Lesson video

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

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

There's lots of exciting things to explore today as well as things to recap that we've looked at before.

So make sure that you're in a nice, quiet place that you can explore this and that you pause the video and have a real think each time I offer you a question, and then I'll help you with any hints, any ideas that will help you develop your mathematics skills.

I'm really excited to explore this topic.

I hope you are, too.

So let's get started.

Welcome to this lesson where we're transforming graphs, y = f of x + a.

So we're gonna look at recognising the effect of applying the transformation of y = f of x + a to a graph.

So we're starting to transform graphs in different ways today.

If you need a reminder as to what a function is or what a transformation is, pause video and do that now.

So let's have a go at exploring y = f of x + a.

There's gonna be links to some Desmos files in this lesson, but you may also want to have some graphing software to explore this yourself.

So we can use the notation f of x to represent a function.

What does f of x + 5 mean? See if you can put this into words.

This means we evaluate the function for a given value of x, then add 5.

So it's the whole function plus 5.

Whatever the function defines, we can plot it on a graph.

How do you think the graphs of f of x and f of x + 5 will look different? Make an early prediction.

We're gonna explore this.

Okay, what about the graphs of f of x + 4 and f of x + 3, what would they look like? So what we're gonna do is we're gonna graph the family of functions f of x + a, where a can be any constant, and we're gonna explore how they are related to each other.

So keep hold of any predictions you have, we're gonna explore that now.

So f of x can be any function.

I have defined a function f of x, you just don't need to know what it is.

I'd like you to click the link below and explore what happens when different values of a are added to the function.

So if you follow that link, you'll be able to drag the slider and describe you notice.

Pause the video and do that now.

So here we have a function.

We don't know what the function is, that's absolutely fine.

And we're adding a, and a is a constant.

So you'll see at the moment that a is -3.

So if I move that slider back to a is 0, so this is now the function f of x.

Now let's look at what happens when I add a positive value of a.

So a is 1, a is 2.

Then if I move that back, and look at what happens when I add negative values of a.

I'd like you to think about what is happening.

How would you describe what is happening when we add a constant to this function? And then we'll look at what some of our Oak pupils think.

So Laura says, "The graph is moving left to right." Andeep says, "I think it's moving up and down." Lucas says, "It looks like it's moving diagonally to me." I wonder which one you've said.

It's actually quite hard to see from the graph I gave you.

We can see that adding a constant is translating the graph, it's moving it to a different position.

But to see the exact movement, we can look at the effect on a natural point.

So click on the new link, and I'd like you to describe the impact of adding a constant to the function now.

Can you improve your description? Maybe you were right the first time.

Follow that link and have a look.

So this time you can see that I've labelled a coordinate, I've labelled the y-intercept 0, 0.

Now let's see what happens when I add a constant.

Right, it's easier to see now that we're watching that coordinate what is happening as we're adding different constants to our function.

So let's look at those positive constants again.

That's adding 1, 2, 3, 4.

And now let's look what happens when we add -1, -2, -3, and so on.

How would you describe that this time? So what you should see now is that Andeep was correct.

Adding a constant to the function moves the graph up or down.

If the constant is positive, it moves the graph up.

The constant is negative, it moves the graph down.

We can see that by fixing that y-intercept, you can see how the y-intercepts moving.

Every other coordinate on that line will also be moving in the same way.

Now using mathematical language, we can say that adding a constant a translates the graph by a in the y direction.

So when we manipulate a function f of x by performing an operation, so if we're doing 5f of x, or f of x, + 5, or f of x + 5, or f of 5x, or all other manipulations, the graph is gonna be transformed to give a new graph.

We call that graph transformation or transforming a graph.

So these operations are gonna change the way the graph looks.

Now different manipulations have different effects on the graph.

Our focus today are graphs of the form f of x + a.

What we've seen is adding a constant a is a translation of the original graph by a in the y direction.

So here we've got f of x = 3x.

So if we do f of x + 5, which would be 3x + 5, you can see our graph has been translated five units in the positive y direction.

If we had f of x as x squared, then f of x - 2 would be x squared - 2.

You can see that's a translation of two units in the negative y direction 'cause it was f of x - 2.

Quick check.

f of x + a transforms the graph f of x by a what? Well done if you said translation of A in the y direction.

So knowing that f of x + a transforms the graph f of x by translation of a in the y direction allows us to draw graphs of transformations.

So here's the graph of the function f of x = sin of x.

What do you think f of x + 1 will look like? There we go.

The graph has been translated one unit in the positive y direction.

So every point on the graph has been translated one unit in the positive y direction.

The new function would be sin of x + 1.

You could call the graph y = sin of x + 1.

We do not need to know what the function is, we can apply this graph transformation to any function.

So here is the graph of a function g of x.

What would g of x - 3 look like? Every point will be translated three units in the negative y direction.

So you can see that I've picked some integer coordinates and then I've translated them three in the negative y direction, and that will be the new graph.

Quick check.

The graph of f of x is shown.

Which of these is the graph of f of x - 2? Well done if you picked D.

We need every point to be translated two in the negative y direction, and that should be that graph there.

We can use our knowledge of graph transformations to identify the new coordinates of a point on a graph after a transformation, and that may well help us do these transformations as well.

So the point 90, 1 is on y = sin of x.

What will be the coordinates of the new point when the graph is transformed to y = sin of x + 10? Laura says, "We're adding 10 so it'll be 100, 11." Andeep says, "I think we add 10 to x so it'll be 100, 1." Lucas says, "I think we add 10 to y so it'll be 90, 11.

Who do you think is correct? I wonder if you've changed your prediction or if one of those fits with what you thought.

Let's have a look.

There's f of x + 10.

It is a translation of 10 units in the positive y direction.

So the x coordinate will remain the same and the y coordinate will be 10 larger.

So there's the coordinate 90, 1.

We translate it 10 units up, it would become 90, 11.

Quick check then.

If the point with coordinates 2, 5 is on the graph y = f of x, which of these must be on the graph y = f of x + 2? Well done if you picked 2, 7.

Some of those other coordinates may be on the graph, but the point 2, 5 will become 2, 7 because it's been translated two units up.

Time for a practise.

Here is the graph y = f of x.

I'd like you to draw the graph y = f of x + 3 and describe the transformation in words, and I want you to think about that mathematical language.

Then I'd like you to do the same for the graph of y of g of x but drawing the graph y of g of x - 5 and describe the transformation.

Questions 3 and 4, you've got some different functions.

Again, look carefully at the transformation you are performing and describe it.

Question 5, here are the graphs y = f of x and the graph of y = f of x + a.

I'd like you to tell me what the value of a is.

For 6, here's the graph with equation y = g of x.

I'd like you to give the coordinates of three points on the graph y = g of x + 4.

For 7, here is the graph with equation y = cos of x.

Can you draw y = cos of x - 1? And for 8, I've sketched the graph y = h of x and I've told you a few key points.

What would the new coordinates be of the four marked points when the graph is transformed to y = h of x - 2? Let's have a look.

So check you've drawn the correct graph.

The graph has been translated three units in the positive y direction.

Well done if your answer is similar to that.

To help me with 2, I picked some integer coordinates and translated them five units in the negative y direction.

So check you've got the correct graph and again the correct description For 3, the graph has been translated two units in the positive y direction.

So again, check those key integer coordinates.

And for 4, we've got a cubic curve.

The graph has been translated two units in the negative y direction.

And again check those key features are in the correct place.

For question 5, the value of a must be 3 'cause this is a translation of three units in the positive y direction.

For 6, you might have picked some integer coordinates that are on the curve.

Then on the translated curve y = g of x + 4, they are gonna become 0,4, 1,7, 2,8, 4,6, 6,4, 7,5, and 8,8.

So you want three of those.

So there is a graph of y = cos of x - 1.

We just need to translate every coordinate on y = cos of x by a one in the negative y direction.

So just check all of your key points, particularly your maximum and minimums and then make sure you've got that nice, smooth graph shape.

For 8, so a will become -1, -2, b will become 0, 1, c will become 1, 2, and d will become 3, -2.

So we've now seen what happens when we apply this transformation.

Let's have a go at using it to sketch transformations of the form y = f of x + a.

So we know that this is a translation of a in the y direction.

Let's look at why this happens first.

So let's pick a function.

I've gone with f of x is 2x + 1, and there's a table of values for 2x + 1.

Now, if I did f of x + 2, that means we evaluate f of x and then add 2 to each output.

So that would be the table of values, so for f of x + 2, and you can see we've added two to every output.

So there's 0, 1, which would be on the graph y = f of x, and there's 0, 3 would be, which would be on the graph y = f of x + 2.

There's the next coordinate on each line, and the next coordinate, and so on.

And you can see for the same x coordinate, the translated graph has a y coordinate two bigger than the original graph.

We have kept the inputs the same, but each output has increased by two.

What would be the equation of the new graph? So the equation of the new graph would be f of x + 2, so 2x + 1, + 2, or 2x + 3, so it'd be y = 2x + 3.

So we can sketch a graph transformation using key features of a graph.

Let's have a go at sketching f of x = x squared - 4x.

We're gonna do a quick recap on how to do this.

So one of the key points we could put on our graph to help us sketch this would be the y-intercept.

Now, the y-intercept has an x coordinate of 0.

So if we substitute that into our equation, y = 0 squared - 4 lots of 0, which would be 0.

So the y-intercept is 0, 0.

And we can actually see that from our equation 'cause we have a constant of 0, and we know the constant is the y-intercept when it's in that form.

So let's plot that in our graph, that's 0, 0.

Other key features we could find would be the x-intercepts or the roots of the graph.

The x-intercepts have y coordinates of 0.

So that'd be 0 = x squared - 4x.

And this is a quadratic equation, so to solve we need to factorise.

So 0 = x lots of x - 4, it's just a single bracket.

And that means we've got two choices, either x has to be 0 or x - 4 or has to be 0.

If x - 4 is 0, then x is 4.

So we have two roots, 0 and 4.

So the x-intercepts are 0, 0 and 4, 0.

Now, we already have 0, 0 'cause it was also our y-intercept, so we can plot 4, 0.

Now we also need to identify any turning points 'cause we need to know the shape of the graph.

What we can do here is complete the square.

So if we look at the expression x squared - 4x.

To complete the square, we need to write that as a binomial squared.

So that would be x - 2 squared.

You half that coefficient of x , that'll help you get that part of the bracket.

But hang on a minute, x - 2 all squared would be x squared - 4x + 4.

We didn't want that, we wanted x squared - 4x, so you need to subtract that extra 4.

Now that it's in completed square form, we can find that minimum or maximum value.

So when you square a value, it's always 0 or greater.

So the smallest value of this function would be 0 squared - 4.

So that means the smallest this could be is -4.

Now that output was when that bracket is 0, so to make that bracket 0 is when the input is 2.

So the turning point must be when x is 2 and y is -4.

Now the coefficient of x squared is positive, so this is a positive parabola, and we can see this as well from the fact that our turning point is a minimum value, so that's our general shape.

Now what would the graph of the function f of x + 5 look like? To sketch it, we just need to add 5 to the y coordinate of each point.

What would be the equation of the new graph? Well, it would be x squared - 4x + 5.

And we can use these skills to help us draw quadratics.

Quick check then.

What is the y-intercept of the graph with equation y = f of x and f of x = x squared + 6x + 5? So the y-intercept is gonna be when x is 0, which gives you 5.

So our y-intercept is 0, 5.

It's also the constant when our equation is in that form.

I'd like you to fill in the gaps to work out the roots of the graph with equation y = f of x when f of x is x squared + 6x + 5.

Can you remember how to find the roots? So factorising gives us x + 1, x + 5, which gives us x = -1 and x = -5.

Your roots then are -1, 0 and -5, 0.

I'd like you to fill in the gaps to work out the turning point of this graph.

Remember the turning point can be found by completing the square.

So if we half the coefficient of x, we've got x + 3 all squared.

But remember that would give us x squared + 6x + 9.

We wanted x squared + 6x + 5.

We need to subtract 4.

The minimum value of this function then is -4.

Now this output occurs when x is -3 'cause the bracket would be 0, -3 add 3 is 0.

The coordinates of the minimum point, -3, -4.

So I'd now like you to use all of that to label this sketch for f of x + 2.

So there's all the key features we've previously calculated, but the graph I've drawn is f of x + 2.

So what are those coordinates that I've marked on? So the y-intercept that was 0, 5 becomes 0, 7.

The roots were -1, 0 and -5, 0.

Those coordinates become -5, 2 and -1, 2.

They're no longer the roots, are they? They're not on our x-axis anymore.

The turning point goes from -3, -4 to -3, -2.

And it's still our turning point, it's just got a new coordinate.

Interesting that our y-intercept became a new y-intercept.

Our turning point became our new turning point, but the roots when they were translated in the y direction were no longer roots, just two coordinates that will be on our curve.

Time to put that all into practise.

I'd like you to draw the graph of f of x = 4x - 2.

Then you've got two translated graphs, and I want to know the equations of your transformed graphs in the form y = mx + c.

Question 2, I'd like you to sketch the graph of f of x = x squared - 8x - 9.

So you need find the y-intercept, the root, and the turning points.

Then I want you to sketch the graph of y = f of x - 6, and I want you to label the new coordinates of any of the key features of the original graph.

For 3, same idea, sketch a graph, transform the graph.

But this time I also want to know what the equation of your new graph is.

Give that one a go.

Let's have a look.

So 4 x - 2 should have a gradient of 4 and a y-intercept of -2.

Notice though that I stretched the scale so that it fit on.

So although it does have a gradient of 4, on our graph it looks like a gradient of 2.

But you can see for every one we move in the x direction, we move four in the y direction.

f of x + 6 then should be a translation of 6 in the y direction.

Again, each grid line represents 2 on the y-axis.

Then f of x - 4 should be a translation of -4 in the y direction.

And again, check your scale for that one.

The equations of the transformed graphs will be y = 4x + 4 and y = 4x - 6.

Question 2, the y-intercept will be 0, -9, the roots -1, 0 and 9, 0, and the turning point 4, -25.

Now my graph got quite busy because I labelled all those coordinates.

Then I also need to translate them six in the negative y direction.

So those coordinates will become <v ->1, -6 and 9, -6,</v> they used to be the roots.

The y-intercept will translate to 0, -15.

The turning point to 4, -31.

For question 3, factorising helps us find the roots.

Completing the square will help you find the turning point.

So you get roots -5, 0 and 3, 0, turning point -1, -16, and a y-intercept at 0, -15.

When we sketch the graph of y = f of x + 5, each of those points translate five in the y direction.

So you get -5, 5 and 3, 5, which used to be the roots, obviously not anymore.

The new turning point is at -1, -11, and the new y-intercept is at 0, -10.

So we have seen then that when we manipulate f of x by performing in an operation, the graph is transformed.

And we looked at one specific type of transformation today, f of x + a transforms the graph of f of x by translation of a in the y direction.

And when sketching the graph, key features need to be labelled.

We practised finding the roots, we practised using completing the square to find the turning point, and then sketching that so we could sketch the transformation.

And that's a skill that you'll need to use for all your transformation of graphs work.

Using graphing software can be an effective way to explore the transformations of graphs.

So I suggest you spend some time, load some graphing software, play around with what happens when you add constants to different functions.

Well done for all your hard work today.

I really hope you enjoyed playing around with that idea and that you'll join us for another lesson in the future.