Lesson video

Hello.

Mr. Broxson here.

Welcome to Maths.

Today, we're checking and securing our understanding of drawing linear graphs.

I'm looking forward to it.

I hope you are too.

Let's get started.

Our learning outcome is I'll be able to draw a linear graph from its equation and relate the features of the graph to its equation.

Lots of keywords you're gonna hear today.

Linear, the relationship between two variables is linear if, when plotted pair of axes, a straight line is formed.

Gradient, the gradient is a measure of how steep a line is.

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

Look out for those keywords throughout our lesson.

Two parts of the lesson today.

We're going to start by plotting linear graphs.

We can plot a linear graph from its equation using a table of values.

For example, y = 2x + 1.

If we populate this table of values, we'll have some coordinates to plot.

We substitute x values into the equation to find the corresponding y values.

I'm gonna start with the easiest x value to substitute in.

That would be x = 0.

When x equals zero, y equals two lots of zero plus one, which is one.

Pop that in the table.

When x equals one, y equals two lots of one plus one.

That's three.

Drop in the table.

When x equals two, y equals five.

And when x equals three, y equals seven.

Pop those in the table.

From here you'll notice something.

It's a linear equation, so the y values follow a linear sequence.

There's a common difference between those y values.

That means there'll be a common difference going this way as well.

So I can take one and subtract two to get to negative one, subtract two to get to negative three, subtract two to get to negative five.

I found the corresponding y values for those negative x values without having to substitute, but it's useful to check our work in mathematics.

I'm just gonna check that when I've substituted in x equals negative three, I should get a y value of negative five.

Two lots of negative three is negative six plus one is negative five.

At that point, I know that I'm correct, which is reassuring.

We can now plot these coordinates.

To draw the graph we'll need an axes.

We'll plot those coordinates.

It's a linear equation so the points form a straight line.

The line through these points represents the linear relationship y = 2x + 1.

We can draw that line.

It's good to appreciate there are infinitely many points on this line and they all satisfy the equation y = 2x + 1.

That point's on the line 1.

54.

When x equals 1.

5, y equals two lots of 1.

5 plus one, which is indeed four.

What about this one? Negative five thirds, negative seven thirds.

Well, when x equals negative five thirds, two lots of negative five thirds plus one is indeed negative seven thirds.

We could keep going, doing point after point after point, and we would go on forever because there are infinitely many points on this line that all satisfy that equation.

Well done to those who you spotted that I have not actually drawn the line y = 2x + 1 yet.

There are infinitely many points on this line beyond the range we can see here.

I've just drawn a line segment.

If I pick this (3.

5, 8) when x equals 3.

5 y equals two lots of 3.

5 plus one, which does indeed equal eight.

That point satisfies our equation.

I could pick infinitely many points beyond that one, which also satisfy this equation.

Therefore, that reason when plotting linear graphs, we draw a line to the edge of our graph.

Quick check you can do this now.

I'd like you to calculate the missing values in this table.

The equation is y = 3x - 4, and there's four missing y values for you to find.

Pause, and do this now.

Welcome back.

See how we did? Hopefully, you started by doing the positive values.

When x equals positive two, y equals three lots of two minus four, which is two.

When x equals three, y equals three lots of three minus four, which is five.

From here you've got a couple of options.

It's a linear graph.

Therefore, the y values are following a linear sequence.

You could use that linear sequence to identify those missing y values or you could use substitution.

Either way works.

As long as you got the coordinates (-1, -7) and (-2, -10), you are correct.

Another check now.

Which point is plotted incorrectly? For the line y = 3x - 4, which is plotted incorrectly? Pause, tell the person next to you, or tell me aloud at the screen.

Welcome back.

You could have gone back and checked that table of values that we just completed and identified it's that coordinate.

(2, 0) is not in our table of values.

I hope you intuitively said, well, it must be that one (2, 0) because it does not align with the other coordinates, y = 3x - 4 is a linear equation.

It will plot a linear graph and that coordinate did not align.

If we plot it correctly and see that it does align, we can draw that line, the y = 3x - 4.

For some equations, a table of values may be inefficient, y = 6 - 1.

25x.

We could calculate all the corresponding y values and they would look like so.

But without a more detailed set of axes than the one we can see here, some of the these coordinates are difficult to plot accurately.

Our number sense should have told us when we saw that equation that we were going to see two integer coordinate pairs in our table of values.

Those two, when x equals zero, y equals six, when x equals four, y equals one.

Because this is a linear graph, we only need two points in order to graph it.

If I plot (0, 6) and (4, 1), I've everything I need to draw the straight line y = 6 - 1.

25x.

We only need two coordinates.

We can see all of these points on our line.

However, we only needed two points in order to draw it.

It's more accurate in this case for us to plot the integer pairs of coordinates.

Quick check you've got that.

If you are plotting this graph, y = 3 - 0.

2x, it'd be useful to have integer value coordinates, so which x values will generate integer y values for this equation? Is it A x = 8, B, x = 5, or C, x = 1, or D, x = -5? Pause, have a think Welcome back.

Let's see how we did.

If you performed all those substitutions, you'd get these values, and you'd notice B and D gave us integer y values.

When x equals five, y equals two.

When x equals negative five, y equals four.

For equations in the form axe + by = c there's a simple method to determine two points on the line.

Here's an equation in that form, 3x + 2y = 12.

And we know we only need two coordinates.

A simple coordinate to find is when x equals zero.

When we substitute that into the equation, three lots of zero ah, wonderful.

That's zero.

Therefore, 2y equals 12.

Y must be equal to six.

We've got our first coordinate.

When x equals zero, y equals six.

We could also find the x value when y equals zero.

If you substitute y equals zero into the equation, two lots of zero.

Wonderful, that's zero.

3x equals 12.

X equals four.

We've got our second coordinate.

When x equals four, y equals zero.

Knowing two points means we can now draw the graph.

There's two coordinates, (0,6) and (4,0).

Straight line through those and now we have it.

The linear graph, 3x + 2y = 12.

Quick check you've got that.

Which two points are on this line, 3x -5y = 15? There's four to choose from.

I'd like you to choose two of them.

Pause, work it out now.

Welcome back.

I hope you said option A because when we substitute in x equals zero and y equals negative three.

It does indeed satisfy the equation.

By contrast, B did not.

If you substitute x equals negative five and y equals zero into the equation, it doesn't equal 15.

It doesn't.

Those values do not satisfy the equation nor do the values in coordinate C.

When x equals zero, y equals three.

Substitute those in.

We do not satisfy the equation.

For D, however, (5,0), absolutely those values satisfy the equation.

We wanted coordinate A and coordinate D.

Once we've plotted those coordinates, we can draw the straight line.

Andeep, is plotting linear graph of y = 7 - 2.

25x.

Andeeep says, "I know that I don't need to do a table of values." And Andeep calculates when x equals zero, y equals seven minus 2.

25 lots of zero.

That's equal to seven.

Good choice to use x equals zero.

And when x equals four, oh, lovely, 2.

25.

0.

25, that's a quarter.

Multiplied by four, we're gonna get an integer value.

Well played, Andeep.

When x equals four, y equals seven minus 2.

25 lots of four, which is two.

Andeep concludes I just plot (0,7) and (4,2).

Plot away.

Draw the line.

Uhoh, Andeep has made an error.

Can you spot it? Pause, pick the error out now.

Welcome back.

Let's see how we did.

Did you notice that it's not quite right? Seven minus 2.

25 lots of four equals two? No it doesn't.

That coordinate is wrong.

Andeep says, "I always check my work using graphing technology." Well done for two things there Andeep.

For always checking your work, that's awesome, and for doing this example with graphing technology.

Andeep's gone to Desmos.

com, clicked on graphing calculator, and created this.

That's awesome.

That's the line y = 7 - 2.

25x.

And Andeep says, "I can go back and correct my work." This is awesome practise.

We're not gonna get everything right first time in mathematics, but when we get something wrong, let's get good at spotting those errors and not be afraid to correct them.

Quick check you've got this, now.

True or false? When using just two points to plot linear graphs, we can use technology to check our work.

Is that true or is it false? Once you've decided, can you pick one of those two statements at the bottom of the page to justify your answer? Pause and do this now.

Welcome back.

I do hope you said true.

We can use technology to check our work.

I hope you justified that with statement A.

Graphing technology like Desmos.

com enables us quickly and easily to check our work.

I like to say use tech to check.

Practise time, now.

Question one.

I'd like you to complete the tables of values and plot these graphs, y = 4x - 1 and y = 4 - 3x.

Pause and do that now.

Question two.

I'd like you to find any two integer value coordinate pairs and plot the graph of this equation.

You're plotting y = 2/3x - 4.

I'd like two integer coordinate pairs that enable you to draw that graph.

Pause, give it a go.

Question three.

Fill in the missing coordinates and plot the graph of each equation.

Each of these equations are in the form axe = by = c, so we're interested in finding the y coordinate when x equals zero, and the x coordinate when y equals zero.

If you fill in those blanks, you'll have enough information to plot all three of those lines.

Pause and do this now.

Feedback time, now.

Question one.

Your table of values y = 4x - 1 should have looked like so.

You can plot those coordinates and you should get that straight line.

Y = 4 - 3x for part B.

Your table of values should have been completed like so, giving you those coordinates and that straight line.

You might wanna pause for a moment.

Just check that your table of values matches mine, your coordinates match mine, and your straight lines look just like mine.

Question two, I asked you to find any two integer value coordinate pairs to plot the graph at this equation, y = 2/3x - 4.

Two thirds, the denominator is the key here.

Two over three.

If we replace x with a multiple of three or zero, we will produce integer value coordinate pairs.

So we could have had (0, -4), (3, -2), (6, 0), (9, 2), or (-3,-6), (-6, -8), (-9, -10).

They were all of the integer value coordinate pairs available on that grid.

You only had to choose two of them.

If you did it accurately, you would've got that line.

For question three, I asked you to fill in the missing coordinates.

Plot the graph of each equation.

For part A, we should have got the coordinates (0, 10) and (2, 0).

When you plot those two coordinates, you get that straight line.

For B, should found the coordinates (0, -10) and (-2, 0).

Those coordinates will be there, and the straight line as so.

For part C, (0, -5), (2, 0), giving you those coordinates and that line.

Again, just pause, check that your coordinates and your graphs match mine.

Onto the second half of the lesson, now.

The gradient and y-intercept.

Completing a table of values can be useful for many reasons.

This is y = 5x - 7.

There's a table of values, and there's a straight line.

A key feature of any linear graph is its gradient.

Gradient.

That's the rate of change the y direction with respect to the positive x direction.

The gradient is visible, not just on our graph, but also in our table of values.

I can see the gradient on the graph.

For every step of positive one in the x direction, I'm stepping positive five in the y direction, and I do it every time.

A change of five in the y direction for every change of positive one in the x direction.

Can you see how that gradient is also visible in our table of values? On the graph, a step of positive one in x sees a step of positive five in y.

In the table of values, a step of positive one in x sees a step of positive five in y.

That table of values is really useful for helping me to spot that gradient.

Another key feature of a linear graph is it's y-intercept.

That's where the line cuts the y axis.

It's not enough to say negative seven, so our y-intercept here is (0, -7) We can see this in our table of values where x equals zero.

Oh look, there it is, (0, -7).

When we intercept the y axis, were always going to have an x coordinate of zero, so we could have picked that y-intercept outta that table of values before we'd even drawn the graph.

These two features are also visible in the equation.

We don't even need the graph or the table of values, as long as that equation is in the form y = mx + c, as this one is, y = 5x - 7.

The key feature is the gradient of five.

That's the coefficient of x.

The y-intercept at (0, -7).

That is the constant.

This is really important that you get this.

You might wanna copy this down or take a screenshot.

This works for any equation in the form y = mx + c.

Y = 6x + 9.

We could fill in the table of values.

We could plot the graph.

Or we can just say that's the equation.

Y = 6x + 9.

I'm gonna get a gradient of six.

That's the coefficient of x, six.

I can see that in my table of values.

I can see it on my graph.

We're also going to get a y-intercept at (0, 9).

That was the constant in our equation.

We can see it in the table of values, and we can see it on the graph.

Quick check you've been following.

What will the gradient of the line for this equation be? The equation, y = 2x -3.

Is that gonna be two, three, or minus two? Pause, take your pick.

Welcome back.

Well done for saying it'll be a gradient of two.

When we have the equation in the form y = mx + c, it's the coefficient of x, which determines the gradient.

Our coefficient of x in this case was two.

Two will be the gradient.

There's the graph.

There, you can see that gradient of two.

For every step of positive one in the x direction we're stepping positive two in the y direction constantly.

Another check, what will the y-intercept of the line for this equation be? Y = 2x - 3.

What's the wine intercept? Three choices.

Pause, take your pick.

Welcome back, and well done for saying it's option C, (0, -3).

We know it from looking at the equation because negative three is the constant.

When we plot that graph, we can see the y-intercept is zero negative three.

Laura has made an error when interpreting the gradient and y-intercept of this equation.

I say, okay Laura, we're going to make errors.

Let's see if we can help Laura understand hers here.

The equation is y = 7 - 4x, and Laura has said, "I think this line will have a gradient of seven and a y-intercept of (0, -4)." Can you spot Laura's error and correct her statement? Pause, see if you can figure this one out, now.

Welcome back.

How do we get on? Let's find out.

The constant in this equation is seven, therefore the y-intercept is actually (0, 7).

The seven wasn't the gradient because it's the constant, and that's what tells us our y-intercept or rather the y coordinate of our y-intercept.

The coefficient of x in this equation is negative four, therefore the gradient is negative four.

What's happened here is the equation is written ever so slightly differently.

Rearranging from y = 7 - 4x to y = -4x + 7 puts it back in the form y = mx + c.

It's bit easier when you have it written as y = -4x + 7 to see, oh, of course, the x coefficient is negative four.

The constant is positive seven, but we have to be aware.

We're also gonna see this in the form y = c - mx.

Now, that we've correctly identified the y-intercept and we know the gradient to be negative four, we can draw this equation.

There we go.

The y-intercept, (0, 7), a gradient of negative four.

For every step of positive one in the x direction, we step negative four in the y direction.

Laura says, "Ah, a gradient of negative four and a y-intercept of (0, 7).

Thank you!" You are welcome, Laura.

When graphing it's important we pay attention to the scale on our axes.

Y = 1.

5x + 2, and here's Laura again, and she says, "I'm confused! I was expecting a gradient of 1.

5 for this equation, but instead I have a gradient of three." Can you see Laura's error this time? Pause.

See if you can spot what's going on here.

Welcome back.

Did you spot it? Laura has.

"Got it! The scale on the two axes is different!" Well spotted Laura.

When we take a closer look, you'll see that the x axis is moving in steps of two, whereas two on the y axis looks like that, the scale on the y axis is moving in steps of one.

We need to pay particular attention to scale because it can be misleading.

When we look at the gradient now, we can see that it is indeed 1.

5, and it's constantly 1.

5.

When in the form y = mx + c, the coefficient of x is the gradient, and it won't always be an integer.

In this case, a gradient of 1.

5.

We won't always have the equation in the form y = mx + c.

Laura and Alex are discussing the line 4x + 5y = 11.

Alex says, "For linear equations, the gradient is the coefficient of x, so this equation has a gradient of four." And I can see where Alex is coming from, but Laura says "This is why we discuss our thinking.

The gradient of this line is negative four fifths." Laura's correct.

Can you see why? Can you see where she got that gradient from? Pause and have a think.

Welcome back.

Let's get to the bottom of what's happening here.

Laura's got a great idea.

She tells Alex, "You need to rearrange to make y the subject of the equation in order to see the gradient." Let's write out our equation.

Let's rearrange.

Add negative 4x to both sides, and we have it in the form 5y = 11 - 4x.

Divide every term by five.

Y = 11/5 - 4/5x.

Wonderful.

This is the form we want.

We can see the gradient here.

Alex says, "Outstanding.

We now have an x coefficient of negative four fifths.

You will write, Laura." It's really important for equations in the form axe + by = c that we can rearrange to reveal the gradient and the y-intercept.

Gradient of negative four fifths, and a y-intercept of (0, 11/5).

Quick check you've got that.

I'd like you to rearrange this equation.

The equation 7x + 6y = 5.

Once you've rearranged it, I'd like you to find the gradient and the y-intercept.

Pause.

Do that now.

Back in a moment with the answers.

Welcome back.

See how we did.

Hopefully, you added negative 7x to both sides to give you 6y = 5 - 7x, and then divided every turn by six.

The equation y = 5/6 - 7/6x.

That gives us a gradient negative seven over six, and a y-intercept of (0, 5/6).

Practise time, now.

Question one.

I'd like you to complete this table.

The left hand column is the equation of the line.

The central column is where we declare the gradient.

The right hand column is the y-intercept.

Pause, fill in these blanks.

Question two, it's a lot like question one.

It's a table.

I'd like you to complete the table.

The added complexity this time is not only do we have the equation in the form of y = mx + c, we've got the left hand column, which is the equation in the form axe + by = 6.

A couple of these are quite tricky, but I'm sure you can figure them out.

Pause, give it a go.

Good luck.

Welcome back.

Let's see how we did.

For question one, your table should now look like this.

It's very busy, so let me show you.

You should have filled in those blanks.

Gradient of four in the first two rows, and y-intercepts of (0, 7) and (0, -7) there.

In the third row, you should have filled in the equation y = 7 - 4x.

In the fourth and fifth rows.

You should have those gradients, and those y-intercepts.

And in the final row, the equation was y = -0.

7x - 0.

4.

For question two, you should have filled in the gradient and y-intercepts in the first two rows.

In the third row, you should have taken a gradient and y-intercept and turned it into those two equations.

Fourth row is a little bit trickier.

If you saw the equation in the form y = mx + c, and multiplied every term by five, you could have found that form, and written it as 3x + 5y = 8, and then you also have to fill in the gradient and the y-intercept and in the final row you needed to fill in those two equations.

It's the end of lesson, now.

We've learned we can draw linear graph from its equation.

Whether the equation is in the form y = mx + c or axe + by = c.

We can use a table of values to plot a graph, but for linear equations only two points are necessary, provided we check our work.

We can relate the key features of gradient and y-intercept to the equation.

For example, the line y = 17x + 15 has a gradient of 17 and a y-intercept at (0, 15).

I hope you've enjoyed this lesson as much as I have, and I shall see you again soon for more maths.

Goodbye for now.