Question Video: Identifying the Graph of a Trigonometric Function after a Period Change | Nagwa Question Video: Identifying the Graph of a Trigonometric Function after a Period Change | Nagwa

Question Video: Identifying the Graph of a Trigonometric Function after a Period Change Mathematics • Second Year of Secondary School

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Which of the following is the graph of 𝑦 = sin 2π‘₯? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

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Video Transcript

Which of the following is the graph of 𝑦 equals sin of two π‘₯?

We will consider each of the five graphs given to determine which one has the same transformations as 𝑦 equals sin of two π‘₯. First, we need to recall the various transformations of the original 𝑦, or 𝑓 of π‘₯, equals sin of π‘₯ function. The first category of transformations are called translations or shifts. A vertical shift up or down π‘Ž units is given by 𝑓 of π‘₯ plus π‘Ž, whereas a horizontal shift left or right by negative 𝑏 units is given by 𝑓 of π‘₯ plus 𝑏.

This kind of translation can be a little challenging to get right, so we will review two quick examples. The notation 𝑓 of π‘₯ plus eight means shift left eight, and the notation 𝑓 of π‘₯ minus eight means shift right eight. The third type of transformation is referred to as a vertical stretch or amplitude change. We recall the standard amplitude of the sine function is one. So, if we multiply the function by a scale factor 𝑐, we get a new amplitude of the absolute value of 𝑐. The last transformation we will review is the horizontal stretch, also referred to as a period change. We recall the original sine function has a period of 360 degrees. If we are given 𝑓 of 𝑑 times π‘₯, this means the period is multiplied by a scale factor of one over 𝑑.

Now we will consider which transformation is given by the function 𝑦 equals sin of two π‘₯. If we start with sin of π‘₯ equals 𝑓 of π‘₯, we should be able to determine what sin of two π‘₯ equals. By replacing π‘₯ with two π‘₯, we determine that sin of two π‘₯ equals 𝑓 of two π‘₯, which we recognize as a period change. In this case, 𝑑 equals two. So the scale factor applied to the period is one over two. So, although the amplitude has not changed, the original period of 360 is multiplied by one-half. Therefore, we are looking for the sine graph that has a period of 180 degrees.

We will now consider option (A). This appears to be the original sine function, with an amplitude of one and a period of 360, which we have highlighted in blue. We recognize the five familiar coordinate points between zero and 360, where π‘₯ equals an angle measure in degrees and 𝑦 equals sin of π‘₯. We will look for these five points in our other sine graphs to determine if there has been a period change or other transformation. We eliminate the first option because it is a graph of the original sine function with no period change.

Moving on to option (B), we can trace one period of sine, starting at the coordinate point 90, zero. This happens when we have a 𝑏-value of negative 90, which causes a horizontal shift of positive 90. We see all five points have been shifted to the right 90 degrees. This horizontal shift preserves the period length of 360 degrees. So this is not the function we are looking for. So, we eliminate option (B) and move on to option (C).

The first thing we notice about option (C) is the period change. The period length has been condensed, but we must determine if it’s been condensed to 180 degrees or not. If we highlight one period of this sine curve, between negative 30 and 90 degrees, we come up with a period length of 120 degrees. So, we eliminate this option based on the fact that the period is not the desired 180 degrees. It also appears this sine curve has been shifted left by 30 degrees.

Moving on to option (D), we notice another period change. We trace one period of this sine curve, which happens to be 180 degrees exactly. This points to a period change with a scale factor of one-half, which means 𝑑 equals two. This is the graph of 𝑓 of two π‘₯, which is our 𝑦 equals sin of two π‘₯ function.

To make sure we’ve considered all the options, we take a look at option (E). The first thing we notice is that this sine curve is no longer centered around the π‘₯-axis. In this case, we have a vertical shift of π‘Ž equals one. So, the period and amplitude stay the same, but all points shift up one unit. Because of the vertical shift and no period change in option (E), we can safely cross this possibility off our list.

Option (D) is the only graph with the correct amplitude of one and period of 180, without any horizontal or vertical shifts. Therefore, (D) is the graph of 𝑦 equals sin of two π‘₯.

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