Lesson video

Hello, Mr. Robson here.

Welcome to maths.

Super choice to join me today.

Problem solving with non-linear graphs.

Do you love problem solving? Do you love non-linear graphs? Then you're gonna love this.

Let's get started.

Our learning outcome is I'll be able to use our knowledge of non-linear graphs to solve problems. Some key words that are gonna come up today.

Quadratic, cubic, and exponential.

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

For example, x squared plus x is a quadratic, but x cubed plus 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 plus bx squared plus cx plus d.

The general form for an exponential equation is y equals ab to the power of x, where a is the coefficient, b is the base, and x is the exponent.

Another keyword you'll hear a lot today is asymptote.

An asymptote is a line that a curve approaches but never touches.

For example, further graph y equals 1/x, the x and y axes are both asymptotes.

We've got two parts to our learning today and we're gonna begin by recognising graphs.

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.

We can see that the graphs of y equals 2x and y equals x squared will be different before drawing their graphs.

Let's look at the table of values for those two equations.

Can you see the difference? For y equals 2x, we got a constant rate of change that tells us it'll be linear in shape when we graph it.

For y equals x squared, it's not a constant rate of change, therefore that one is gonna be non-linear.

There's other things we can see here.

For y equals 2x, the y values are constantly increasing.

But for y equals x squared, the y values are decreasing initially in our table then they start increasing.

That tells us there will be a turning point on this graph.

Another feature we can pick out from this table of values both have a y-intercept at the origin 0,0.

Another interesting feature, y equal x squared has only positive y values excluding zero of course, whereas y equals 2x has negative y values.

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

Let's look at those graphs now.

There's a graph of y equals x squared and the graph of y equals 2x.

Y equals 2x, well it was a linear equation, therefore we have a linear graph.

Y equals x squared was a quadratic equation, therefore it's parabolic in shape.

The graph of that is a parabola.

We'll see more complex equations whereby the graphs look different.

For example these two, y equals 7 minus a 1/2 of x, y equals negative x squared minus x plus 6.

But their essence does not change.

In the case of y equals 7 minus 1/2 of x, it's still a linear equation.

The highest exponent of x is 1.

We've got a y exponent of 1 also.

It's a linear equation, therefore we get a linear graph.

In the case of y equals negative x squared minus x plus 6, there's a lot more going on in that equation.

But it's still a quadratic equation, therefore the graph is still a parabola.

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

For example, in y equals 7 minus 1/2 x, that 7 is a constant.

Because the 7s a constant when x equals 0, y equals 7, so we get a y-intercept at 0,7.

We see something similar in that quadratic equation too.

With a constant of 6, we get a y-intercept at 0,6.

We could have seen those two features before we even plotted these graphs.

In addition, in the case of y equals 7 minus 1/2 x, we got a negative coefficient of x, hence the line has a negative gradient.

In y equals negative x squared minus x plus 6, we got a negative coefficient of x squared.

That tells us that the turning point is the maximum value on the graph.

There'll be no higher y value than that one.

Quick check, you've got all that so far.

If you've got it, you should be able to match these equations to these graphs.

Four equations, four graphs, pause this video now and match them up.

See you in a moment.

Welcome back.

Let's see how you did.

Hopefully you paired them up like so.

The fourth graph was equation A, y equals 3x minus 4.

The third graph was equation B, y equals x squared minus 5x plus 4.

The second graph was D, y equals 2x plus 8.

And the first graph was C, y equals x squared plus 3x minus 4.

How did we know? Well, the first graph was a parabola with a negative y-intercept, or shall I say a negative y-coordinate at the y-intercept.

Second graph was linear, positive y coordinate at the y-intercept.

The third graph was a parabola also, so we knew we were looking for a quadratic equation, but this time the constant positive 4 in the equation gave us a positive y-coordinate at y-intercept.

And in the case of the fourth graph, a linear graph, so we're looking for a linear equation with again a negative y-coordinate at the y-intercept.

We didn't need scales on the axes 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 equals x squared and y equals x cubed will be different before drawing their graphs.

Look at the y values.

You can see that y equals x cubed, like y equals x squared will also be non-linear.

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

That's gonna make these two graphs very distinguished from one another once we've drawn them.

Oh look, y equals x squared, y equals x cubed, very different graphs.

A quadratic equation forms a parabola, no surprise there.

But in the case of y equals x cubed, a cubic equation, our graph forms a cubic curve.

In the case of the parabola, we've got constant of zero, so the graph is present in quadrants I and II.

That's because when we square those negative x values, we get a positive y value.

But by contrast, when the constant is zero for y equals x cubed, the graph is present in quadrants I and III.

That's because cubing the positive gave us a positive for quadrant I, whereas cubing a negative gives us a negative, hence the graph is present in quadrant III.

Another feature of y equal x cubed, for increasingly positive x values.

We see our y values typically increase, and vice versa.

Let's change those graphs up a bit.

What if we graph y equals negative x squared and y equals negative x cubed? Well, y equals negative x squared, we have a negative coefficient, but it's still a parabola.

But at this time as the absolute value of x increases, y values decrease.

That turning point on the parabola is now a maximum value.

So we see the graph in quadrants III and IV.

In the case of y equal negative x cubed, you have a negative coefficient x cubed, but we still get a cubic curve.

The difference this time being, as the value of x increases, y values typically decrease, and vice versa.

So we see this graph in quadrants II and IV.

What if we add a constant to those equations? We're now looking at y equals negative x squared plus 5, y equals negative x cubed plus 5.

When we add a constant, the shape of the graphs are the same.

We still got a parabola with a maximum value and a cubic curve with y values typically decreasing as x values increase.

But we've translated the y-intercept by adding a constant.

Let's check if you've got all that.

For the graph of y equals 5x cubed, what will be true of the coordinate pairs we plot? Four statements there, some are true, some are not, I'll leave you to decide which are which.

Pause now, see you in a moment.

Welcome back, let's see how we did.

Hopefully you said A was true, positive x values will have positive y values.

Absolutely.

Keeping a positive value results in a positive value.

B was not true.

C was not true.

D was true because when we cube a negative value we will get a negative value.

So negative x values in this equation will have corresponding negative y values.

Well done.

Next check, I'd like to match the equations to their respective graphs.

Four graphs, four equations, pause and pair them up.

Welcome back, let's see how you did.

Hopefully you paired them up like so.

The fourth graph being y equals negative x cubed plus 5.

The third graph being y equals x cubed minus 5.

Our second graph being y equals x squared minus 8.

And our first graph being y equals x cubed plus 5.

How did we know which was which? Well, in the case of the first graph, we were looking for a cubic curve or we had a cubic curve, so we're looking for a cubic equation with a positive x cubed coefficient and a positive y value at the y-intercept.

So our constant had to be positive, hence that one's x cubed plus 5.

The case of the second graph, it's quadratic, it's parabola.

It was the only parabola of those three graphs, so we matched it to the only quadratic equation.

Case of the third graph, it's another cubic curve with a positive x cubed coefficient.

This time a negative y value at the y-intercept.

Look at the constant in the equation, negative 5.

That's how we knew that was the graph of x cubed minus 5.

That leaves us with a fourth graph which was a cubic curve, but this time with a negative x cubed coefficient.

And for further reassurance we had a positive y value at the y-intercept.

Next, we also need to recognise reciprocal graphs like y equals 1/x.

Some features can be seen in a table of values.

Let's look at the corresponding y values for these x values.

Some things should jump out at you like negative x values have negative y values.

Well of course, because when we take a positive divided by a negative, the result will be negative.

You can also see positive x values have positive y values.

We're taking positive 1, dividing it by a positive x value, we must get a positive y value.

Another interesting feature, dividing by zero is undefined.

So when we try to use x equals zero, well it's undefined.

So neither x nor y can be zero for this graph.

When we plot the graph, we get that beautiful shape.

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

That's y equals 1/x.

What about y equals 1/x plus 2? Well, it looks like that.

It's the same shape, it's just in a different position.

How about y equals negative 1/x? Oh, look, it's the same shape, but this time in different quadrants.

Y equals 1/x was present in quadrant I and III.

Y equals negative 1/x is present in quadrants II and IV.

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

For example, when x equals positive 1, we substitute that into the equation y equals negative 1 divided by positive 1, which is negative 1.

That's the coordinate 1, negative 1.

which would be present in quadrant IV.

How about when x equals negative 1? Well we'd have y equals negative 1 divided by negative 1, that's positive 1.

That'll give us the coordinate pair negative 1, 1, which would be present in quadrant II.

Quick check you can repeat that skill.

Sofia is sketching the reciprocal graph of y equals negative 12/x.

Sofia has the right shape but the wrong quadrants.

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

Pause, have a go at this problem now.

Welcome back, let's see how you did.

I've substituted in when x equals positive 4, y equals negative 12 divided by 4, that equals negative 4.

We have the coordinate pair 4, negative 3, which would be present in quadrant IV.

So that's how Sofia knows of those curves should have been in quadrant IV.

Next, I've substituted in x equals negative 2, because y will then equal negative 12 divided by negative 2, that's positive 6.

I'll have another quadrant pair negative 2, positive 6, which will be present in quadrant II.

You could have substituted in any two x values, one positive and one negative to have shown Sofia the correct two quadrants.

Oh, look, Sofia's reflected on that feedback and corrected the graph.

Sofia says, that's awesome, thank you.

My graph must look like this.

I'll use that checking mechanism next time.

Well done, Sofia.

We also need to recognise exponential graphs like y equals 2 to the power of x.

In the form y equals b to the power of x, where b is not equal to zero, the x-axis will always be an asymptote, because b to the power of x can never equal zero.

There's no x value we can substitute in to make b to the power of x zero, provided b is not zero itself.

The form y equals b to the power of x, will also have coordinate pairs 0,1 and 1,b.

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

There's y equals 2 to the power of x.

It's an exponential curve.

What about y equals 2 to the power of x minus 3? Well, it's still the same shape.

We've just got a different asymptote.

Instead of having the x- axis as an asymptote, we've now got y equals negative 3 as a horizontal asymptote.

How about y equals 1/2 to the power of x? Well, it's still an exponential curve, but because that base is now a fraction between zero and 1, we've got a downward curve instead of an upward curve.

How about y equals negative 2 to the power of x? How will that one look different to the exponential graphs we've seen so far? Well, it'll look like that.

Do you notice it's the same shape as y equals 2 to the power of x, but we've now got negative y values.

We've still the x-axis as an asymptote, but all the y values are negative and as the x values increase, the y values decrease.

The graphs all show an exponential curve and a horizontal asymptote with y values only appearing on one side of that asymptote.

Let's check you've got that.

Which of these is not an exponential graph? One of them is not an exponential graph.

Once you've decided which one it is, I'd like you to write a sentence to justify your answer.

Pause and do that now.

Welcome back.

Hopefully you said A and C are exponential and it was B that was not an exponential graph.

How might you have justified it? Well, you might have written, graph B has horizontal and vertical asymptotes, contrast that with graphs A and C where we've got one horizontal asymptote.

That's how we knew B was not an exponential graph.

Another graph you have to recognise is the graph of the equation of a circle.

The equation of a circle with origin at centre stands out from the rest.

Look at those four equations.

Y equals x squared, that's clearly a quadratic.

Y equals 2 to the power of x, the exponent is the variable, well that's an exponential equation.

Y equals 2/x, that's a reciprocal.

X squared plus y squared equals 2 squared is the equation of a circle with radius 2.

That's 'cause the general form of the equation of a circle centred at the origin is x squared plus y squared equals r squared.

This image helps us to remember that general form.

Remembering that the equation of a circle comes from using Pythagoras' theorem is a useful way to remember.

If you look at that image, if we take x, square it, take y, square it, and add those two results, we'll get the square of our radius, hence the general form is x squared plus y squared equals r squared.

Quick check you've got that.

Which of these equations could represent a circle? Three equations there, which of them are circles? Pause and have a think.

Welcome back.

Hopefully you said A could represent a circle.

If we compare x squared plus y squared equals 150 to the general form, you'll see r squared must be equal to 150.

Well, that must be a circle with a radius the square root of 150.

B was not.

Why not? Because the y exponent is not 2.

If it read x squared plus y squared equals 15 absolutely would have a circle.

But in the case of that equation, the y exponent is 1 not 2, so that is not going to be a circle.

C, absolutely.

You might have been fooled by the fact that the r squared position read 4 cubed, but 4 cubed is 64.

64 is a perfect square.

We could call it 8 squared.

In the form x squared plus y squared equals 8 squared, you can clearly see that's a circle with a radius of 8.

Practise time now.

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

Five graphs, five equations, match them up.

Pause and do that now.

Question two is another matching exercise.

Just a slightly different variety of graphs and equations now.

Still five graphs, five equations, pause and match them up.

Feedback time, let's see how we did.

For question one, we should have matched them up like so.

How did we know? Well, we've got a reciprocal, a cubic, a quadratic, a cubic, and a quadratic respectively.

The difference between the cubics being we've got a negative y-intercept in x cubed minus 15, a negative y-intercept in x squared minus 15, positive y-intercept in 15 minus x cubed, and a negative y-intercept in negative 15 minus x squared.

You'd also notice the shape of the negative x cubed and negative x squared graphs.

Question two, we should have matched them up like so.

How did we know? Well, we've got a reciprocal, an exponential, an exponential, a quadratic, and a circle.

The circle you'd need to rearrange to get it in the form x squared plus y squared equals r squared.

If you rearrange it to read x squared plus y squared equals 4, that's how you know that's the equation of a circle.

Onto the second half of our lesson now, problem solving with the graph of a circle We can use our knowledge of the graph of a circle to solve trickier problems. By trickier problems, what do I mean? I mean this problem.

A circle has equation x squared plus y squared equals 3 squared and its graph is drawn.

The graph of x equals 4 and y equals negative 5 are also drawn the same axes.

Considering the fourth quadrant only, what is the size of the area between the graph of the circle and lines x equals 4 and y equals negative 5? Wow, there's a lot going on there.

So let's break it down a little.

Thankfully, we've got a visual representation of this problem to work with.

First things first, it says considering the fourth quadrant only.

Ah, we're interested in that rectangle to begin with.

The area of that rectangle is 4 by 5, it's 20 square units.

What are we looking for next? Well, we need to subtract 1/4 of the circle.

We're looking for the area between the graph of the circle and those two vertical and horizontal lines.

So let subtract 1/4 of that circle.

We know the circle's got a radius of 3, therefore 1/4 of that circle that'll be a 1/4 of pi r squared, which in this case simplifies to 9 pi over 4.

So and I need to do the area of rectangle and subtract that 1/4 circle.

I can't do that one in my head, so I'm gonna use my calculator.

And I'm absolutely gonna plug in the exacts value 9 pi over 4.

When I do 20 minus 9 pi over 4 in my calculator, I get answer 12.

93141, et cetera.

I'm gonna round that to something sensible, like one decimal place.

So I'll communicate my answer as the area equals 12.

9 square units to one decimal place.

Quick check you've got that.

A circle has equation x squared plus y squared equals 25 and its graph is drawn.

The graph of x equals 9, and x equals negative 5, and y equals 6 are also drawn.

Considering the first and second quadrants only, what is the size of the area between the graph of the circle and the three lines? Give your answer to two decimal places.

Pause, see you in a moment.

Welcome back, let's see how we did.

Tricky problem this one.

So don't worry if you don't get it right first time.

We're not gonna get everything right first time when we do mathematics, but it's really important we reflect on any errors and be willing to try again.

So a solution.

Well if we look at the problem like this, it's the rectangle minus 1/2 of that circle.

So the area rectangle is 14 by 6, which is 84 square units.

1/2 of that circle will be 1/2 of pi r squared, which in this case is 25 pi over 2.

And then we need to do the rectangle, subtract that 1/2 of circle, and let my calculator take the strain on that one and I'll do the rounding.

The area is 44.

73 square units to 2 decimal places.

You might wanna pause and check that you're working out looks just like mine and your answer is as accurate as mine.

Sometimes we're not given a visual representation.

For these cases, we can draw a sketch to help us.

For example, a circle has the equation, x squared plus y squared equals 4 and its graph is drawn.

The graph of y equals x is also drawn in the same axes.

What is the area of the region bound by the circumference of the circle, the line y equals x and the x-axis? We can't make sense of all those words and equations without a visual representation.

So I'll do that sketch and that's all I need.

I don't need to label those axes, I just need to note a circle, centred at the origin with the radius of 2.

And you can see the line y equals x going through it.

And you can see I've labelled the x and y-axis.

From here, we can start to answer this problem.

Oh, look.

The diagram, even though it's just a sketch, helps to see that the answer is just 1/8 of the circle.

We're looking for the region bound by the circumference of the circle, the line y equals x and the x-axis.

That's my shaded region there.

So we need 1/8 of the area of this circle.

Again, I'm gonna let my calculator do that one for me.

1/8 of pi r squared, it's going to be 1.

5707 et cetera.

We'll round that to something sensible.

That area will be 1.

57 square units to two decimal places.

So let's check that you've paid attention there.

True or false.

When we're not given a diagram, we should sketch our own to support our understanding of the problem.

Is that true or is it false? Once you've decided, use one of the statements at the bottom of the screen to justify your answer.

Pause and do this now.

Welcome back.

I hope you said absolutely true.

And that visual representations help us make sense of complex problems in maths.

You definitely shouldn't have said statement B, visual representations are not helpful in maths.

What nonsense.

They are really, really helpful.

Practise time now.

Question one, a circle has the equation x squared plus y squared equals 25 squared and its graph is drawn.

The graphs of x equals negative 34 and y equals negative 29 are also drawn in the same axes.

Considering the third quadrant only, what the size of the area between the graph of the circle and the two straight lines? I'd like you to give your answer to one decimal place.

Pause and have fun with that problem now.

Question two, a circle has the equation x squared plus y squared equals 20 and its graph is drawn.

The graph of y equals 10 minus x is also drawn on the same axes.

What is the area of the region bound by the circumference of the circle, the line y equals 10 minus x and the x and y axes.

Your answer should be given exactly.

Little hint for you on this one.

I don't currently see a visual representation, but you'll need one.

Do a sketch, pause, give this problem a go now.

Welcome back, feedback time.

Hope you enjoyed wrestling with those problems. Let's have a look at problem one.

Hopefully you did something like that with the diagram and identified there's a rectangle there, dimension 34 by 29, therefore 986 square units.

And we need to subtract 1/4 of the area of the circle from that.

In this case, 1/4 of the area of this circle is 625 pi over 4.

We'll let our calculate to do the next bit and we'll round up answer to one decimal place.

So we should have got an area of 495.

1 square units to one decimal place.

Question two, we didn't have a visual representation.

We need one.

It should have looked something like that.

I can see the linear graph y equals 10 minus x.

I can see a circle centred at the origin with a radius square root of 20.

We're ready to work on the problem now.

We need to work out the area of the region, which is bound by the circumference of the circle, the line y equals 10 minus x and the x and y axes.

I see that as a triangle minus 1/4 of the circle.

The triangle has an area of 10 by 10 over 2, which is 50 square units, and we're subtracting 1/4 of the circle, which in this case simplifies beautifully to 5 pi.

So our answer, the area of our region is 50 minus 5 pi.

We were told to leave the answer exact so we don't turn that into a decimal, we don't round, we just leave it as 50 minus 5 pi square units.

Sadly, we're at the end of the lesson now.

But we've learned, knowledge of non-linear graphs can used to solve problems and understanding how the parts of an equation relate to its graph can help us with matching non-linear graphs to their equations.

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

Good bye for now.