Lesson video

Thank you for joining us for this Transformations of Graphs lesson.

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

I'm really excited to explore this topic, I hope you are too.

So let's get started.

Welcome to this lesson on problem solving With graph transformations, we're gonna have a look at all the graph transformations you know and apply them to different problems. So we know a transformation is a process that may change the size, orientation, or position of a shape or a graph.

And of course today we are looking specifically at graphs.

We're gonna start by looking at transformations with asymptotes.

We're gonna do a lot of exploring in this first part of the lesson.

You may wish to use some graphing software alongside me, there will be different links to take you to different graphs of functions as we go through.

When we manipulate a function f(x) by performing an operation, the graph of f(x) is transformed to give the new graph.

Different manipulations, transform the graph in different ways.

Pause, video.

Can you describe the effect these transformations have on the graph f(x)? A bit of a recap.

Let's have a look.

How precise were you with your language? F(x) plus A is a translation of A in the Y direction.

F(x) plus A is a translation of negative A in the X direction.

Af(x) is a stretch by scale factor of A in the Y direction.

F(x) is a stretch by scale factor of 1/A in the X direction.

Negative f(x) is a reflection in the X-axis and f(-x) is a reflection in the Y-axis.

Pause video if you need to make any notes on that or correct any of your answers.

So we're gonna explore what happens to graphs with asymptotes when they are transformed.

Each of the links below apply a different transformation to the graph of the function f(x) equals 1/X with a restricted domain of X, where X is greater than zero.

So we're just looking at inputs greater than zero.

Click on each one.

Where are the asymptotes for Y equals 1/X, and how does each transformation affect those asymptotes? Pause the video and explore those four transformations.

So here's our graph of 1/X, when X is positive.

At the moment we've got A is zero.

Let's see what happens when we add A.

Look what's happening to that asymptote as well.

Don't forget you've got two asymptotes on this graph.

Then we've got our negative values of A.

Again, think about what's happening to each of those asymptotes.

Keep that in your head.

We're going to explore another transformation.

So here we've got f(x) plus A.

Again, we've got two asymptotes, look what happens as we increase A.

Think about the transformation that's being applied, what is happening to those two asymptotes.

And then here we are multiplying our function by A.

So we can see that happening with those coordinates we fixed.

Look at what happens to the asymptotes.

And finally here, we're multiplying our input before putting it into the function.

If we multiply by a value greater than one, you'll see what's happening to our coordinates.

Notice what's happening to our asymptotes.

If I go back, I put A as 1/2, you'll see what happens as well.

So that coordinate goes from one, one to two, one 'cause it's a stretch of two in the X direction.

Notice what happens to the asymptotes.

Absolutely take your time to load those files and try that yourself.

So the graph Y equals 1/X, has asymptotes at X equals zero and Y equals zero.

Restricting the domain to X where X is greater than zero does not change the asymptotes.

When we applied the transformation of f(x) plus A, each point had a added to the Y coordinate.

The asymptote at X equals zero was unaffected, but the asymptote at Y equals zero was translated as well.

The new asymptote was at Y equals A.

When we applied the transformation f(x) plus A, each point had negative A added to the X coordinate the asymptote at X equals zero was translated as well.

The new asymptote was at X equals negative A.

The asymptote at Y equals zero was unaffected.

Now when we applied the transformation Af(x), the Y coordinate of each point was multiplied by A, the asymptote to X equals zero was unaffected.

The asymptote at Y equals zero was unaffected but there's an interesting reason why and that's because zero times A is zero.

So the Y coordinates were already zero multiplying by A doesn't change them.

We applied the transformation f(ax), the X coordinate of each point was multiplied by 1/A, and the asymptote to X equals zero was unaffected.

Well again, if all the X coordinates are zero, multiplying them by 1/A will not gonna affect them, and the asymptote at Y equals zero was unaffected.

Would you think the graphs of negative f(x) and f(-x) will look like, the f(x) equals 1/X? And again, we're just restricting that domain.

So X has to be greater than zero.

Let's have a look.

Well, negative f(x) is a reflection in the X-axis.

The asymptotes are not affected this time.

Have a think about why? F(-x) is a reflection in the Y-axis and again, the asymptotes not affected.

Sofia says, "So the only transformations which affect the asymptotes are translations." Jen says, "That was just that example, I found one where stretching the Y direction also changes the asymptote." Now I'd like you to click on the link and explore Jun's function.

Why does the asymptote change the ah(x) this time, do you think? So with our original function, Sofia is right.

The only transformations that affected the asymptotes were translations, but there's a reason why.

So why does multiplying by A this time affect the asymptote whereas it didn't for our previous example.

Have a play around.

See if you can spot this then we'll look at it together.

Right, it's because the asymptote is at Y equal one.

So when this is multiplied by A, it changes to Y equals A.

Now this did not happen in the previous example because our asymptote was Y equals zero, and multiplying by zero doesn't change the value.

What do you think the graphs of negative h(x) and h(-x) will look like? Right, you might wanna type this into graphing software and see what happens.

So there's negative h(x), it's a reflection in the X-axis.

So the asymptote at Y equals one has also been reflected in the X-axis, with a new asymptote at Y equals negative one.

The asymptote X equals zero is unaffected.

It's because this transformation changes the Y coordinates only right? For h(-x), the asymptote to Y equals one is unaffected, 'cause this transformation changes the X coordinates only.

The asymptote asymptote at X equals zero is also unaffected, as this is the line of symmetry, multiplying zero by negative one is still zero.

So the graph of Y equals f(x), when f(x) equals two to the power of X, as an asymptote at Y equals zero.

I'd like you to tell me the equations of the asymptotes for fach of these transformed graphs.

So for this first translation, there's now an asymptote at Y equals two.

All the coordinates have had two added to the Y value.

So all the coordinates on that asymptote will have two added to the Y value.

So the new asymptote will move to Y equals two.

For the other three, the asymptote stays at Y equals zero for f(x) plus two, that's because this only affects the X coordinates.

And although the X coordinates are shifting left two, that's not gonna affect the asymptote.

It stays on that line, Y equals zero.

The negative f(x) because the asymptote is the line of symmetry, that's not gonna affect the asymptote.

An f(2x) that's a stretch of 1/2 in the X direction.

And again, that's only gonna impact the X values not the Y values.

So that asymptote stays where it is.

This is a new function, f(x) is two to the power of X plus two and it has an asymptote at Y equals two.

What are the equations of the asymptote for each of these transformed graphs? Give this a go.

The asymptote stays at Y equals two.

We're only impacting the X coordinates.

The same for f(3x) and the same for f(-x).

The asymptotes are staying the same, 'cause the asymptote is the equation Y equals two, and these transformations only affect the X coordinates.

Right, how about these three? Where's the new asymptote this time? Right, these three transformations impact the Y coordinates.

So we've now got an asymptote at Y equals five, 'cause it's been translated three in the positive Y direction.

Three f(x), we now have an asymptote at Y equals six, 'cause we've multiplied each of the Y values by three.

So Y equals two, becomes Y equals six.

And for negative f(x), we now have an asymptote at Y equals negative two, 'cause we've reflected it in the X-axis.

We can use what we know about asymptote to transform the graph of tan(x).

So here's part of the graph f(x) equals tan(x).

Where are the asymptotes? Right, the ones we can see on this graph, X equals 90, X equals -90, X equals 270, and X equals -270.

These also reoccur every 180 degrees.

I'd like you to make a prediction which of these will change the position of the asymptotes the f(x) equals tan(x).

Let's have a look.

There's f(x) plus three, There's 3f(x), there's negative f(x).

These three transformations affect the Y coordinates of the points on the graph and they also affect the Y coordinates of the asymptotes.

Because the asymptotes are vertical lines X equals 90 and X equals -90, these do not move.

So those three transformations do not impact the asymptotes for tan(x).

However, there's f(x) plus 30, f(3x) and f(-x).

They're asymptotes now at X equals 60 and X equals -120 for the first one.

X equals 30 and X equals -30 for that second one.

And the asymptotes have reflected in the Y-axis, but because there are symmetrical, they do not look to have moved but they have moved.

The asymptote at X equals 90 has now become X equals -90.

But they still have the same equations, they've just swapped.

So these transformations affect the X coordinates of the points on the graph and the asymptotes.

The f(x) plus 30, the asymptotes have translated <v ->30 in the X direction.

</v> The f(3x) they've been stretched by a third in the X direction, and we've already talked about f(-x).

Quick check.

This is part of the graph Y equals tan(x).

Sketch the graph with equation Y equals tan(x) minus 90.

Mark on any asymptotes.

Let's have a look.

So every coordinate and every asymptote should be translated positive 90 in the X direction.

Now got asymptotes at X equals zero and X equals 180.

Of course if we draw more of the graph, we would see more asymptotes.

That'd also be one at negative 180.

Time for practise.

For this task you may wish to use graphing software to help you.

I've told you what the function is for each so you can use that to then transform them using graphing software.

Have a go at those questions.

You've got the same function for each of these questions, again, feel free to utilise graphing software.

Give those a go.

For this one, you've got a new function again, I'd want you to draw the graph after each transformation and write the equations of any asymptotes.

Use graphing software if you want to help you or to check your answers.

And for seven, you've got a section of the tan graph.

Again, feel free to use graphing software to help you.

And a little challenge looking at the graph of tan(x).

If you can always check your answers by using that technology.

Right, for the first one, those are the new coordinates.

asymptotes at Y equals three and the one at X equals zero has stayed.

The second one there are new coordinates, new asymptotes at Y equals three.

The third one, there's our new coordinates asymptotes at stayed at X equals -90 and X equals 90.

Pause the video and check your sketches.

For three, again those are the new coordinates.

You've got asymptotes X equals negative four, but Y equals three has remained the same.

There are new coordinates for B, X equals zero has stayed the same.

Y equals six as the new asymptote, and there are new coordinates but our asymptotes have stayed at X equals zero and Y equal three.

For five, we've got a new asymptote at Y equal negative two after a reflection in the X-axis, the asymptote has stayed at Y equals two with a reflection in the Y-axis.

For our tan graphs, pause the video, check your sketches and check your new asymptotes.

For question eight, you'll still have two for f(x) plus two, and for two f(x), 'cause they're only impacting the Y values.

For C, you are now gonna have four because that's a stretch of 1/2 in the X direction.

So you'll be able to fit in two more asymptotes in that range.

For D, you're gonna get six.

For E one, and for F, 20.

You could always plot these and check.

There were some really complicated ideas there, so absolutely fine if you use that graphing software to help you.

We're just playing about with this at this stage and seeing how we can apply our knowledge.

Don't worry if you didn't get everything spot on correct.

We're now gonna look at sketching a transformed graph.

We can use our knowledge of transformations to help us sketch different graphs.

Here's the graph of Y equal X squared and you can see I've labelled some key points.

What would the graph of Y equal X plus five squared minus three look like? Well if f(x) equals X squared, then our new graph is the transformed function f(x) plus five, minus three.

You can see how X squared has changed to X plus five squared minus three.

So this is a translation of five units in the negative X direction and three units in the negative Y direction.

So we can use Y equals X squared to draw this other graph.

If we translate every coordinate and draw our new graph, we've now got the graph of Y equals X plus five squared minus three using our transformation skills.

How could we use this to sketch X squared minus 4x, plus four? What do you think? It's not easy to see the relationship between these expressions at the moment.

What we can do is complete the square to make this easier.

So this would be X minus two, all squared.

So if f(x) is X squared, then our new graph is f(x-2).

Well now we can see this is a translation of two units in the positive X direction.

So don't worry if you looked at Y equals X squared minus 4x, plus four, and couldn't see the relationship between that and X squared.

Completing the square allows us to see that relationship.

Let's translate that.

And that's Y equals X squared, minus 4x, plus four.

Andeep wants to sketch the graph of Y equals X squared minus 6x, plus eight, by transforming Y equals X squared.

He says, "I have factorised expression and I'm not sure what to do now." Aisha's gonna help.

"This doesn't help us see the transformation.

We actually want it in completed square form." So remember it's not factorising that we want to do to see the transformation, it's completing the square.

That's gonna put it in the form that we want.

Let's do that then.

So we've got X minus three all squared, minus one.

Now we can see that it's a translation of X squared.

It's a translation.

Three units in the positive X direction, negative one in the Y direction.

Lucas says, "We have been looking at Y equal 4x squared, minus three, but we have two different answers." Jacob thinks, "If f(x) equals X squared, then this is gonna be four f(x) minus three." Lucas thinks you can write it as 2x squared minus three.

So this is f(2x) minus three.

What transformations of X squared would each of these ideas involve? So Lucas' would be a stretch of 1/2 in the X direction and then a translation of negative three in the Y direction.

Jacob's would be a stretch of four in the Y direction, then a translation of negative three in the Y direction.

There's f(x) equals X squared, there's f(2x), there's four f(x).

Interesting result, if we're starting with X squared.

And there's f(2x) minus three, and there's four f(x) minus three.

So for this graph, these transformations are equivalent.

That's not always gonna be the case, but because of the shape of Y equals X squared, a stretch of four in the Y direction was the same as a stretch of 1/2 in the X direction.

Which of these is equivalent to 2x squared minus 16 X plus 32.

Think about your completing the square skills.

What if you said A, if we take out a factor of two, we get two lots of X squared, minus 8x, plus 16.

Completing the square gets us two lots of X minus four all squared because that's the perfect square.

What transformations would map the graph of Y equal X squared to the graph of the equation Y equal 2x squared, minus 16x, plus 32.

Think about what we just did.

There's a completed square form.

So our new graph is the transformed function 2f(x) minus four.

This is a translation of four in the positive X direction and a stretch of two in the Y direction.

So if we know one graph has been transformed to form a new graph, we can use this to write the equation of the new graph.

Izzy drew the graph of Y, it was X squared minus 4x and applied two transformations to get the new graph.

What transformations do you think Izzy performed? Right, I think you might have said reflection in the X-axis and that's how we've then got that negative X squared parabola and then a translation of negative three in the X direction.

If we define f(x) as X squared minus 4x, and the new graph is the transformed function negative f(x) plus three.

What is the equation of this new parabola? Right, there's a lot going on here, we do negative f(x) plus three, so we need to replace X with X plus three.

We also need to multiply the whole thing by negative one.

So we've got a bit of rearranging to do.

We've got to square the binomial, X squared, plus 6x plus nine.

Then we need to subtract four lots of X and four lots of three.

So subtract 4x, subtract 12.

And remember this was a reflection in the X-axis.

We need to multiply everything by negative one.

So then we've simplified and multiplied by negative one.

The new graph is Y equal negative X squared minus 2x plus three.

This is quite a complicated concept so we can use tech to check.

So there is f(x) and then I've typed in negative f(x) plus three.

Those are the two graphs that Izzy had.

Then I could also check the equation of my new parabola by typing it in.

So I thought it was negative X squared, minus two X, plus three.

I type that in and I can see that that sits over my transformed function, so I must have the right equation.

You can absolutely use that to help you with the next task.

What two transformations map the graph of Y equals X squared, minus 4x, plus four to the transformed graph.

What do you think? Well let's look at that shape.

We can see that we've got a stretch in the X direction and we can see that there's a translation.

So it's a stretch of 1/2 in the X direction, translation of negative one in the Y direction.

It doesn't matter which order we do those in.

Right, which of these is the function of the new graph? Think about what we just said.

Very good, if it's a stretch of 1/2 in the X direction, that's f(2x) but of course we need to translate that negative one in the Y direction.

We need to subtract one from the whole function.

Right, so let's use that idea to write the equation of the graph.

You definitely want to show your working for this.

There's quite a bit of algebra manipulation, Right, so we need to do 2x all squared, minus four lots of 2x, plus four.

Then we need to subtract one from the whole thing.

We get 4x squared, minus 8x, plus three and again I've put in my original function and then I've done f(2x), minus one, which is my transformed function.

And that looks like the graph we had before.

So it looks like we were right and if I type in 4x squared, minus 8x, plus three, we can see that graph lies over my transformed graph.

So it must have the same equation.

Time to have a practise, I'd like you to use the graph of Y equals X squared to draw each of these.

Feel free to use graph and software to help you.

You could type in Y equals X squared and then type in these new equations and see if they look like yours.

Try doing it yourself first though.

For two you're gonna want to complete the square or rearrange some of these formats to help you, off you go.

And you've got three more to do.

Alright, Sam has drawn the graph of Y equals X squared and applied two transformations to get this new graph, I'd like you to describe the transformation Sam could have applied to get the new graph.

What will be the equation for the new parabola? Laura has drawn the graph of Y equal X squared plus two x and applied two transformations.

Can you describe the transformations and then write the equation of the new parabola.

And another one to do, give this one a go.

Let's have a look, pause the video and check your graphs.

You should have a translation of three in the Y direction, a translation of three in the X direction, and a stretch of two in the Y direction, and a translation of negative four in the X direction, and a reflection in the X-axis.

For two, if you complete the square, you can see that's a translation of two in the negative X direction, three in the Y direction.

Then for B, that's a translation of negative three in the X direction and negative three in the Y direction.

And then we've got a stretch of two in the Y direction and a translation of negative three in the Y direction.

For D, we've got a stretch of three in the Y direction and a translation of negative four in the Y direction.

For E, we've got a stretch of two in the Y direction, a translation of negative two in the X direction.

You can do that in either order.

For F, we've got a reflection in the X-axis and then a translation of three in the X direction.

Again, you can do those in either order.

So Sam could have a translation of two units in the positive X direction and five units in the negative Y direction.

Have a look at my working.

Your new equation will be Y equals X squared, minus 4x, minus one.

For Laura's, this is a reflection in the X-axis and a stretch of two in the Y direction.

The new equation then will be Y equals negative 2x squared, minus 4x.

And for Alex's, I think you've got a choice.

You have a translation of four units in the positive X direction and a stretch of 1/2 in the Y direction or a reflection in the Y-axis and a stretch of 1/2 in the Y direction.

So if you use those two different ideas, you should get the same equation.

You get Y equals X squared minus 4x.

Please take your time to look at my working for each of those options.

Fantastic, I hope you enjoyed playing around with those ideas today.

There were some tough ideas, but at this stage we can use that technology to help us explore what is happening.

It doesn't matter if we can't see it straight away at the moment, we're just beginning to explore all these ideas.

So thank you for persevering and I hope that you are seeing how these things can be really useful and seeing how you can develop your skills even further than you already have.

We are all amazing mathematicians and we know that part of being an amazing mathematician is making predictions that don't turn out to be correct, but knowing how to then check that and changing our ideas to explore something new and that is really powerful.

Thank you for joining me.

I hope you have a fantastic rest of your day and you enjoy using this math in the future.