[Music] in this video we're going to look at transformations of graphs we're going to start by drawing the graph of y = x^2 we'll do this by substituting X values to work out the corresponding y values so let's start with xal 0 or 0 squ is also 0er so we have the point 0 0 then when X is 1 we have 1 squ which is also 1 so we get the0 1 1 if x is 2 we do 2^ 2 which is 4 so we get the 2 4 and then when X is 3 3^2 is 9 so the 39 when we substitute in the negative numbers we actually get the same values since - 1^2 is also 1 -2 is also 4 and -32 is also 9 so this is the graph of y = x^2 now let's think about the graph of y = X2 + 3 so the equation of the graphs is the same we've just added three to it so when we substitute in 0 we'll do 0 squ like we did before which is 0o but then we need to add three so the point is 03 when we substitute in one we do one squar just like before but then add three which is four so we get the 0.14 if we carry this on with x = 2 we get the 27 and then when X is 3 we get 312 and as before when we substitute in the negative values we get the same yv values out so the graph of y = X2 + 3 looks like this notice how the point 0 0 moved up three units to the 03 and in fact all of the points on the red graph moved up three units to the corresponding point on the blue graph this means that if we have a graph and its equation and then we add a constant number to that equation the resulting graph will be the same shape but it will be shifted up by a number of units equal to the number that we added the word we use for this is a translation in maths a translation just means to move something so we could say that the blue graph here is a translation of the red graph it's been translated three units upwards in the yde Direction it's a similar idea if you subtract a constant number from a graph as well so let's now have a look at the graph of y = x^2 subtract four so this is the same graph as the red graph apart from we've subtracted a constant number in this case four if we're going to subtract a number rather than moving upwards we move downwards so we need to take all of the points on the red graph and move them down four units so if we take 0 0 first that'll move down to 0-4 then if we look at 1 1 and -1 1 they'll move down here to 1 3 and13 then if we move all of the points down by four units this will be the graph of y = x^2 subtract 4 we can summarize what we've learned here using function notation so if we have a graph y = f ofx then the graph y = f ofx plus a where a is that constant number is a translation of a units up and also if we have the graph of y = f ofx then y = f ofx subtract a is also a translation but we go a units down instead so if you add or subtract a constant number to the equation of a graph this will move the graph upwards or downwards by that many units but what about if we take a function and we add or subtract a number inside the function so we alter the input rather than the output we're now going to look at f ofx plus a and F ofx subtract a let's go back to our original graph of y = x^2 which look like this if we to write this as a function we would say F ofx = x^2 we now want to consider the function f ofx + 2 and this is not the same as adding two to the function we're adding two to the input of the function so the X has changed to an X+ 2 to work out this function we replace all of the X's with X+ 2s so instead of x^2 we've got x + 2^ 2 so let's now draw the graph of y = x + 2 all 2 to do do this we're going to substitute in some points again so let's start with x = 0 so we need to do 0 + 2^ 2 well 0 + 2 is 2 and if you square two you get four so we get the point 04 when X is 1 we do 1 + 2 all squar 1 + 2 is 3 and 3 squ is 9 so we get the0 1 n when we go to x = 2 we have 2 + 2 all 2 2 + 2 is 4 and 4 2 is 16 this one unfortunately won't fit on the graph so let's go backwards and look at x = -1 -1 + 2 well that's just 1 and 1^ 2 is 1 so when X is - 1 Y is positive 1 now X = NE -2 well -2 + 2 is actually 0 and 0^ 2 is 0 so we get the point -2 0 then X is -3 -3 + 2 is -1 and the square of1 is 1 so we get the point -3 1 two more points here when X is -4 we get a positive4 so -44 and when X is5 we we get a positive 9 so 59 so the resulting graph looks like this this time each of the points on the graph has been translated horizontally it's gone to the left by two units most people find it initially surprising that the function moves to the left here and not to the right we added two to the input of the function and horizontally we tend to think of adding is moving to the right rather than to the left but the function does actually move to the left by two units if you give it some thought though this does actually make sense each of the inputs is going to have two added to it before we Square it so if you take an output from the original equation for example 9 9 will have come from 3 squared so we need to reduce the input of three by two to account for that extra two that we're about to add so if you input one and then add two to it you end up with three which when you square is nine so the input needs to be two lower than it was before to achieve the same output value which is why the graph moves to the left you can probably guess what happens if we subtract a value from the the input of the function so if we did x - 3 all 2 when we added two the function translated two units to the left if we subtract three it's going to translate three units but this time to the right so if we take each of these crosses and move them three units to the right we end up with the graph of y = x - 3 all 2 you can see here that each of these points has been moved three units to the right if we now return to our previous statements we've got two more which we can add so if yal f ofx then the function y = f ofx + a is a translation of a units to the left if yal F ofx then the function y = f of xus a is a translation of a units to the right now let's take a look at some exam questions on this topic sometimes an exam question will give you a graph for example here is the graph of yal sin of X and ask you to draw some graphs using this one so we might be asked to sketch the graph of y = sin of x + 2 for X values between 0 and 300 60° you can think of the original graph the red one as yal F ofx in which case this graph must be yal f of x + 2 since it's the same graph we've just added two to the function we know that when we add two to a function we translate the graph vertically and it goes up by two units so all we need to do is Mark on all of the key points of the original graph and then move each of these points up two units so the point at 0 0 here will go up to 02 the point that's at 91 will go up to 93 the one that's at 180 will go up to 182 2701 will go up to 270 positive 1 and 360 0 becomes 362 we can now connect up these points with that nice move curve and this is the graph of y = sin of x + 2 what about if we were asked to sketch this graph y = sin of x - 1 so if y sin of X is the F ofx this is f ofx - 1 since we've just subtracted one from that function this means we translate the graph down by one unit so if we take those key points again and move all of those points down by one unit we'll find the resulting graph I would always recommend marking on these points and doing them one at a time before you draw the final graph at the end so something like this let's have a look at some more examples on this graph so what if we were asked to sketch the graph of y = sin of x + 90 this time we're talking about changing the input of the function the input X has been changed to x + 90 so we have f of x + 990 we know that when we add a number inside the function it seems to do the opposite of what we might normally expect so instead of moving to the right 90 we're going to move to the left 90 let's take all of these key points and move them to the left 90 so this point here 0 0 will move to 9 0 but I don't actually need this point because it asks me to sketch values from 0 to 360 so we can remove that one the point at the top at 91 will move across to 01 then 180 0 will move to 90 0 2701 will become 1801 and 360 0 will become 270 0 so we end up with this graph here now we haven't actually finished because we need to go all the way up to 360 so if we imagine continuing the red graph like this then this point here would move to the left 90° so we should draw in this bit as well and that would be the completed graph of y = sin of x + 90 what about if we did the graph of y of x - 45 in this one we've also changed the input but we've subtracted a number so this is f of x - 45 when we subtract a number inside the input of the function we're going to move the function right so we need to go right 45° so if we now Mark on all of the points of the original function and we're going to move those right 45° 45 is half of 90 so we need to go to the right 1 square for each of these now you can see the final Point goes beyond 360 so we don't need that one and if we we connect up these points with a nice wave shape we end up with this graph here so this is the graph of y = sin of x - 45 sometimes you could be asked to do more than one transformation in the same question so let's take a set of axes and here is the function yal f ofx which looks like this notice this time we haven't been given the equation of the function we've just been told it's f ofx a question could ask us to sketch the graph of y = f of x + 1 and then subtract 3 this + one here is inside the IM so this is going to translate the function in the horizontal Direction it does the opposite of what we expect so when we add one we actually move to the left one unit so this tells us to move one left and this -3 here is outside of the function so it means we're going to move in the y direction and it's going to go down three since it's negative so all we need to do with this function is move it one left and down three so what we want to do is Mark on some of the key points from the original function preferably ones that have integer values so these ones here then once we've marked on all of these key points we're going to move that whole function to the left one unit and then down three so this would be the resulting function of F ofx + 1 subtract 3 so far we've learned about four different Transformations you know how to translate a graph upwards downwards to the left and to the right but there are two more Transformations that you need to know about and we're going to look at those now let's consider the graph of y = 2x - 4 that would be a straight line that looks like this so if we said that this was our function f of X = 2x - 4 we're now going to consider the function minus F ofx if you put a minus in front of a function here you're multiplying it by -1 so we need to take the original function and then multiply this by1 so1 LS of 2x subtract 4 if you multiply this bracket out you do -1 * 2x which is -2X and then -1 * -4 which is actually a positive 4 so the graph of negative f ofx is - 2x + 4 notice how this is the same as the original function just all of the signs have changed so instead of positive 2x -2X instead of -4 pos4 let's now draw the graph of Y = -2x + 4 that graph would look like this if you look closely at this you can see that to get the blue graph from the red graph we've just reflected it in the x-axis so if we take a point on the original one 1 -2 this has been reflected up to 1 pos2 if if we take another point above the axis like 4 4 this has been reflected in the x-axis down to 4 -4 this makes sense based on what we said earlier the functions are exactly the same apart from the signs are different so when you input a value you get exactly the same output value apart from the sign has been changed so if you ever have the function negative f ofx it's a reflection in the x-axis and you can do this by moving all of the points that are below the x-axis the same distance above and all of the points that are above the same distance below below now we have just one more type of function to look at this time the negative is not going to be applied to the whole function but just to the input so F ofx to do F of negx we take the original function which was 2x - 4 but we replace all of the X's with negative X like this if you expanded this out we do 2 * X which is -2X and then the -4 remains unchanged so if I draw on the graph for you of -2X subtract 4 it looks like this this time the graph has been reflected in the Y AIS rather than the xaxis you can see this because if we take a point on the original graph like 32 this has been reflected in the y- axis across to -32 and if we take a point to the left of the y axis -1 -6 this has been reflected across to POS 1 -6 so when we do the graph of f ofx we reflect the points in the y- AIS we can add these to our previous statements so if yal F ofx then the the graph of y = f ofx is a reflection in the x- axis if y = f ofx then the graph of y = f ofx is a reflection in the y- axis these are all of the transformations of graphs you need to know at gcsc level now let's have a look at some exam questions using these Transformations so here we have the graph y = f ofx and we're going to be asked to draw the graph of y = minus F ofx since this negative here is applied to the whole function this is a reflection in the x-axis so we need to take each of the points on the red function and reflect them in the x-axis I'm going to do this by starting at the left hand side of the function over here at -43 to reflect this in the x-axis I need to count down how many squares it takes to get to the x-axis and then go this many squares once more so to get from this point to the x-axis I go down three squares so I need to go past the xaxis by three squares then for this point here we're only one away from the xaxis so I go one point past it like this now this point here is quite interesting because it's on the x-axis if we reflect in the x-axis then this point won't change this point here is one square below so it'll move one square above this point here is on the axis once again so it won't change this point here is to above so it will go to below this one here is three above so it'll go three below this one here is almost one square above so it'll go almost one square below this one's on the x-axis so it will remain unchanged and this point here is one square below so it'll go one square above there's no rule that tells you how many of these points you need to do but I would try and do all of the points that have integer valued coordinates and also all of the ones that are on the axis so if we join these up with a smooth curve we end up with this graph here and that's the answer to the question now let's use this graph to do a second question so this time we're going to sketch the graph of y = f ofx the negative this time is on the input of the function so we're going to reflect in the y- axis this is really similar to what we did before but we reflect in the y- axis instead so let's start with this point here rather than going down we go horizontally across four squares until we hit the y- axis so we must go four squares past this point to here then for this point here we're three squares away from the axis so we go three squares across this point down here is one square away from the y-axis so one square across this point here is on the axxis so it remains unchanged this point here is two squares away from the axis so we go two squares across this one is four so we go four across and this one all the way down here is five squares so we go five squares across if you join all of these up with a nice moveth curve like this you end up with this as the answer to the question sometimes rather than translating the whole graph we only need to consider what happens to one of the coordinates let's have a look at one of these now so here's the graph of y = f ofx and we're told that the point a is on this graph and it's -13 so that's this point here the question might say the graph of y f ofx is transformed to the graph y = f ofx + 1 + 2 we're told the point a is going to be mapped to point B under this transformation and we need to work out the coordinates of point B so if we look at this transformation here we've got a plus one on the input of the function we know this moves it left one unit this plus two that's been added to the function moves it up two units so all we need to do is take the point a and move it one left and two up and that will be the point B so we didn't need to translate the entire graph we just did the point that we were interested in so the coordinates of point B are -25 let's try a second example of this so what if this time the function was yal F ofx + 2 this time point a is mapped to point C and we need to find its coordinates so if we look at at the function we've been given this negative X here tells us we need to do a reflection in the y- AIS this plus two here tells us to move the graph up to units now for this function it doesn't matter which order we apply these in we'll still end up in the same place so let's start by doing the reflection in the y- AIS and then move up two notice how this would be exactly the same as if we moved up two and then reflected in the y-axis we end up at the same point so this must be the coordinates of C which are at 15 sometimes the order in which you apply the Transformations does matter so for this question here we're going to sketch the graph of y = f ofx + 3 this negative here tells us to do a reflection in the xaxis and this plus three tells us to move the graph up three but it will make a difference which order we apply these in if we take the original graph and reflect it in the xaxis first we end up with this graph here and then if we move it up through units this is the final position so we have this graph here or if we take the original graph but move it up three first and then reflect it in the y- AIS we end up with this graph here the reason we end up with two different answers here is because both of these Transformations apply to the Y direction if you think about it a reflection in the x-axis is happening in the y direction and so is moving the graph three upwards so we do need to take particular care in which order we apply these the graph is actually correct in this scenario is the purple one so we should have done the reflection in the x-axis first so if you end up with a situation where you have netive F ofx + 3 you do the reflection first and then translate there's a second example where we also need to take care so if we had to do the graph of f ofx + 3 so in this one the negative inside the function here is going to be a reflection in the y- axis and this plus three here is going to move it three units to the left so if we take the graph and do the reflection first the reflection in the y- axis looks like this and then move it three units to the left we end up here If instead we take the graph and move at three units to the left first and then reflect this in the y- axis we end up with this graph here so once again you can see there are two different graphs this time it's the green graph that's correct so if we end up with a function like this we do the translation first and then we do the reflection second so translate then reflect so in these situations the order in which you apply the Transformations does matter sometimes exam questions don't have a diagram of the graph at all for example graph C has equation y = f ofx graph C is transformed to give graph D graph D has the equation y = f of x - 4 - 6 the point on graph C with coordinates 31 is mapped to the point p and we need to work out the coordinates of the point P so we don't actually need the drawn graph to do this question if we look at the resulting transformation here we've got a NE -4 inside the input which means four units to the right and then we subtract six from the whole function so this means to go down six so we can apply these two translations to the coordinate here if we were to move this coordinate right four the x coordinate would increaseed by four so instead of being three it will go up to seven if we move it down six the y coordinate will go down six so we do 1 subtract 6 which is -5 let's try a second example like this so this time we're going to have a different transformation F ofx + 1 and we have the point 59 which is going to be mapped to the point p and we need to work out the coordinates of P again so this time we have F ofx which is a reflection in the y- axis and then plus one which is moving up one so we can just apply both of these to the coordinate 59 if we're going to reflect in the y- axis the y-coordinate remains unchanged but the x coordinate will take the opposite sign so when the x coordinate was originally five it will now become -5 then we've got to do this translation so we need to move up one which will mean the y-coordinate will increase by 1 from 9 to 10 sometimes in exam questions we don't need to deal with a graph or coordinates we're just dealing with the function itself so let's take a look at this one graph P has the equation Y = x^2 - 5 graph p is reflected in the x-axis to give graph Q we need to find the equation of graph Q so in this one graph p is reflected in the x- axis we know this is the transformation y = f ofx if you think of the original function as being y = FX then we've got FX = x^2 - 5 if we're going to do Nega F ofx we just take the original function but apply a negative to it so we have y = x^2 - 5 if you expand out this bracket you get y = x^2 then -1 * -5 is POS 5 so the equation of graph Q is x^2 + 5 let's try another example like this so graph P has the equation Y = X Cub + 2x + 1 graph p is translated three units to the left to give graph q and we need to find the equation of graph q and we've been asked to give it in a particular form y = x Cub + a X2 + B X+ C so this time the transformation is a translation three units to the left we know this is f of x + 3 since it does the opposite of what you might expect to do f of x + 3 we're going to take this original function up here but replace all of the X's with x + 3s so we end up with Y = instead of X cubed x + 3 cubed instead of 2x two lots of x + 3 and a plus one at the end so we need to expand out all of this because they've asked for it in this particular form when we expand out x + 3 or cubed this is x + 3 * x + 3 * x + 3 so we have a triple bracket to expand here I'm not going to go into detail about how you expand this triple bracket that's a whole another video I'll put a link to that one in this video's description but the result from the expansion would be y = x Cub + 9 x^2 + 27 x + 27 we can expand this bracket here quite easily though 2 * X is 2X and 2 * 3 is 6 and of course we've got this plus one at the end now all we need to do is simplify we've got y = x Cub + 9 x^2 then we can collect together these like terms here 27x + 2x is 29x and then also the constant terms here 27 + 6 + 1 is 34 and that's your answer to the question you can see it now matches the form they asked for X Cub + ax2 plus BX plus C thank you for watching this video I hope you found it useful check out the one I think you should watch next And subscribe so you don't miss out on future videos and check out the exam question booklets on this topic that I put in this video's description