Video: Transforming Graphs with Arithmetic Operations

The graph of 𝑓(π‘₯) is given by the dashed red line in the following figure. Also shown are graphs of 𝑓(π‘₯/2), 𝑓(2π‘₯), and 2𝑓(π‘₯). Match up these functions to their graphs.

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Video Transcript

The graph of 𝑓 π‘₯ is given by the dashed red line in the following figure. Also shown are graphs of 𝑓 π‘₯ over two, 𝑓 two π‘₯, and two 𝑓 π‘₯. Match up these functions to their graphs.

To help us solve this problem, what I’m gonna look at is some of our transformation rules. The first ones I’m gonna start with are our translation rules. The first of these is the translation 𝑓 π‘₯ plus π‘Ž. So we actually have π‘Ž outside the parentheses. And this gives us the vector zero π‘Ž. And what this actually means is a shift by π‘Ž units in the 𝑦-direction. Then our next translation rule is 𝑓 π‘₯ plus π‘Ž. It’s worth noting here that the π‘Ž is inside the parentheses this time. And this gives us the vector negative π‘Ž, zero. And what this means in practice is actually a shift of negative π‘Ž in the π‘₯-direction.

And now, we’re gonna take a look at our stretches. First one is π‘Žπ‘“ π‘₯. And this is the stretch parallel to the 𝑦-axis, if the scale factor of π‘Ž. And again, as you note here, the π‘Ž is actually outside the function itself. And then our second stretch is actually 𝑓 π‘Žπ‘₯. Again this time, the π‘Ž is inside the parentheses. And this is the stretch parallel to the π‘₯-axis with a scale factor of one over π‘Ž. Okay, Fab. We now have our transformation rules. Let’s see if we can use them to actually solve the problem.

So what I’m gonna do is I’m gonna actually start with our first function. And this one is 𝑓 π‘₯ over two. And to help us decide which one of our transformation rules is used, we can actually think of this as 𝑓 a half π‘₯. And when we do that, we can see that it’s clearly gonna be our second stretch transformation. So therefore, if we can actually use our second stretch information rule, we can see that if we have a function π‘Žπ‘₯, in this case half π‘₯, then this is gonna be stretch parallel to the π‘₯-axis by a scale factor of one over π‘Ž. In the- that case, it’s gonna be a scale factor of one over a half, which is the same as multiplying by two. So therefore, we can say that the π‘₯-coordinates are gonna be multiplied by two.

So therefore, if I pick a point on our original function β€” so I’m gonna pick this point here β€” what we can see is that, on the original function, this point is gonna be three, one. So therefore, if we look at the function that we have when it’s transformed, which is 𝑓 π‘₯ over two, then we know that the same coordinates should be six, one. And this is because our π‘₯-coordinates have been multiplied by two. Okay, great, so we can now use this to actually identify which one of our graphs is gonna be the graph for the function 𝑓 π‘₯ over two. Well, if we look in the graph, we can see the point six, one. We can see that it actually lies on graph c. And that’s our corresponding point. So therefore, we can say that function π‘₯ over two is equal to c.

Okay, great, we can now move on to our other graphs. Well, if we move on to the next transformation, we’ve got 𝑓 two π‘₯. And again, we can actually use the second stretch rule to help us with this one. Well, for this function, we can actually see that our π‘₯-coordinates are gonna be multiplied by one over two, so a half. So then, if we take the starting point the same as previously β€” so we had three, one β€” then if we multiply our π‘₯ coordinate three by half, we get three over two. And then our 𝑦-coordinate stays the same. So that means the corresponding point is gonna be three over two, one. So therefore, we can say that the function two π‘₯ is gonna be equal to a, so graph a.

So fabulous, we’ve actually done two of the graphs now, so two of our transformations. We can move on to our final transformation. Well, our final transformation is two 𝑓 π‘₯. And we can see this time the two or so our π‘Ž is actually outside the parenthesis or outside the function itself. So therefore, this one actually represents our stretch rule number one. And it tells us that it’s gonna be a stretch parallel to the 𝑦-axis with scale factor of two. So what this means in practices that our 𝑦-coordinates are gonna be multiplied by two. Well, as graph b is the only graph left, we assume that graph b is going to be the correct graph for two 𝑓 π‘₯. But we’re just gonna check it now by doing the same processes before. Well, we’re gonna choose the same starting point as before. So we’ve got the point three, one on the original function. So therefore, the corresponding point on our new function, so our transformed function, is gonna be three, two. And this is because we’ve actually multiplied our 𝑦-coordinate which was one by two, which gives us two.

So when we check the graph, yes, this is correct because actually, the corresponding point that we mark just now on graph b was actually at three, two. So therefore, we can say that two 𝑓 of π‘₯ is equal to b. So great, what we’ve actually done now is matched up all the functions to the graphs. So we can actually say that we’ve solved the problem. I’m just gonna recap how I did that. 𝑓 π‘₯ over two was equally graph c. And that’s because what we did was we actually multiplied each of the π‘₯ terms in our original function by a scale factor of two. 𝑓 two π‘₯ was equal to graph a. And again, this time we used the same stretch role. But what we actually did is we multiplied each of our π‘₯-coordinates by a scale factor of one over two, so a half. And then finally, two 𝑓 π‘₯ is equal to graph b. And the reason it was like that is because we used our first stretch rule. And we actually multiplied each of the 𝑦-coordinates by a scale factor of two.

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