Video: Graph Transformations of Exponential Functions

William has graphed three exponential functions all in the form 𝑦 = π‘Žπ‘’^𝑏π‘₯ or 𝑦 = π‘Žπ‘’^((1/𝑏)π‘₯), where π‘Ž and 𝑏 are positive integers. The red graph is 𝑦 = 𝑒^π‘₯. By considering properties of graphical transformations, determine the equation of the blue graph. By considering properties of graphical transformations, determine the equation of the green graph.

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Video Transcript

William has graphed three exponential functions all in the form 𝑦 equals π‘Ž times 𝑒 to the power of 𝑏π‘₯ or 𝑦 equals π‘Ž times 𝑒 to the power of one over 𝑏π‘₯, where π‘Ž and 𝑏 are positive integers. The red graph is 𝑦 equals 𝑒 to the power of π‘₯. By considering properties of graphical transformations, determine the equation of the blue graph.

Let’s label the red graph as 𝑦 equals 𝑒 to the π‘₯. Our task is to determine the equation of the blue graph. And we’re told to do this by considering properties of graphical transformations, so transformations of graphs. The first question that we’re going to ask ourselves then on the way to finding the equation of the blue graph is what transformation takes the red graph to the blue graph.

It might be a good idea to pause the video and think about the different kinds of graph transformations you know about. It may take a while to convince ourselves that the transformation we looking for isn’t a translation or a rotation or a reflection, but a dilation or stretch. In particular, this is a dilation from the 𝑦-axis.

A point on our original red curve is transformed towards the 𝑦-axis so that its distance from the 𝑦-axis halves. This new point therefore has coordinates π‘₯ over two 𝑦. Let’s call the function represented by the blue graph 𝑓 of π‘₯. The value of 𝑓 of π‘₯ over two is the 𝑦 coordinate of the pinky purple point that we plotted.

That’s this 𝑦-coordinate. And this 𝑦-coordinate is the same as the 𝑦-coordinate of the other point, the orange point, on the red curve 𝑦 is 𝑒 to the power of π‘₯. And so 𝑓 of π‘₯ over two is 𝑒 to the power of π‘₯. This function is defined a bit strangely because we have an input of π‘₯ over two instead of π‘₯. But it still gives us the rule of the function.

We take our input π‘₯ over two and double it to get just π‘₯. And then we raise 𝑒 to the power of that doubled input. We can rewrite this in a more familiar form where the input is π‘₯. We double it to get two π‘₯ and then raise 𝑒 to the power of that. So 𝑓 of π‘₯ is equal to 𝑒 to the power of two π‘₯.

This means that the equation of the blue graph, which is after all what we’re looking for, is 𝑦 equals 𝑒 to the power of two π‘₯. To summarize, we saw that the graph transformation that we were looking for was a dilation from the 𝑦-axis by a factor of a half.

And this took the graph of 𝑦 equals 𝑒 to the π‘₯ to that of 𝑦 equals 𝑒 to the two π‘₯. This is a special case of the general rule, that a dilation from the 𝑦-axis by a factor of an arbitrary constant π‘˜ takes a graph of 𝑦 equals 𝑓of π‘₯ to that of 𝑦 equals 𝑓 of π‘₯ over π‘˜.

In our question, π‘˜ was a half and 𝑓 of π‘₯ was 𝑒 to the power of π‘₯. And so our transformed graph had equation 𝑦 equals 𝑒 to the power of π‘₯ over a half or 𝑦 equals 𝑒 to the power of two π‘₯. We’re not quite done. There’s another part to our question.

By considering properties of graphical transformations, determine the equation of the green graph. The only difference from the previous part is that we’re finding the equation of the green graph and not the blue graph. As before it might make sense to pause the video and think about what transformation takes the red graph to the green graph.

Spoiler alert! I left the general rule about dilations from the 𝑦-axis on the screen. And this is because the transformation is another dilation from the 𝑦-axis. This dilation has a different factor, a different value of π‘˜, a point π‘₯, 𝑦 on the red graph is taken to the point with coordinates two π‘₯, 𝑦 from which we see that π‘˜ is equal to two.

We could follow the process we used in the first part. But now that we have this general rule, we might as well use it. The equation that we’re looking for is 𝑦 equals 𝑓 of π‘₯ over π‘˜. 𝑓 of π‘₯ is the function corresponding to the untransformed red graph, which is still 𝑒 to the power of π‘₯.

And you’ve seen already that in this part of the question the value of π‘˜ is two. Putting these together, we get that 𝑦 is equal to 𝑒 to the power of π‘₯ over two or 𝑒 to the power of a half π‘₯.

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