Lesson video

Hello, Mr. Robson here.

Welcome to Maths: Problem solving with non-linear graphs today.

I love problem solving.

I know that you love problem solving, so let's get solving problems. Our learning outcome is I'll be able to use our knowledge of non-linear graphs to solve problems. Key words that we'll see throughout this lesson: linear, quadratic, and cubic.

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

A quadratic is an equation, graph, or sequence whereby the highest exponent to the variable is 2.

For example, x squared + x is a quadratic, but x cubed + x is not.

A cubic is an equation, graph, or sequence whereby the highest exponent of the variable is 3.

The general form for a cubic is axe cubed + bx squared + cx + d.

Two parts to our learning today, we're gonna begin by looking at the shape of a graph.

In mathematics, graphs are used to model many real-life situations.

Therefore, it's really useful to be able to look at a graph and recognise the key features.

Doing so can help us understand the nature of the equation behind it.

In this example, this graph is linear.

That means there's a constant rate of change between the two variables.

It's also intercepts both axes at the origin.

Therefore, it's also a graph showing direct proportion.

What do we mean by that? We can pick any pair of coordinate pairs.

And we'll see constant multiplicative relationships between them.

That's what happens on a graph of direct proportion.

This graph is also linear, but it's not an entirely straight line.

There is a constant rate of change, which then changes to a new constant rate of change.

If we were to label the axes as d for distance and t for time, this moment would represent an increase in speed.

We also use linear graphs for abstract modelling.

This means including negative values and using all four quadrants of the coordinate grid.

For example, this graph y = 2x.

This is the graph y = 2x and we see it in the first and third quadrants intercepting the y-axis at the origin.

The graph of y = 2x - 3 has a different y-intercept and is in three quadrants of the grid, but its linear shape has not changed.

We can see that the graphs of y = 2x and y = x squared will be different before we even draw their graphs.

There's a table of values for y = 2x and a table of values for y = x squared.

Oh, look, do you see how I noticed they're going to be different shapes? y = 2x has a constant rate of change, therefore it's going to be linear in shape.

y = x squared, however, does not have a constant rate of change.

Therefore, that graph must be non-linear.

In the case of y = 2x, the y values are constantly increasing.

In the case of y = x squared, however, the y values are decreasing, then they start increasing.

There will be a turning point on that graph.

I know that before I've even plotted it.

Both graphs have a y-intercept at 0, 0.

We can see that in the table of values.

However, y = x squared has only positive y values excluding 0, 0 of course.

The fact that the square of a negative value is positive has a dramatic effect on the shape of the graph, y = x squared.

There are the two graphs: y = 2x and y = x squared.

In the case of y = 2x, a linear equation, therefore a linear graph.

In the case of y = x squared, we had a quadratic equation.

The exponent of x there is two and that's the highest exponent in the equation, so it's a quadratic, therefore it's parabolic in shape.

We'll see more complex examples of equations like these whereby the graphs will look different.

Take these two examples, y = 7 - 1/2x and y = -x squared - x + 6.

But their essence does not change.

y = 7 - 1/2x is a linear equation, so no surprise, it's a linear graph.

y = -x squared - x + 6.

It's a quadratic equation, so no surprise that the graph is a parabola.

There are other key features we know from just looking at the equations.

The constant in y = 7 - 1/2x is 7.

Therefore there's a y-intercept at 0, 7.

That means when x = 0, y = 7.

Well, of course it does, because if x = 0, <v ->1/2 of 0 is nothing.

</v> We're just left with y = 7.

You can do something similar for that quadratic equation, y = -x squared - x + 6.

What about when x = 0 in that equation? Well, of course, y = 6.

The constant in the equation is 6.

Therefore we've got y-intercept at 0, 6.

Other features, in y = 7 - 1/2 of x, we've got a negative coefficient of x.

The coefficient of x is negative of half.

If you've got a negative coefficient, we've got a negative gradient.

In the case of y = -x squared - x + 6, we've got a negative coefficient of x squared.

Therefore, the turning point on the parabola is the maximum value.

Quick check you've got all that.

If you've got all that, you will be able to match the equations to the graphs.

Four equations, four graphs, but which is which? I'll leave you to figure that one out.

Pause and do that now.

Welcome back.

Let's see how we got on.

Hopefully, you match the equations up to the graphs like so.

Now, why would we match them like that? Well, in the case of equation C, y = x squared + 3x - 4? We know that's a quadratic equation, so we're looking for a parabola.

We also know in that equation there's a constant of -4.

It's gonna have a y-intercept of 0, -4.

The y-coordinate and the y-intercept is negative.

For the second graph, we match that to D.

y = 3x + 8.

Why? y = 3x + 8's a linear graph, so no surprise, we see a straight line.

It's got a positive y-intercept because it's a positive constant in the equation.

For B, we were always gonna get a parabola because it's a quadratic equation, y = x squared - 5x + 4.

And we have a positive y value in the y-intercept that matches with our positive 4 as the constant in the equation.

Therefore, y = 3x - 4 must be our last graph there.

Well, it's linear.

It's a linear equation, a linear graph, and we've got -x, -y coordinate for the y-intercept.

Now, do take note, we didn't need scales on the axis to be able to tell these graphs apart.

The shape and the key features gave them away.

We can see that the graphs of y = x squared or y = x cubed will be different before drawing their graphs.

Look at the table of values for y = x squared and the table of values for y = x cubed.

You can see some key differences.

y = x cubed will also be nonlinear.

There's not a constant rate of change in those y values.

But, whereas the square of a negative value is positive, the cube of a negative value is negative.

Let's look at these two graphs now, y = -x squared and y = -x cubed.

We've got a negative coefficient on x squared, but it's still a parabola.

We've got a negative coefficient on x cubed, but it still forms a cubic curve.

In the case of y = -x squared, as the absolute value of x increases, y values decrease.

That means our turning point on this parabola is a maximum point on the graph and the graph is present in quadrants three and four.

In the case of y = -x cubed as the value of x increases, y values typically decrease and vice versa, meaning this graph is present in quadrants two and four.

Let's take a look at what happens to those two graphs when we add a constant.

Instead of y = -x squared, we've now got y = -x squared + 5.

Instead of y = -x cubed, we've got y = -x cubed + 5.

In both cases, when we add a constant, the shape of the graphs of the same, we haven't altered their essence, however, we've translated the y-intercept.

Well, of course we have by adding a constant when x equals 0, y now equals 5.

in both cases.

Let's check you've got this.

For the graph of y = 5, execute what will be true of the coordinate pairs we plot.

Four statements, which ones are true? Which ones are false? Pause and sort that out.

Welcome back.

I wonder what you said.

Hopefully, you said A is true.

Positive x values without positive y values.

Cubing a positive value results in a positive value.

Hopefully, you said B was not true, C was not true, but D was true.

Negative x values will have negative y values? Why is that? Because cubing a negative value results in a negative value.

Another quick check.

I'd like you to match the equations to the graphs.

Four equations, four graphs.

By picking out the essence of their shape and the key feature that is the y-intercept, you'll be able to decide which is which.

Pause and do that now.

Welcome back.

Let's see how we did.

Hopefully, you match the equations to the graphs like so.

So, why did we match them like that? Well, the first graph was always gonna be y = x cubed + 5 because it was a cubic curve with a positive x cubed coefficient.

That means its x values are increasing, y values are typically increasing.

And we had a positive y-intercept or a positive y-coordinate in the y-intercept.

The second graph was always going to be D, y = x squared - 8 because y = x squared - 8, out the four, that was the only quadratic equation, therefore was the only one that was going to graph a parabola.

The third graph is y = x cubed - 5.

It's a cubic equation, a cubic curve, positive x cubed coefficient, which means its x values are increasing.

y values are typically increasing with a negative y-intercept in this case or a negative y-coordinate.

So the y-intercept 1x = 0, y = -5, so that y-intercept was always gonna be below the x-axis.

The fourth graph was slightly different.

It was a cubic curve, but the -x cubed coefficient meant as x values are increasing, y values were typically decreasing.

Another key feature you might have picked out is the fact it had a positive constant in the equation.

Therefore the y-intercept is gonna be above the x-axis at 0, 5.

We also need to recognise reciprocal graphs like y = 1/x.

Some features of that can be seen in a table of values.

If I populate this table, it looks like that.

Some key features we can pick out.

Negative x values have negative y values.

Well, of course, 1 divided by -1 is going to give us -1 so that makes sense.

Negative x values have negative y values.

The opposite is true.

Positive x values have positive y values.

Another curiosity is dividing by 0 is undefined.

If you try to substitute x = 0 into that equation, 1 divided by 0, it's undefined, so neither x nor y can be 0.

That gives this graph a rather special shape.

Take note of that.

That is the graph of y = 1/x.

It's a reciprocal graph.

We can change the equation but we won't change the nature of the graph.

y = 1/x.

That's a reciprocal graph that looks like that.

y = 1/x + 2.

Oh look, it's the same shape.

It's just that constant puts it in a different position.

Y = -1/x.

I wonder what that will look like.

It'll look like that.

It's the same shape.

It's just this time it's in different quadrants.

You know which quadrants this one will appear in by testing a few simple values.

y = -1/x.

Well, what about when x = 1? The y value would be -1 divided by 1, which is -1.

So you knew 1, -1 was going to be a coordinate pair.

That is gonna be present in quadrant four.

How about when x equals -1? Well, y would be -1 divided by -1, which is positive 1.

So you knew that -1, positive 1 was going to be a coordinate pair.

That is in quadrant two.

Quick check.

You've got this.

Sofia is sketching the reciprocal graph of y = -12/x.

Sofia has the right shape but the wrong quadrants.

Substitute any two x values to show her which quadrants her graph should be in.

Pause and do this now.

Welcome back.

Substitute any two x values to show Sofia which quadrants her graph should be in.

I'm gonna substitute x equals positive 4 to show her that y would equal -12 divided by 4, which is -3.

If positive 4, - 3 is a coordinate pair on this graph, that will be there in quadrant four.

I'm also going to show her when x = -2.

y = -12 divided by -2 is positive 6.

<v ->2, positive 6 will be in quadrant two.

</v> Any two x values, one being positive, one being negative, could be used to show Sofia this.

Sofia has taken our feedback on board and redrawn her graph.

That's wonderful, Sofia.

Sofia says, "That's awesome.

Thank you.

My graph must look like this.

I'll use that checking mechanism next time." Great idea, Sofia.

Practise time now.

Question one.

I'd like you to match the graphs to their equations.

Four graphs, four equations, match them up.

Pause.

Do that now.

Question two.

I'd like you to match these graphs to their equations.

Five graphs, five equations.

Pause and get matching now.

Feedback time.

Let's see how we did.

We should have matched them up like so.

How did we know? Well, in the case of the first graph, it was always gonna be C, y = x, 'cause it's a linear equation.

Therefore a linear graph.

D, y = x squared was always going to match that graph 'cause it's a quadratic equation, therefore it's getting to graph a parabola.

Our third graph is B, y = 1/x, because it's a reciprocal graph, and that is the shape of a reciprocal graph without a constant there.

We just see that graph in two quadrants.

The fourth graph is y = x cubed 'cause it's a cubic curve.

Question two, matching graphs to their equations.

We should have matched the five equations to the graphs like so.

How did we know? Because you know the difference between a reciprocal equation and the shape of a reciprocal graph, a cubic equation, and the shape of a cubic curve and a quadratic equation, and the shape of a parabola.

That was clue number one.

We can then look at constants in the equations to identify those which have a negative y-intercept.

Negative y-intercept, positive y-intercept, negative y-intercept.

What else did we know? y = 15 - x cubed.

That's a -x cubed coefficient.

So as x values increase, y will be typically decreasing.

We also had a -x squared coefficient in equation D, y = -15 - x squared.

That's how you could tell the graphs apart.

Onto the second half of our learning.

Now, we're going to be using graphs to model.

Understanding how various graphs apply to various models in a real-life context is a powerful skill.

For example, water is poured into this vase at a steady rate.

How will the graph of depth versus time look? What graph of depth versus time? That graph of depth versus time.

What do you think that graph is going to look like as water is poured into that vase at a steady rate? Pause and have a good think about that.

Welcome back.

I wonder what you thought.

Did you think, well, water is added at a steady rate and as water is added at a steady rate, the graph starts to look like so.

Two parallel sides to that vase means the depth will change at a constant rate, so we get a linear graph when plotting depth versus time.

The only way we'd see anything other than a straight line graph is if we altered the rate at which water is poured.

So we have to be certain to assume that the water was added at a steady rate.

How will the graph for this vase be different? And again, assume water is poured at a steady rate.

What do you think? Pause.

See if you can sketch that graph.

Welcome back.

Wonder what you thought.

Hopefully, you noticed that the depth will increase at a constant rate until this point.

And then beyond that, the depth will start to rise more quickly because the vase is narrower.

Our graph will look like that.

It's a constant rate of change, just a different faster one at the end.

Quick check.

You've got this.

Which vase would be modelled by this graph? Is it vase A, vase B, or vase C? Again, assume that water is poured at a steady speed.

Pause.

Match that graph now.

Welcome back.

I hope you said vase A.

Why is it vase A? Well, we get a rapid, constant rate of change at the start when the vase is really narrow, followed by a slower constant rate of change at the end.

Let's compare this vase to a slightly different type.

That vase.

How is that one different? What do you think will be different about the graph of depth versus time when we fill this vase? Pause.

Have a think about that.

Maybe try and sketch that graph.

See you in a moment.

Welcome back.

Wonder what you thought.

Well, what's happening is at the beginning, the vase is narrow.

Look at our fast rates of change between depth and time, but at the end, the vase is wider.

We'll get a slower rate of change of depth with respect to time.

It will constantly be getting slower.

At no point will depth be changing at a constant rate.

So what does this graph look like if there's not a constant rate of change? It looks like that.

The rate of change is constantly slowing.

Let's check you've got that.

Which vase will be modelled by this graph? Is it A, B, or C? Once again, we have the assumption that water is border to steady speed.

Pause.

Have a think about that problem.

Welcome back.

Did you say vase B? If so, well done.

Why is it vase B? Well, it's a quickening rate of change.

The rate at which depth changes with respect to time is getting faster and faster.

Why? Because the vase is getting narrower and narrower.

Crucially, in that graph, you should have recognised that there's no constant rate of change.

Let's check this one now.

C's an unusual vase.

Do you think you could sketch the graph of it for depth versus time? Pause and give this problem a go.

Welcome back.

Let's see how we did.

Hopefully you noticed at the start it's a wide vase so we've got a slow constant rate of change.

In the middle, we get a slightly faster constant rate of change.

So we've still got a straight line on our graph.

It's just a bit steeper.

Then at the end, it's a constant rate of change but even faster.

So another third section of straight line, but even steeper.

Practise time now.

I'd like you to match each vase to its depth-time graph.

Pause and do that now.

Question two.

I'd like you to draw a depth-time graph for this vase.

That's a lovely vase.

I wonder if the graph's going to be that lovely.

Pause and give it a go.

Feedback time now.

Let's see how we did.

Hopefully, you matched the vases to the graphs like so.

You might wanna pause now and just check how you did.

Question two, I asked you to draw a depth-time graph for this beautiful vase.

At the beginning we had an increasing rate of change.

The vase was getting narrower.

Then in the middle, we had a steep constant rate of change.

That's as narrow as the vase gets.

So that's the steepest moment your graph should show.

Then at the end, as the vase gets wider, the rate of change of depth with respect to time gets slower.

So a graph curves off like so.

Sadly, that's the end of our lesson now, but we've learned that our knowledge of linear graphs can be used to solve problems. Understanding how the parts of an equation relate to its graph can help with matching non-linear graphs to their equations and non-linear graphs in a real-life context can be sketched.

I hope you've enjoyed this lesson as much as I've enjoyed it and I'll look forward to seeing you again soon for more mathematics.

Goodbye for now.