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Lesson Video: Combining Function Transformations Mathematics

In this video, we will learn how to identify function transformations that include a combination of translation, dilation, and reflection.

16:22

Video Transcript

In this video, we’ll learn how to identify function transformations which include a combination of translations, dilations, and reflections. We’ll begin by recalling the key function transformations that we should know. These can be grouped as transformations that cause horizontal changes to the function, and they’re achieved by implementing changes to the π‘₯-variable, and those that cause vertical changes, which are achieved by changing the function itself.

Given a function 𝑓 of π‘₯, the function 𝑓 of π‘₯ plus π‘Ž for some real number π‘Ž is a translation π‘Ž units to the left or by the vector negative π‘Ž, zero. And the function 𝑓 of π‘₯ plus 𝑏 is a translation 𝑏 units up or by the vector zero, 𝑏. Next, we can multiply π‘₯ by some constant 𝑐 giving us the function 𝑓 of 𝑐π‘₯. This leads to a horizontal stretch or compression by the scale factor one over 𝑐. And multiplying the entire function by some constant 𝑑 to give us 𝑑𝑓 of π‘₯ is a vertical stretch by a scale factor of 𝑑. Now, in fact, in our previous transformation, if 𝑐 is less than zero, this also results in a reflection in the 𝑦-axis. And if 𝑑 is less than zero, this results in a reflection in the π‘₯-axis.

In the simplest terms, we can separate the situation where 𝑐 and 𝑑 are equal to negative one. The function 𝑓 of negative π‘₯ is a reflection in the 𝑦-axis, and the function negative 𝑓 of π‘₯ is a reflection in the π‘₯-axis. As an aside, we notice that for horizontal transformations, the effects of additional multiplication are the opposite of what we would expect. For example, 𝑓 of π‘₯ plus three, that is, adding three to the π‘₯ before inputting it to the function, results in a geometric transformation which shifts the graph three units to the left. A multiplication by two of the value of π‘₯ results in a horizontal stretch by a scale factor of one-half. On the other hand, algebraic transformations on the function itself result in the expected geometric transformations. So with this in mind, let’s recap how to identify function transformations.

If the following graph is reflected in the π‘₯-axis and then shifted to the left by one unit and down by three units, what is the equation of the new graph?

Remember, given some function 𝑓 of π‘₯, the corresponding function negative 𝑓 of π‘₯ represents a reflection of the original graph in the π‘₯-axis. In order to represent a shift or translation left by one unit, we need to convert 𝑓 of π‘₯ onto the function 𝑓 of π‘₯ plus one. Finally, given some function 𝑓 of π‘₯, the corresponding function 𝑓 of π‘₯ minus three is found by translating the original function down by three units. So we’re going to take the original function, which is given by 𝑦 is equal to the absolute value of π‘₯, and then complete each of these transformations in turn.

To perform a reflection in the π‘₯-axis, the function 𝑦 equals the absolute value of π‘₯ needs to be changed to 𝑦 is equal to negative the absolute value of π‘₯. And the corresponding graph of this function is shown. Then, in order to shift this to the left one unit, we add one to the value of π‘₯. And we do this to the new function. So the function we have now is 𝑦 is equal to the negative of the modulus of π‘₯ plus one, and the corresponding graph of this function is shown.

Finally, in order to translate this graph three units down, we subtract three from the entire function. So we take the function we have and we subtract three from it, giving us 𝑦 equals negative the modulus of π‘₯ plus one minus three. And the corresponding graph of this function is shown. So after this series of transformations, the equation of the new graph is 𝑦 equals the negative modulus of π‘₯ plus one minus three.

In our next example, we’ll apply a combination of a vertical dilation or stretch and a reflection to a given function.

The function 𝑓 of π‘₯ equals π‘₯ plus two cubed plus three is stretched in the vertical direction by a scale factor of two then reflected in the 𝑦-axis. Write the equation of the transformed function 𝑔 of π‘₯.

There are two transformations that have been applied to our function 𝑓 of π‘₯. Firstly, it’s stretched in the vertical direction; then it’s reflected. Now, we might recall that in order to stretch a function in the vertical direction, we need to multiply that entire function by the scale factor. So 𝑓 of π‘₯ will be mapped onto two times 𝑓 of π‘₯ to achieve this. And given some function 𝑓 of π‘₯, the corresponding function 𝑓 of negative π‘₯ represents a reflection in the 𝑦-axis of that original function. Now, the order in which we apply these is important. So we will follow the order in the question, starting with the vertical stretch.

We have 𝑓 of π‘₯ equals π‘₯ plus two cubed plus three, and we’re going to multiply the entire function by two. When we do, we find that two 𝑓 of π‘₯ is equal to two times π‘₯ plus two cubed plus three. Then we can distribute this two across the parentheses and we have that two times 𝑓 of π‘₯ is two times π‘₯ plus two cubed plus six. Now that we’ve performed the stretch in the vertical direction, we’re going to perform a reflection in the 𝑦-axis.

To achieve this, we change the π‘₯ to a negative. In other words, we essentially multiply the value of π‘₯ by negative one. Since we’re starting off with our already transformed function, we need to change two 𝑓 of π‘₯ to two 𝑓 of negative π‘₯. And all we do here is we multiply the value of π‘₯ by negative one. So two 𝑓 of negative π‘₯ is two times negative π‘₯ plus two cubed plus six. And so we have the equation of the transformed function, which we can now write as 𝑔 of π‘₯. 𝑔 of π‘₯ is two times negative π‘₯ plus two cubed plus six.

In our previous examples, we found algebraic expressions of functions after a combination of transformations were applied in a given order. We’re now going to consider examples in the reverse. So given an algebraic expression of a function which is a result of a series of transformations, how do we retrace the steps to identify all the transformations and obtain the given expression? Well, we need to know the order in which we complete a series of several transformations if we’re given the algebraic expression. And just as there’s an order of operations for calculations, for instance, we can apply an order for our combination of transformations.

Firstly, we look for any horizontal translations and we apply those first. Then we look for any stretches or, of course, compressions. Remember, we can have stretches in both the horizontal and vertical direction, and it really doesn’t matter which order we apply those. Then, we complete any reflections in either the π‘₯-, 𝑦-axis, or both. And then a vertical translation will be the last thing that we do. Now, of course, this is hugely important when applying a combination of transformations. But there will be occasions where different combinations and orders of transformations may result in fact the same effect on the graph of a function.

At this stage, though, we don’t have any way to determine whether that will be the case. So we’ll follow the order as shown. Let’s demonstrate this in an example.

This is the graph of the exponential function 𝑦 equals 𝑓 of π‘₯. Which of the following is the graph of 𝑦 equals four minus 𝑓 of two π‘₯?

Let’s compare the original equation to the equation of our transformed function. Three distinct things have happened here. We’ve added four, we’ve multiplied the entire function by negative one, and we’ve multiplied the π‘₯ by two. But we need to be careful to ensure that we’ve done these in the correct order. The order is as follows. We begin by looking for any horizontal translations. Is the graph going to move left or right? Then we stretch it, and we can do that in both directions if necessary. We then perform any reflections either across the π‘₯- or 𝑦-axis or both. And finally, we deal with any vertical translations. Does the graph move up or down?

So a horizontal translation occurs when we add some constant to the value of π‘₯. So 𝑓 of π‘₯ maps onto 𝑓 of π‘₯ plus π‘Ž by a translation π‘Ž units left. Now, if we look at our function, nothing has actually been added to the value of π‘₯. And so we aren’t going to be applying any horizontal translations. But we are going to be applying a stretch. If we have 𝑓 of π‘₯, 𝑓 of 𝑏 times π‘₯ is a stretch in the horizontal direction by a scale factor of one over 𝑏. We have an expression of this form. In fact, since the coefficient of π‘₯ is two, since we’re multiplying our value of π‘₯ by two, we have a horizontal stretch by a scale factor of one-half. We can add this interim graph to our diagram.

We know it will still pass through the 𝑦-axis at two. Then the π‘₯-value of the point with coordinates one, approximately 0.7 will half, and we get the coordinate 0.5, 0.7. Similarly, the curve passes through the point two and approximately 0.2. The π‘₯-value will halve at this coordinate. And we get one, 0.2. Up here, we have a point which has coordinates negative 0.6 roughly and four. And halving the π‘₯-coordinate, that becomes negative 0.3, four. And so the graph of 𝑦 equals 𝑓 of two π‘₯ looks as shown, and we can see that that’s got that horizontal compression. It’s a stretch by a scale factor of one-half. And what about reflections? Well, these occur when we either change the coefficient of π‘₯ to be negative or we multiply the whole function by negative one.

Well, we do indeed have negative 𝑓 of two π‘₯. This is going to result then in a reflection in the 𝑦-axis. And so whilst we can’t quite fit the whole graph on, we know it’s going to pass through negative two on the 𝑦-axis as shown. So we’ve performed a stretch. We’ve performed a reflection. What about a vertical translation? This would be of the form 𝑓 of π‘₯ plus 𝑑. So given an original function 𝑓 of π‘₯, we map it onto 𝑓 of π‘₯ plus 𝑑 by moving it 𝑑 units up.

If we rewrite our function as 𝑦 equals negative 𝑓 of two π‘₯, which are the transformations we’ve performed, plus four, we should now be able to see that we need to move our graph four units up. And so it will now pass through the 𝑦-axis at two. And out of our options for (a), (b), and (c), this is graph (b). So graph (b) is the graph of 𝑦 equals four minus 𝑓 of two π‘₯.

So we’ve now seen how to apply a series of transformations and the order in which they should be applied if necessary. And of course, it follows that to reverse a series of transformations, in other words, to work out the original function, we’d need to work backwards through these steps. And if we’re given the order in which a series of transformations is applied, we can also work out the original function by reversing those. So let’s see what this would look like.

The graph represents a function 𝑔 of π‘₯ after a vertical positive shift by three units followed by a reflection on the π‘₯-axis. Which of the following represents the original function 𝑓 of π‘₯? Is it (A) 𝑓 of π‘₯ equals negative π‘₯ plus two cubed minus two? Is it (B) 𝑓 of π‘₯ equals negative three π‘₯ minus one cubed minus six? Is it (C) 𝑓 of π‘₯ equals π‘₯ minus one cubed plus two? (D) 𝑓 of π‘₯ equals negative π‘₯ minus four cubed minus two. Or (E) 𝑓 of π‘₯ equals negative three π‘₯ minus one cubed plus two.

We’re told that 𝑔 of π‘₯ comes from the original function 𝑓 of π‘₯ after applying two transformations. We can recover the original function 𝑓 of π‘₯ from 𝑔 of π‘₯ by reversing the transformations but beginning from the second one. The process will give 𝑓 of π‘₯ as an expression involving 𝑔 of π‘₯. And then we can finish this problem by finding an equation for 𝑔 of π‘₯.

So let’s begin with the second transformation which is a reflection over the π‘₯-axis. And we might remember that given some function β„Ž of π‘₯, the corresponding negative β„Ž of π‘₯ is a reflection of the original graph across the π‘₯-axis. We reverse it by applying the same transformation. So we take 𝑔 of π‘₯ and we map it onto negative 𝑔 of π‘₯.

Next, we’ll reverse the first transformation. That was a vertical positive shift by three units. The opposite to this is a vertical negative shift, in other words, shifting or translating the graph three units down. Now, we achieve that by taking the function β„Ž of π‘₯ and subtracting three from it. Or, in the case of our transformed function negative 𝑔 of π‘₯, we subtract three from that, so we have negative 𝑔 of π‘₯ minus three. And since we’ve reversed two transformations, this must be equal to our original function 𝑓 of π‘₯.

So all we need to do now is find the equation for 𝑔 of π‘₯. We simply look at the shape and we can already see that it must come from the parent function 𝑦 equals π‘₯ cubed. It’s definitely a cubic graph. But of course if the coefficient of π‘₯ cubed is positive, the cubic graph has the opposite shape. In other words, it looks like this graph but reflected across the 𝑦-axis.

So to achieve a reflection across the 𝑦-axis, we make the entire function negative. So 𝑦 equals negative π‘₯ cubed definitely has the correct shape and orientation. But of course, 𝑦 equals π‘₯ cubed and 𝑦 equals negative π‘₯ cubed intersect the 𝑦-axis at zero. In fact, this is also the stationary point of the function or the point of inflection. On our curve, that corresponds to the point with coordinates one, negative five. So we take 𝑦 equals negative π‘₯ cubed and translate it one unit right and five down. Translating it one unit right occurs when we subtract one from the value of π‘₯. And to move it five units down, we simply subtract five from the whole function. So it certainly appears as if we have the equation for our function 𝑔 of π‘₯.

It might be sensible just to check this by substituting a couple of the coordinates in. For instance, when π‘₯ is equal to negative one, 𝑦 is equal to negative negative one minus one cubed minus five, which equals three as we expected. Similarly, when π‘₯ is equal to zero, 𝑦 is equal to negative zero minus one cubed minus four, which is negative four also as we expected. So with some certainty, we can be assured that we have the correct equation for this graph. So 𝑔 of π‘₯ is negative π‘₯ minus one cubed minus five, and that means we can now find an expression for 𝑓 of π‘₯.

By replacing 𝑔 of π‘₯ with this expression in our equation for 𝑓 of π‘₯, we have 𝑓 of π‘₯ is negative negative π‘₯ minus one cubed minus five minus three. And then we distribute the parentheses to get π‘₯ minus one cubed plus five minus three, which is π‘₯ minus one cubed plus two. And we can now look at our options and see that that corresponds to option (C) 𝑓 of π‘₯ is equal to π‘₯ minus one cubed plus two.

We’ve now demonstrated several examples of how to apply a series of transformations. We’ve used graphs and we’ve used algebraic expressions. And we’ve also seen how to reverse such processes.

Let’s recap the key points. The order in which we apply a combination of transformations is followed. We find horizontal translations first, then perform any stretches before performing reflections, finally identify any vertical translations. And whilst this order is important because applying them in a different order can result in a different final graph, sometimes different combinations and orders of transformations may actually result in the same effect of the graph of a function.

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