Lesson video

Thank you for joining us for this transformations of graphs lesson.

My name is Ms. Davies and I'm going to 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 video on transforming graphs of the form F of X plus A.

By the end of this lesson, you'll be able to recognise the effect of applying this transformation to a graph.

If you'd like a recap as to what a function or what a transformation is, pause the video and read through those now.

So we're going to explore then this transformation.

So what happens when we add a value to X before putting it into a function? So if we have a graph of the function F of X, the graph of the function F of X plus 5 is a translation of five units in the positive Y direction.

If you haven't looked at a lesson on that already, it may be something you want to explore first, what do you think will happen if we transform the graph of F of X into the graph of F of X plus 5? Why don't you try and make a prediction first.

Andeep says the new function is the original function plus five.

So I think this will still be a translation of five in the positive Y direction.

Lucas says we are adding five to X rather than the whole function.

So I think this will be a translation of five in the positive X direction.

Now, I'd like you to think about those two predictions and your own prediction and then click the link below to explore this.

You'll see that I've drawn a function.

If you drag the slider, it'll change the value of A in that bracket so it will- So we've got a function F of X at the moment, A is zero.

If we change A, you can see what's happening to the graph.

I've also fixed a coordinate there so you can see how that coordinate is changing.

Let's move the slider back and let's look at what happens when A is negative, right? Think about how you would describe what was happening.

Was either of our pupils correct? This time we can see the graph is moving left and right as a constant is added to the value of X within the function.

This is still a translation, but in the X direction, not the Y direction.

However, Lucas's statement was also incorrect.

What happened when you changed the value of A to 5? You might wanna look at the graph again and try this.

What happened when you changed the value of A to -2? Can you describe what is happening? Right.

I've now plotted a different function.

I'd like you to use the link below to see if your idea works for this different function.

Right.

We've got a quadratic function this time.

So let's see what happens when A is positive.

There's A is 1, 2, 3, 4, there's A is 5.

Did that fit with your prediction? You can see that coordinate that was 0,0 is now negative 5,0.

Let's take A back to 0.

There's our original function.

Let's see what happens when A is negative again, did that fit with your prediction? So when we change the value of A to 5, you should see that the graph moves five units left.

When we changed the value of A to -2, the original graph was translated two units right.

So adding a positive constant to the input of the function moved the graph in then negative X direction.

Adding a negative constant to the input, moves the graph in the positive X direction and you'll see that that worked for both of the graphs that we drew.

So when a constant is added to the input of the function, e.

g.

F of X plus A, the graph is translated by negative A in the X direction.

So there is F of X, I've defined it as 2X plus 2.

F of X plus 3, you'll see is a translation of three left.

So a translation of three in the negative X direction.

If I do F of X minus 3, that's a translation of three in the positive X direction.

And this is something we will explore why this happens in a later part of the lesson.

Quick check, F of X plus A transforms the graph of F of X by what? So this is a translation of negative A in the X direction.

Laura has drawn a function and a transformation.

She says these are the graphs of F of X and F of X minus 5.

Do you agree? No.

F of X minus five will be a translation of -5 in the Y direction because we're subtracting five once we've evaluated the function, this we can see is a translation of five units in the negative X direction.

She's changed her prediction.

She says it's F of X and F of X minus 5.

Is she correct now? No.

F of X minus 5 would be a translation of five units in the positive X direction.

If we're going for the function that she's labelled F of X and getting to the other function that I've drawn in purple, you can see that we're going in that opposite direction.

We're moving in the negative X direction, so that'll be F of X plus 5.

It's really important that you think carefully about which direction you're going in because we're affecting the input if we're adding a positive constant, we're moving left, we're adding a negative constant, we're moving right.

So knowing that F of X plus A transforms the graph F of X by translation of negative A in the X direction allows us to draw graphs of these transformations.

So here is a graph of the function F of X equals sin of X.

What do you think F of X plus 90 would look like? There you go.

I've drawn F of X plus 90.

The important part is the graph has been translated 90 units in the negative X direction.

This is the graph of the function, sin of X plus 90.

What do you think the graph of sign of X minus 180 would look like? Right.

That would be a translation of 180 in the positive X direction.

We can apply this graph transformation to any function.

We just need to translate each point by negative A in the X direction.

Here's the graph with equation Y equals F of X, dunno what the function is.

What would the graph of Y equal F of X plus 4 look like? Well, it'll be a translation of four in the negative X direction.

So I've picked all the integer coordinates that I can see, including the turning points and the Y-intercept, I have translated them all left four, There's going to be my new coordinates, there's my new graph.

I'd like you to match up the transformations with their graphs.

So for A, that's going to be G because a translation of two in the negative X direction.

B, you'll see we're adding two to the whole function.

So that's a translation of two in the positive Y direction.

So that's going to be E.

C is going to be a translation of two in the positive X direction.

So that's going to be F.

And D, that's going to be a translation of two in the negative Y direction.

So we're seeing here the difference between transforming A function F of X plus A and transforming a function of F of X plus A.

Here's the graph of sin of X, which of these is a graph of sin of X minus 90? So remember that's going to be a translation of 90 in the positive X direction.

So fix some points, work out where they're going to translate to and it's A.

Time for practise.

Here's the graph Y equal F of X.

I'd like you to draw the graph Y equals F of X minus 3 and then describe the transformation.

Then you've got the same function, I'd like you to draw the graph of Y equals F of X plus 2 and describe that transformation.

For two, you've got a different function this time we can see that it is a quadratic because it's formed the parabola.

Can you draw the graph Y equals G of X minus 4, and then the graph Y equals G of X minus 4.

For each one, describe the transformation.

Question three.

You've got Y equals H of X, you've got cubic function this time.

Can you draw the graph Y equals H of X plus 5 and then I've drawn a new graph, Y equals H of X plus A.

What is the value of A? For four, you've got the graph of Y equals cos of X and Y equals cos of X plus A.

What could the value of A be? Then draw the graph with equation Y equal cos of X minus 180 on the same axes.

Let's have a look.

So the graph has been translated three units in the positive X direction.

For C, we need to be translating two units in the negative X direction.

So check your graph.

For 2A, we have a translation of four units in the positive X direction, so you should have a turning point now on 1,0.

For C, that's a translation four units in the negative Y direction.

So make sure that you've got the distinction between those two.

For three, H of X plus 5 should be a translation five units in the negative X direction.

So five units left, just check your key coordinates.

And B, A should be positive 2, that's the graph H of X plus 2.

So there were some options here, but because the graph of cos of X repeats, it's actually impossible to know exactly what A is.

Sometimes different translations can produce graphs that look the same, but we don't know if the coordinate 0,1, for example, has moved to the coordinate 271 or if it's moved to the coordinate -91 or plenty of other choices.

If we don't know which one that coordinate has moved to, we don't know what the translation is.

Some suggestions you might have come up with, you could have said that it was Y equals cause of X plus 90.

So that A is 90 and that would move it 90 left or you might have said you thought that Y was cos of X minus 270, which is 270 in the positive X direction.

But actually it's impossible to know which one this transformation is because all those would look the same.

Cos of X minus 180 then will look like this.

I've drawn it in green, key coordinates 0,-1 90,0, 180,1, 270,0, and so on.

Let's have a go then at sketching these transformations we have discovered that F of X per se transforms the graph of F of X by a translation of negative A in the X direction.

Let's have a look at why this happens.

And the key point here is Y is adding a positive constant to the input, move it in the negative X direction, but adding a negative constant to the input, moves it in the positive X direction and that's what we're going to look at.

Now if we return to F of X plus 1, that means we evaluate the function and then add one to it.

We're adding one to the output of F of X, but here with F of X plus 1, we add one to the value of X and then substitute this new value into the function.

We're adding one to the input of F of X.

So let's look at a table of values.

So this is a table of values for 2X plus 1, and now we're going to evaluate a table of values for F of X plus 1.

That means our first input is -2, and -2 plus 1 is -1.

So that first value is actually going to be F of -1, which is -1.

We saw that in our previous table.

Right.

Well, our next value is F of -1 plus 1, which means F of 0.

So if we evaluate F of 0, two dots of zero plus 1 is 1.

And again we can see that from our previous table.

Our next one is actually going to be F of 1, which is three and so on.

And you should now be able to see the relationship between our two tables.

The values in the first table being shifted to get the values in the next table.

Let's see what they look like when plotted.

So I'm going to start by plotting -2, -3.

Then I'm going to plot -1, -1, and -2, -1.

You can see they have the same output, but for a different input.

Then here I'm going to plot 0,1 and -1, 1.

Again, they've got the same output for a different input and keep going.

And I'm also going to plot 3,9.

That would be 4,9 if I was plotting F of X.

So for the new function, the X value one less than the original function will have the same output, whereas negative one produced the output of negative one in our original function, negative two one less produced the output of negative one in our new function.

So what would the equation of the new graph be? Well, if we're evaluating F of X plus 1, I need to substitute X plus 1 instead of X.

So I'm going to do two lots of X plus 1 plus 1, expanding gives me 2X plus 2 plus 1, which is 2X plus 3.

Right.

Your turn, I'd like you to populate the table of values for F of X minus 2 and then draw the transformed graph.

So for the first one we're going to do F of -2 minus 2, which is F of -4.

So three lots of -4 plus 4 is -8.

For the next one we're going to do -1 minus 2, which is -3.

So we're doing an F of -3, which is -5.

Then we've got F of 0 minus 2, so that's F of -2.

We can actually see what F of -2 is for the original table, that's going to be -2.

Then we've got F of 1 minus 2, which is F of -1.

And again we can see that from the table F of -1 is 1.

Then I've got F of 0, which is 4.

Then my last one was F of 3 minus 2, which is F of 1, which was 7.

And then we can draw that in.

And again you can see that every coordinate has shifted right to this time.

So we can use our knowledge of graph transformations to identify the new coordinates of a point on a graph after a transformation.

We've started doing this a little bit already to help us draw these transformations.

So the 0.

48 is on Y equals F of X.

What will be the coordinates of the new point when the graph is transformed to Y equals F of X plus 10? What do you think? Andy says we are adding 10 to X, so 14,8.

Lucas says this is a translation in the negative X direction, so -6,8.

Are either of those correct? Do they fit with what you thought was going to happen? Right.

Lucas is correct.

If we think about F of -6 plus 10, that means we're actually doing F of 4.

So the output of the new function for an input of -6 is the same as the output of the original function for an input of 4.

So you can see that the Y coordinate has stayed the same, but the X coordinate has had 10 subtracted from it to give us the new X coordinate that produces the same Y value.

So here's a sketch of a cubic curve, the roots and the turn points are labelled.

If we wanted to sketch the graph of Y equals F of X plus 5, we would need to subtract five from the X coordinate of each point.

So 0,0 would become -5,0.

3,12 would become -2, 12, 9,0 would become 4,0, and that would be our new graph.

We can see clearly where the roots and the turning points are.

However, we no longer know the Y intercept.

So the roots because they've just moved left five are still roots.

The turning point is still a turning point, but the value that was the Y intercept is no longer the Y intercept, it's moved off the Y axis.

We can sketch a graph transformation using key features of a graph.

So we've gotta sketch the graph of this function.

To get the Y intercept, we just need to know what happens when X is 0 and to get a Y value of -12, you can see it's also the constant in our equation.

The roots we need to factorise.

So we get X plus 6 X minus 2.

The roots are when X is -6 and when X is 2.

So there are coordinates and the turning point, we need to complete the square to be half the coefficient of X, that would give us x plus two all squared, but that would give us X squared plus 4X plus 4, we want X squared plus 4X minus 12, so we need to subtract 16.

Our turning point then is -2, -16 because when X is -2, that bracket is 0.

So the whole function would be -16.

And then we can plot those and it's a positive parabola, so we are expecting it to have that shape there.

So what would the graph of F of X minus 7 look like? Well, this is going to be a translation of seven in the positive X direction.

So we need to add seven to the X coordinate of each.

This is going to look quite busy on our graph because we've got lots of coordinates on here now, and that's our new graph.

And again, you'll see that the roots have shifted seven right.

And the new routes are 1,0 and 9,0.

The turning point has shifted seven right now our new turning point is 5,-16 and the value that was the Y-intercept 0,-12 is now 7,-12 and no longer the Y-intercept right.

Let's look at the equation of the new graph.

Well, this was F of X minus 7.

So that means we've done X minus 7 squared plus 4, lots of X minus 7 minus 12.

And this is going to take a little bit of algebra manipulation to simplify.

So product of two binomials, square in that bracket gives us X squared minus 14X plus 49.

Then we've got plus 4X minus 28 minus 12.

Bit of simplifying gives us X squared minus 10X plus 9.

Your turn, I'd like you to sketch F of X equals X squared minus 10X.

Think about the Y-intercept, the roots and the turning points.

There we go.

So our Y intercept is 0,0, our root 0,0 and 10,0, and our turning point, 5,-25.

There it is.

Okay, I'd like you to use your sketch to sketch F of X plus 5.

So every coordinate is going to move five in the negative X direction.

So 0,0 becomes -5, 0, 10,0 becomes 5,0, 5,25 becomes 0,-25.

So let's plot those and again, we've got our two routes.

We've got our turning point.

Because of the function we had this time, we can actually see that our new turning point is actually the Y-intercept.

Right, your turn.

I'd like you to draw the graph of F of X equals 3X minus 6.

Then draw the two transformations and tell me what the equations of the transformed graphs are.

Question two, here's a graph with equation Y equals G of X.

I'd like you to give the coordinates of three points on the graph Y equals G of X plus 4.

You don't necessarily need to be able to plot 'em on the graph, just need to know what the coordinates would be.

For three, exactly the same, I've drawn you a sketch and I've labelled some of the key features.

Where would those coordinates move to? The graph was transformed to Y equals H of X minus 2.

For question four, I'd like you to sketch the graph of F of X equals X squared minus 6X plus 8.

I put in some structure to help you.

Can you find the Y-intercept, the roots and the turning point? And then also sketch the graph of Y equals F of X plus 6.

I'd like you to label the new coordinates of any of the key features.

Question five, you've got a new quadratic to sketch this time, then sketch the graph of Y equals F of X minus 10.

I'd like to know the equation of the new graph.

Be careful with your algebra manipulation.

And finally, can you sketch F of X equals negative X squared plus 6X? Think carefully, particularly when you are factorising and completing the square.

When you've done that, sketch the graph of Y equals F of X plus 4.

Making sure you label the new coordinates of any key features.

Let's have a look.

So you've got 3X minus six has a Y-intercept of -6 and a gradient of three, and you've got F of X plus 6, which is a translation of six in the negative X direction and F of X minus 2, which is a translation of two in the positive X direction.

The new equations then three lots of X plus 6 minus 6 is 3X plus 12, and three lots of X minus 2 minus 6 is 3X minus 12.

Then to help you, I've labelled the integer coordinate.

So you might have had any three of -4,0, -3,3, -2,4, 0,2, 2,0, 3,1, and 4,4.

I've just subtracted four from each of those X coordinates.

The Y coordinates staying the same.

For three, A is going to move to 1,0, we're adding two.

B, 2,3.

C, 3,4, and D, 5,0.

For four, our Y-intercept is 0,8, roots 4,0 and 2,0, and the turning point, 3,-1 The translation of F of X plus 6, they should all move six in the negative X direction.

So the Y-intercept has now moved to -6,8, no longer a Y-intercept.

Our roots have moved to -4,0, and -2,0.

They're still roots and our turning point has moved to -3,-1.

So this time if we factorised, we can see roots at -6,-2.

Completing the square gives us a turning point of -4,-4.

The Y-intercept is 0,12.

Then we need to add 10 to every X-coordinate.

So our roots become 4,0 and 8,0.

Our turning point becomes 6,-4.

The coordinate that was the Y-intercept is now at 10,12.

We don't know what the Y-intercept is from this diagram.

The equation of the new graph then X minus 10 all squared plus 8 lots of X minus 10 plus 12.

So making sure we're being careful when we're squaring our binomial and simplifying gives us X squared minus 12X plus 32.

That Y-intercept then is actually going to be at 0,32 now.

And finally the Y-intercept is going to be 0,0.

There's no constant in our equation.

Factorise, if I take out factor of negative X, I can write that as negative X multiplied by X minus 6.

So either negative x is zero or X minus 6 is 0 to get our roots, which gives you 0,0 and 6,0.

Turning point was a little bit trickier.

If you remove a factor of -1 and just look at X squared minus 6X, so it's negative X minus 3 all squared plus 9.

Little bit tricky that completing the square with a negative coefficient of X-squared.

The turning point then is at 3,9, it's a maximum as negative X minus 3 all squared has to be less than or equal to zero because X minus 3 all squared has to be greater than or equal to zero.

But multiply that by -1, it must be less than or equal to zero.

Right.

Sketch in the graph are key features are now -4,0 and 2,0 and our maximum is negative -1,9.

Fantastic.

We're really exploring that sketching quadratic skills here.

Thank you for joining us today.

I hope you enjoyed playing around with that transformation.

You might want to spend some time at learning some graphing software.

You can plot your own functions and then transform them in any way you like.

Think about what happens when you add certain constants to the input, what happens when you add certain constants to the output and just play around with that idea.

I hope you choose to learn with us again.

Have a great rest of your day.