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Question Video: Finding the Equation of a Trigonometric Function from a Graph Mathematics • 10th Grade

The figure shows the graph of a function. Which of the following equations represents the graph? [A] 𝑦 = sin (2𝑥) [B] 𝑦 = sin (𝑥 + 2) [C] 𝑦 = 2sin (𝑥) [D] 𝑦 = sin (𝑥) − 2 [E] 𝑦 = sin (𝑥) + 2

02:38

Video Transcript

The figure shows the graph of a function. Which of the following equations represents the graph? (A) 𝑦 equals sin two 𝑥, (B) 𝑦 equals sin 𝑥 plus two, (C) 𝑦 equals two sin 𝑥, (D) 𝑦 equals sin 𝑥 minus two, or (E) 𝑦 equals sin 𝑥 plus two.

So to answer this problem, what we’ve done is drawn a quick sketch of a part of our sine graph. That’s the sort of first part that we have that’s positive. And what we can see is that the peak of 𝑦 equals sin 𝑥 is at one. And if we carried it on a little bit, so we actually took it back from zero to the negative side, we can see that our trough or one of our troughs would be at negative one. However, if we take a look at the graph that we’ve got here, then we can see that the peak is at negative one. And in fact, the trough is at negative three. So we can see that there’s been a vertical shift downwards two units.

So what we can do is remind ourselves about our translations with graphs. So we know that 𝑓 of 𝑥 plus 𝑎 is a vertical shift of 𝑎 units or a shift in the 𝑦-direction. Then we’ve got 𝑓 of 𝑥 plus 𝑎 is a phase shift or a shift in the 𝑥-direction of negative 𝑎 units. So therefore, if we’re just considering the shift, we could say that it would be 𝑦 equals sin 𝑥 minus two because we’re looking at our first translation. But what we could think is, well, hold on, is there going to be a stretch as well?

Well, in fact, we don’t know what the 𝑥-coordinates are actually in because they’re not in degrees. So it’s difficult to tell whether a stretch has occurred or not. Well, if we take a look at (C), we can see for (C) we’ve got a stretch here. And it’s gonna be a stretch in the 𝑦-direction. So therefore, we know that this cannot be the correct answer because actually the amplitudes of both of our graphs are exactly the same, cause the amplitudes are both one.

Well, if we take a look at our other stretch, we can see that this is a stretch parallel to the 𝑥-axis. And here there isn’t a shift applied as well. So we’ve just got the stretch. We’ve already identified that a shift has taken place. So this cannot be the correct answer. So therefore, we can say that (D) is the equation which represents our graph. And if we have a look at (B) and (E), well, we can see that (B) would be incorrect because that’d be a shift in the 𝑥-direction. So it’d be a phase shift. But we said that ours is a vertical shift. And (E) would be incorrect cause for (E) what we would’ve done is actually moved the graph two units up instead of two units down. So this would’ve been incorrect as well.

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