Question Video: Recognizing the Effect of a Phase Shift in Radians on the Graph of the Sine Function Mathematics • 10th Grade

The figure shows the graph of 𝑦 = sin π‘₯. Which of the following is a graph of 𝑦 = sin (π‘₯ + 2)? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

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Video Transcript

The figure shows the graph of 𝑦 equals sin of π‘₯. Which of the following is a graph of 𝑦 equals sin of π‘₯ plus two?

We will begin by examining the original sine function graph, before any transformations are applied. We notice the five key points spanning one period of the sine curve. It appears that the scaling on the π‘₯-axis is in radians. And we know that sine has a period length of two πœ‹ radians, which is approximately 6.28 radians. The main sine curve also has an amplitude of one, which is the distance from the center line to the peak or the trough.

We will now review the four transformations of the sine function graph, beginning with vertical and horizontal translations. A vertical translation, or shift, of 𝑓 of π‘₯ by π‘Ž is 𝑓 of π‘₯ plus π‘Ž. A horizontal translation, or phase shift, of 𝑓 of π‘₯ by negative 𝑏 is 𝑓 of π‘₯ plus 𝑏. To clarify what is meant by this rule, let’s look at two examples. 𝑓 of π‘₯ plus eight is a translation left eight, and 𝑓 of π‘₯ minus eight is a translation right eight.

The sine graph is stretched parallel to the 𝑦-axis by a scale factor of 𝑐 if 𝑓 of π‘₯ is multiplied by 𝑐. We may also refer to this as an amplitude change. The sine graph is stretched parallel to the π‘₯-axis by a scale factor of one over 𝑑 if we have 𝑓 of 𝑑 times π‘₯. We may also refer to this as a period change. We will now clear some space in order to determine which transformation is being applied to the sin of π‘₯ function to get sin of π‘₯ plus two and how each of the five graphs given compares to the sin of π‘₯ function.

If sin of π‘₯ equals 𝑓 of π‘₯, then sin of π‘₯ plus two equals 𝑓 of π‘₯ plus two, where the 𝑏-value is two. Therefore, we are looking for the sine graph with a horizontal translation of negative two, which does not affect the original period of two πœ‹ or the amplitude of one. This transformation is just a simple shift of two radians to the left.

Now we are ready to consider the graph from option (A). If we highlight one period of the given sine curve, we notice the amplitude is one, but the period has changed from approximately 6.28 to approximately 3.14, or in other words from a period of two πœ‹ to a period of πœ‹. This curve appears to be the graph of 𝑓 of two π‘₯, which effectively cuts the period in half. We also do not observe any horizontal shift. Therefore, we eliminate option (A).

Now we are ready to consider option (B). If we again highlight one period of the sine curve, we notice the amplitude is one. One period of the curve extends from negative two to just past positive four. So this graph appears to have the desired period length of two πœ‹, or approximately 6.28. And compared to the main sine graph, which passes through the origin zero, zero, this curve passes through the point negative two, zero, which is evidence of the horizontal shift of negative two. So we conclude that option (B) is the graph of 𝑓 of π‘₯ plus two.

Let’s take a quick look at the other options to make sure that we have not missed anything. Here is the sine curve given as option (C). This graph appears to have the right period, but the amplitude is two instead of one, and there is no horizontal translation of two either. Because of the amplitude change, this appears to be the graph of two times 𝑓 of π‘₯. So we can confidently eliminate option (C).

The graph given in option (D) is noticeably different from any of the sine graphs we have seen so far because this sine curve does not touch the π‘₯-axis at all. This curve still has a period of two πœ‹ and an amplitude of one, but the most prominent difference is the vertical translation up two. So this is a graph of 𝑓 of π‘₯ plus two outside of the parentheses, not to be confused with 𝑓 of π‘₯ plus two inside the parentheses. Since this graph is translated vertically instead of horizontally, we can eliminate option (D).

The last graph to check is given by option (E). This sine curve has the correct amplitude and period, but it appears to be shifted to the right two, rather than to the left two. So this is the graph of 𝑓 of π‘₯ minus two. It is important to remember with horizontal translations, π‘₯ plus two moves the graph to the left and π‘₯ minus two moves the graph to the right. So we eliminate option (E).

In summary, only option (B) showed a sine graph with a period of two πœ‹, an amplitude of one, and a horizontal translation of negative two. Therefore, option (B) is the graph of 𝑦 equals sin of π‘₯ plus two.

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