One form that you've probably seen it in before is Y equals MX plus C.

When equations of the line are written in this form, the gradient and the Y are easy to identify.

The coefficient of x, which is represented by M, is the gradient of the line.

The constant C is the Y intercept.

So if we know the gradient of the Y intercept, we can write an equation for the line.

So here is a line, in order to find the equation, we need the gradient and the Y intercept.

So we can see that this has a gradient of three and a Y intercept of zero, negative five.

So the equation would be Y equals 3X minus five.

Is it possible to draw a different line with a gradient to three and a Y intercept of negative five? What do you think? No it's not, this graph is uniquely described by its gradient and its Y intercept.

Any equation of aligned with the same features will actually be an equivalent equation.

So for example, if you drew 3X minus Y equals five, that's the exact same line.

It has a gradient of three and a Y intercept of negative five.

It's just a rearrangement of the equation Y equals 3X minus five that we've got above.

So we can work out the equation of a line from any graph.

Let's do one together.

First we need to calculate the gradient.

So for an increase of one, our Y values decrease by two.

So our gradient is negative two.

The graph of Y equal negative 2X looks like that.

Our line is a translation of the graph four squares up.

So it's Y equals negative 2X but it's been translated four squares up.

So it must be y equals negative 2X plus four.

Remember that constant in our equation is the Y incept.

As long as we're writing in the form Y equals MX plus C.

Of course, that can be written as Y equals negative 2X plus four or Y equals four minus 2X.

It doesn't matter which way round we write the terms. And then what's really great about this is we can check our equation is correct by substituting some known coordinates.

So we can see that two zero should be on our line.

So we can check, is the Y coordinate, X coordinate multiplied by negative two plus four? Yes, it is, if we wanted to we could try another one, is our Y coordinate the X coordinate multiplied by negative two, add four.

Yes it is.

So we can see that that equation is gonna work for the coordinates on our line.

We've seen before that it's important to check the scales when reading from graphs.

So let's try this one together.

In order to get to another point with an integer coordinate, I'm gonna need to go three in the positive X direction and then look carefully at your Y axis.

That's minus eight.

So the gradient is gonna be calculated by doing negative eight divided by three.

So change in Y divided by the change in X.

We can see that in our ratio table, so we can write that as negative eight over three.

We definitely wouldn't wanna keep that as a fraction.

We can see from our graph that the Y intercept is zero, 10.

The equation of our line is Y equals negative eight over 3X plus 10.

We can rearrange our equation of the straight line to write it without any fractional terms. This is what Jacob would like us to do.

You can multiply each term by three so that none of the terms have a fractional coefficient.

So let's try that.

3Y equals negative eight X plus 30.

And then you can see that Jacob also wants both of his variables on the left hand side.

So if we add 8X to both sides, 3Y plus 8X equals 30.

So we're just practising our rearranging skills so that we can write these equations in any form we want.

So what we can do now is we can check that we've got the right equation by substituting known points.

There are a few different coordinates on there we can try.

I've picked three, two and six, negative six.

So first let's check our first equation.

Y equals negative eight over 3X plus 10.

So if I substitute Y for two and X for three, I can check that works and two equals negative eight plus 10.

That's true, let's try our other coordinate.

Negative six equals negative 16 plus 10.

That is true.

So that first equation is correct and let's just check our rearrangement as well.

So three lots of two plus eight lots of three.

Is that 30? Yes it is, and three lots of negative six plus eight lots of six.

Is that 30? Yes it is.

So that second one is correct as well.

So true or false, this line has equation Y equals eight minus four X.

What do you think? Can you justify your answer? That one is true, we have a gradient of negative four.

Look carefully at the Y axis.

It has a step of two, have a gradient of negative four and a Y in step of zero, eight.

How about this one? This line has equation Y equals negative 2X plus two.

True or false? Think about your justification.

It is false because the gradient of this line is clearly positive.

So the coefficient of X should also be positive.

Which of these equations is equivalent to Y equals three minus a half X? So think about your rearranging skills.

Which of these are equivalent, off you go.

In fact, the only one that's equivalent is X plus 2Y equals six.

We multiply all the terms by two and then add X to both sides.

We get X plus 2Y equals six.

Time to have a practise then, I'd like you to match each graph to the correct equation.

Off you go, well done.

I'd like you to work out the equation of each line in the form Y equals MX plus C.

Think about calculating the gradient and the Y concept to help you, give that one a go.

Okay, a few more for you to try.

Again, I want our equations in the form Y equals MX plus C.

Take care to look at the scales on your axes.

Feel free to use ratio tables to help you with your gradient as well.

Off you go.

Okay, we've got a few slightly more challenging ones here to test you out, exactly the same as what we did before.

Just be really careful about your gradients, off you go.

And finally, I'd like you to work out the equation of this line, but I'd like you to write it in the form AXE plus BY equals C where A, B and C are integers.

And then I'd like to show me that you know how to check your equation is correct by substituting two points which are on the line.

You might have to pay particular attention to the smaller grid lines.

Think about what each little square is representing in the X direction and in the Y direction 'cause they might be different.

Give that one a go.

Let's have a look then.

So pause the video and check that you've matched the right graph to the right equation.

Particularly pay attention to your negative and positive gradients.

Make sure you've got your quarters and your negative quarters the right way round.

Okay, so this first one, you should have the equation Y equals X plus four.

So it's a grading of one and a Y intercept of four.

For B, Y equals negative 2X.

It doesn't have a constant because the Y is zero, zero.

If you wrote Y equal negative 2X plus zero, there's nothing wrong with that.

It's just not the simplest form.

And C, you've got Y equals 3X minus five.

Then we've got a little bit trickier, for D, we've got a gradient of negative half, so Y equals negative a half X minus two.

For E, you have to pay attention to your scales.

So we've got an increase of five on the Y axis.

Our equation is Y equals 5X minus five and F, you can see as our graph increases by five in the X direction, it decreases by 10 in the Y direction.

So we've got Y equals negative 2X plus 10.

And then for G, again pay attention to the scales.

Our X axis is increasing by two.

Our equation then is y equals negative a quarter X plus one, for H, we've got an easy scale here, X and Y are both increasing by one, which is helpful.

We might wanna use a ratio table to help us get our gradient as X increases by two, Y increases by three, so that's a gradient of 3/2.

So Y equals three over 2X minus three.

And finally we've got coordinates of zero, negative two and one, four.

That's gonna be the easiest ones to use here.

We can see then that as X increases by one, Y increases by six.

So the equation is Y equals 6X minus two.

Okay, well done for giving this one a good go.

Let's see how close we got.

Now we're gonna need to use the smaller grid lines to help us here.

I picked an increase of two in the X direction and then if I use the small grid lines, I can see that's an increase of five in the Y direction.

If you didn't want to use the smaller grid lines, you can see that there's two easy coordinates, negative two, zero and the other one at six, 20.

If you use that, you can get an increase of eight in the X direction, which is an increase of 20 in the Y direction.

Either way, you get a gradient of 5/2 if you simplify your fraction.

We want it in the form AXE plus BY equals C.

So if you multiply everything by two, you get 2Y equals 5X plus 10.

Then rearrange that you can have 2Y minus 5X equals 10 or negative 5X plus 2Y equals 10.

It doesn't really matter which way around you write those terms. And then we want to check our equation is correct by substituting two points.

As we've just said, the easiest ones might have been negative two, zero and six, 20, but you could have picked any other point that you could find on that line using the smaller grid lines.

So substitute your coordinates in and show that you get a balanced equation.

If you want to read through my working there, just pause the video.

Fantastic work then.

So we've really checked that we know how to find the equation of the line and we've challenged ourselves today by looking at some trickier gradients and thought about some common mistakes and how we can spot those in our working.

Thank you for joining us today and I hope you can now use those skills when you are looking at linear graphs and that you'll join us again to have a look at some more algebra.