Lesson Video: Function Transformations: Translations | Nagwa Lesson Video: Function Transformations: Translations | Nagwa

Lesson Video: Function Transformations: Translations Mathematics

In this video, we will learn how to identify function transformations involving horizontal and vertical shifts.

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Video Transcript

In this video, we will learn how to identify transformations involving horizontal and vertical shifts or translations. Let’s say that we’re given this function 𝑓 of π‘₯ equals two π‘₯ squared plus three. It’s important to remember that a function is just a series of inputs and outputs. In this case, the π‘₯ is the input. And when we substitute a specific input value into the expression on the right-hand side and evaluate it, then we get a particular output. We’ll now take a look at the different ways in which we can change the input or the output values of a function. And we’ll see how those changes affect the way the function looks.

When we first started learning about functions, we may have considered function machines. In a function machine, when π‘₯ is the input, then 𝑓 of π‘₯ or the function of π‘₯ represents the output. We can think of any of the transformations in one of two ways. The first way is by changing the output values. For example, we might add on three to every one of the outputs. This means that the input values stay the same, but every 𝑦-value would have three added on. Alternatively, we might change the input values, for example, by adding on three. That means that instead of just having π‘₯ as the input, we would add on three to every one of the original input values and see what their corresponding output values would be.

It might be tempting to think that both of these options would produce the same result. However, we’ll start by looking at how changing the output changes a function. And then we’ll compare that with how changing the input changes a function.

Let’s say that we’re given this graph of 𝑦 equals 𝑓 of π‘₯. We can transform this function by adding four onto the output or 𝑓 of π‘₯ values. Let’s start by taking this coordinate zero, negative two on the original function. And we can even consider this in terms of a function machine. The coordinate zero, negative two means that for an input of zero, the output would be negative two. On the transformed function, however, we need to add four to each of the outputs. So an input of zero would give an output of negative two plus four. As negative two plus four is two, then that means that our transformed function has the coordinate zero, two.

We can take a different point on the original function. This coordinate of negative two, two means that for an input of negative two, the output is two. Transforming this by adding four means that we’ll have a new 𝑦-value of six. So we have the coordinate negative two, six. Taking a third coordinate, this time the point two, two, we can see how the transformed function would have the coordinate two, six. We can draw the transformed function 𝑦 equals 𝑓 of π‘₯ plus four by joining these three points with a curve.

So what exactly do we notice with this transformed function? Firstly, we notice that every coordinate has been shifted upwards by four units. We should also note that the transformed function is exactly the same shape and size as the original function. It’s just being translated. We could make a general note then that if we transform a function by adding π‘Ž units to it for any real value of π‘Ž, then the new function can be written as 𝑦 equals 𝑓 of π‘₯ plus π‘Ž and the graph shifts upwards by π‘Ž units.

Let’s see how this function would change if instead of adding four to the output, we subtract four. We can start once again by looking at this coordinate zero, negative two, remembering that that means for an input of zero, there’s a corresponding output of negative two. And so we can write the coordinate on our transformed function as zero, negative six. Looking at a different point then, the coordinate negative two, two, when that’s transformed, we subtract four from the output. Two minus four is negative two. So we know that the coordinate negative two, negative two will lie on the transformed function.

We might notice the pattern here that this time the function is being translated four units downwards. Verifying this with another coordinate, this time the point three, seven, we can see how this new coordinate of three, three will appear on the transformed function. We can then see what this transformed function would look like by drawing a smooth curve through the points. And a note about drawing functions, don’t worry if you can’t get your curve wonderfully smooth. What examiners are really looking for are the key points marked and the correct general idea of the shape of the function.

We can now add to the previous note if we’re transforming a function by subtracting π‘Ž for any real value of π‘Ž, then the new equation can be written as 𝑦 equals 𝑓 of π‘₯ minus π‘Ž. And the graph will shift downwards by π‘Ž units. In general then, if we’re transforming a function by adding or subtracting values to the output, then the graph will translate in a vertical direction. Now that we have looked at changing the outputs, let’s see what happens when we change the inputs of a function. We can keep the same representation of 𝑦 equals 𝑓 of π‘₯. And this time, we’re going to change the input values by adding on three.

We can take this coordinate of zero, negative two remembering that that means for this function an input of zero gives a corresponding output of negative two. When the input was π‘₯, this output was nice and simple to figure out. However, let’s see what happens when the input is actually π‘₯ plus three. When the input of zero is actually equal to π‘₯ plus three, then that means that the value of π‘₯ must be negative three since negative three plus three gives us zero. This means that we can plot one of the coordinates of the transformed function at negative three, negative two.

We can now check a different coordinate. Let’s look at the point negative two, two. Remember, that means that when the input is negative two, the output is two. So when π‘₯ plus three is equal to the input negative two, that means that the value of π‘₯ must be negative five. This time, the coordinate that we’re plotting will be negative five, two. You might already be starting to spot a pattern. It looks as though the transformed graph is three units to the left of the original function. We can check another coordinate just to be sure. We can take the coordinate two, two. When we’re transforming the function, we’re saying that this input of π‘₯ plus three is two. So that means that π‘₯ must be equal to negative one.

Plotting this coordinate on our graph, we can confirm that there’s been a translation of three units to the left. The transformed function 𝑦 equals 𝑓 of π‘₯ plus three would look like this. We can make a note then that if we transform a function by adding π‘Ž to the input values for any positive real value of π‘Ž, then the new function can be written as 𝑦 equals 𝑓 of π‘₯ plus π‘Ž and the graph will shift left by π‘Ž units.

We can now consider what happens when we transform a function by subtracting three units from the input. We can start once more with this coordinate of zero, negative two. Remember that once again we don’t have this nice, simple situation where π‘₯ equals zero is considered as the input, but instead we have π‘₯ minus three going into the function machine. So when π‘₯ minus three is equal to zero, that means that π‘₯ must be equal to three. And that’s our first coordinate: three, negative two.

Taking another coordinate, this time the coordinate three, seven, remembering that when we’re transforming it, we’re saying that the input three is actually equal to π‘₯ minus three. So the π‘₯-value is six, and it’s giving us the coordinate of six, seven. So this time we have a translation of three units to the right. And that’s true for any coordinate we might choose. We can join the points to get the graph of the transformed function 𝑦 equals 𝑓 of π‘₯ minus three like this.

For our notes then, we can say that when we transform a function by subtracting π‘Ž from the input for any real value of π‘Ž, then the new equation can be written as 𝑦 equals 𝑓 of π‘₯ minus π‘Ž and the graph shifts right by π‘Ž units. This transformation can actually feel a little bit unsettling. We’ll think that if we add π‘Ž units, then the transformation will be right. But in fact the reverse is true. It will shift left. And when we subtract π‘Ž from the input, we’ll not be translating left; we’ll be translating to the right.

It can be worthwhile learning these transformations, but it’s also very important to understand what’s happening and why. We’ll now look at some questions involving the transformation of functions.

Which of the following is the graph of 𝑓 of π‘₯ equals the cube root of π‘₯ minus one?

There are several different ways in which we can approach a question like this. One way is to take several different input values or π‘₯-values and see what their corresponding output or 𝑓 of π‘₯ values would be. Let’s say that we take the values of π‘₯ of negative one, zero, and one. When π‘₯ is equal to negative one, the cube root of negative one is negative one. And then subtracting one from that will give us negative two. When π‘₯ is equal to zero, the cube root of zero is zero, and subtracting one will give us negative one. Finally, when π‘₯ is equal to one, the cube root of one is one, and subtracting one will give us an output of zero.

We then know that the graph must contain the coordinates negative one, negative two; zero, negative one; and one, zero. The only one of these graphs which passes through these points is that given in option (b). And so that would be the answer. An alternative method of solving this question would be one involving the transformation of functions. We can consider the function 𝑓 of π‘₯ equals the cube root of π‘₯. And notice that the function that we were given is different because one has been subtracted from the output.

We can draw a quick sketch of the graph of 𝑓 of π‘₯ equals the cube root of π‘₯. We can then use the fact that if we transform a graph by subtracting π‘Ž units from the output for any real value of π‘Ž, then the graph will shift down by π‘Ž units. In this case, we were subtracting one from the output, so our graph of 𝑓 of π‘₯ equals the cube root of π‘₯ will shift downwards by one unit. The graph which demonstrates this is the one given in option (b). And so that is the answer.

Let’s take a look at another question.

The function 𝑦 equals 𝑓 of π‘₯ is translated eight down. Write in terms of 𝑓 of π‘₯ the equation of the translated graph.

In these questions, we’re given a function 𝑦 equals 𝑓 of π‘₯. We don’t know what it looks like. For example, it could be a straight line, a quadratic, or even something like a sine wave. However, it doesn’t matter what this function looks like as we’ll still be able to write the equation of this graph as we’re given the information about its translation.

Because we’re told that the transition is downwards, then we can use the fact that when we change the output of a function by subtracting π‘Ž for any real value of π‘Ž, then the graph shifts π‘Ž units down. And the new equation can be written as 𝑦 equals 𝑓 of π‘₯ minus π‘Ž. In this question, as we’re told that we’re shifting eight units down, so we subtract eight from the output. And so the equation of the translated graph is 𝑦 equals 𝑓 of π‘₯ minus eight.

We’ll now look at a similar question. Only, this time, the translation will be in a horizontal direction.

The function 𝑦 equals π‘₯ minus one times two π‘₯ minus three times four minus π‘₯ is translated two units in the direction of the positive π‘₯-axis. What is the equation of the resulting function?

In this question, we’re given this function and the translation. But don’t worry; we don’t need to actually graph this function in order to be able to answer the question. We can instead apply some logic to this transformation given that we’re told that it’s a translation in the direction of the positive π‘₯-axis. The direction of the positive π‘₯-axis is just another way of saying that this translation occurs to the right.

When a graph has shifted right by π‘Ž units, that means that the input has changed to π‘₯ minus π‘Ž for any real value of π‘Ž. So this graph, which has shifted right by two units, means that the input becomes π‘₯ minus two. To find the equation of the transformed function, then we take every value of π‘₯ and replace it with π‘₯ minus two. So that’s going to be relatively simple. For the first set of parentheses, we’ll have π‘₯ minus two minus one. For the second set of parentheses, we’ll need to be a little more careful. The original input of π‘₯ was multiplied by two. So that means that all of π‘₯ minus two will also need to be multiplied by two.

For the final expression, because we subtracted π‘₯ from four, we’ll also need to subtract all of π‘₯ minus two from four. We can then simplify our equation. The first set of parentheses will simplify to π‘₯ minus three. Next, when we distribute two across the parentheses containing π‘₯ minus two, we’ll have two π‘₯ minus four. And subtracting three from that will give us two π‘₯ minus seven.

In the final expression, distributing the negative across the parentheses gives us four minus π‘₯ plus two, which simplifies to six minus π‘₯. We can then give the answer for the equation as 𝑦 equals π‘₯ minus three times two π‘₯ minus seven times six minus π‘₯.

We can now look at a question involving more than one transformation.

The following graph is a transformation of the graph of 𝑦 equals the absolute value of π‘₯. What is the function it represents? Write your answer in a form related to the transformation.

In this question, we’re given the graph of a function which we’re told is the transformation of the graph of 𝑦 equals the absolute value of π‘₯. So it might be worth recalling what the graph of 𝑦 equals the absolute value of π‘₯ actually looks like. We can consider a similar range of π‘₯-values. When π‘₯ is a positive value, then its absolute value will be equivalent. However, when π‘₯ is a negative value, its absolute value will simply be its magnitude.

And so the graph of 𝑦 equals the absolute value of π‘₯ will look like this. So how is this graph different to the one that we were given? The first thing we might notice is that this graph looks upside down. In other words, there’s been a reflection in the π‘₯-axis. However, if we just had a reflection, then the graph would look like this in orange. So some other transformations must also have occurred.

Let’s consider how the top of each of these graphs is different. We can compare the coordinates zero, zero and negative one, four. This means that there must have been a shift of one left and four up, remembering that we’re transforming this graph of 𝑦 equals the absolute value of π‘₯ to the unknown function and not the other way round. We now need to write these three transformations in terms of an actual equation. A reflection in the π‘₯-axis occurs when the output 𝑓 of π‘₯ is mapped to negative 𝑓 of π‘₯. This is because every 𝑦-value on a transformed function is simply the negative of its original value.

A shift of one left means that the input value has changed to π‘₯ plus one. And the function can be written as 𝑦 equals 𝑓 of π‘₯ plus one. A shift of four up means that the output has increased by four. And the new function can be written as 𝑦 equals 𝑓 of π‘₯ plus four. We can now apply these three changes to the graph of 𝑦 equals the absolute value of π‘₯. Let’s start by applying the reflection. This gives us 𝑦 equals negative the absolute value of π‘₯. Then instead of having π‘₯ as the input, we’ll have π‘₯ plus one. Once we’ve done that, we can add four to the output. We can then give our answer in the slightly nicer form as 𝑦 equals four minus the absolute value of π‘₯ plus one. But either of these two function forms would be valid.

We can now summarize some of the key points of this video. We saw how we can transform a function by changing the input π‘₯ or the output 𝑓 of π‘₯. We can summarize the changes to the input or the output in this table. We can remember that the changes to the output are the ones that perform in the way we expect them to. By that we mean that when we add π‘Ž to the output, the graph shifts upwards by π‘Ž. When we subtract π‘Ž, the graph shifts downwards by π‘Ž.

The changes to the input, however, are the ones that are more tricky. For example, when we add π‘Ž to the input, we’ll think that the graph will shift right in a positive direction. But actually it shifts left. Subtracting π‘Ž means that the graph will shift right. So we need to be careful when we’re translating functions.

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