Question Video: Identifying the Image of a Point on a Trigonometric Graph following a Transformation Mathematics

The figure shows the graph of 𝑓(π‘₯). A transformation maps 𝑓(π‘₯) to 2𝑓(π‘₯). Determine the coordinates of 𝐴 following this transformation.

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Video Transcript

The figure shows the graph of 𝑓 of π‘₯. A transformation maps 𝑓 of π‘₯ to two 𝑓 of π‘₯. Determine the coordinates of 𝐴 following this transformation.

Now, we’re going to look at this question in two ways. The first method is to think about the algebraic representation of each of our transformations. Specifically, suppose we have the graph of a function 𝑦 equals 𝑓 of π‘₯. This is mapped onto the graph of 𝑦 equals π‘Ž 𝑓 of π‘₯ for some constant π‘Ž by a vertical stretch scale factor π‘Ž. So, for the transformation that maps 𝑓 of π‘₯ to two 𝑓 of π‘₯, we’re performing a vertical stretch by a scale factor of two. Now, unfortunately, we can’t fully draw this on the diagram given, but we can assume that it might look a little something like this. All of the 𝑦-values in the coordinates are doubled. This means it does intersect the π‘₯-axis at the exact same points.

Now, specifically, we’re interested in coordinate 𝐴, which is 180, negative one. We said that the π‘₯-values remain the same, but that the 𝑦-values are doubled. So, the image of 𝐴, the coordinates of 𝐴, after its transformation, which we’ll call 𝐴 prime, has a 𝑦-value of negative one times two. So 𝐴 prime is 180, negative one times two or 180, negative two. So we’ve shown an algebraic interpretation of this transformation.

But we said there was another way. Now, that way is to think about the actual equation of the original graph. We have that sinusoidal shape. It has a 𝑦-intercept of one, maxima and minima at one and negative one, respectively, and it appears to repeat, be periodic, with a period of 360. We can therefore say that 𝑓 of π‘₯ must be cos of π‘₯. Now, in fact, we can check this by substituting π‘₯ equals 180 into this function and check in we get negative one out. Well, 𝑓 of 180 is cos of 180, which is in fact negative one. So, we’ve identified the equation of 𝑓 of π‘₯. It’s cos of π‘₯. This means the function that we’re interested in after the transformation must be two 𝑓 of π‘₯, and that’s two cos of π‘₯.

We could then plot the graph of 𝑦 equals two cos of π‘₯ on our axes. We could do so using a table or any other suitable method. Either way, we once again see that we can map 𝑓 of π‘₯ under two 𝑓 of π‘₯ by a vertical stretch with a scale factor of two. Either way, we find that the coordinates of 𝐴 following our transformation is 180, negative two.

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