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 maps𝑓(π‘₯) to 4𝑓(3π‘₯ βˆ’ 45) + 1. 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 four 𝑓 of three π‘₯ minus 45 plus one. Determine the coordinates of 𝐴 following this transformation.

And then, we have a graph, which shows point 𝐴 to have coordinates 180, negative one. So, in order to determine where point 𝐴 ends up after a transformation, let’s think about all the transformations that have happened to our function 𝑓 of π‘₯. Since there seemed to be a few transformations, we do need to be a little bit careful on the order in which we apply these. Now whilst the order in which we apply these transformations doesn’t always affect what happens, we can ensure that no mistakes are made by first performing any horizontal shifts.

Consider the function 𝑓 of π‘₯. We can find a horizontal shift by π‘Ž units to the left by applying the function 𝑓 of π‘₯ plus π‘Ž. If we look carefully inside the parentheses in our expression, we see that we’re subtracting 45 degrees. This means we’re moving 45 degrees to the right for this part of the transformation. Next, we look for any stretches. In particular, the function 𝑓 of π‘₯ is mapped onto the function 𝑏 times 𝑓 of π‘₯ by a vertical stretch scale factor 𝑏, whilst we map onto the function 𝑓 of 𝑐 times π‘₯ by a horizontal stretch scale factor one over 𝑐. This is essentially a compression.

We actually have two stretches here. We’re going to stretch by a scale factor of four in the vertical direction, but a scale factor of one-third in the horizontal direction. This is because we’re multiplying the entire function by four, but we’re multiplying just the π‘₯ by three. Now, normally, we would apply any reflections, that is, mapping 𝑓 of π‘₯ onto negative 𝑓 of π‘₯ for a reflection in the π‘₯-axis or 𝑓 of π‘₯ onto 𝑓 of negative π‘₯ for a reflection in the 𝑦-axis. But in our transformation, this isn’t necessary. So, we’ll move on to the final one.

The final one is the vertical shift or translation. 𝑓 of π‘₯ maps onto 𝑓 of π‘₯ plus 𝑑 by a translation 𝑑 units up. In this case, we see that we’re adding one to the entire function, so we need to move it one unit up. With all this in mind, let’s apply these transformations to the point 𝐴 with coordinates 180, negative one.

We begin by moving 45 degrees to the right. When we do this with point 𝐴, its image has coordinates 225, negative one. This is found by simply adding 45 to 180. Then, we stretch by a scale factor of four vertically in the 𝑦-direction and a scale factor of one-third in the horizontal π‘₯-direction. This means the 𝑦-coordinate of the new point is multiplied by four to give us negative four. The π‘₯-coordinate is multiplied by one-third or divided by three. 225 divided by three is 75. Finally, step three tells us to move the entire point one unit up. To achieve this, we add one to the value of the 𝑦-coordinate. Negative four plus one is negative three. And so, this takes us to the new point with coordinates 75, negative three.

And so, the coordinates of point 𝐴 following the transformation that maps 𝑓 of π‘₯ onto four 𝑓 of three π‘₯ minus 45 plus one is 75, negative three.

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