Question Video: Finding the Equation of a Trigonometric Function from a Graph Mathematics

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

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Video Transcript

The figure shows the graph of a function. Which of the following equations represents the graph? Is it (A) 𝑦 equals sin of two π‘₯? (B) 𝑦 equals sin of π‘₯ plus two. Is it (C) 𝑦 equals two sin π‘₯? (D) 𝑦 equals sin of π‘₯ plus two. Or (E) 𝑦 equals sin of π‘₯ minus two.

Let’s begin by inspecting the graph of our function. It has that really recognizable wave shape that we know generally corresponds to the graph of the sine and cosine functions. But of course in the options here, we’ve been given only sine functions. In fact, these are transformations of sine functions. So let’s begin by drawing the graph 𝑦 equals sin of π‘₯ on the same set of axes and identify the transformation or transformations that map that graph onto the graph we’ve been given.

We know that the function 𝑦 equals sin of π‘₯ is periodic and it has a period of two πœ‹ radians. It reaches maxima and minima at one and negative one, respectively. And it passes through the π‘₯-axis at zero, πœ‹, two πœ‹, and so on. So we can draw part of the sine function as shown.

We can now look at some of the key features of this graph in order to identify how it is mapped onto the graph we’ve been given. For instance, let’s consider the point of intersection with the 𝑦-axis. On the graph of 𝑦 equals sin of π‘₯, that has coordinates zero, zero, whereas on the graph we’ve been given, that has coordinates zero, two. It certainly appears as if this point is mapped simply two units up by a single translation.

But let’s check by looking at some of the maxima and minima on our graph. For instance, on the graph of 𝑦 equals sin of π‘₯, we have a relative maxima at the point with coordinates πœ‹ by two and one. On the graph we’ve been given, this appears to have the same π‘₯-value, but the 𝑦-value appears to be two units larger. Once again, this corresponds to a translation by two units up.

Let’s just triple check this by considering one of the relative minima. The graph of 𝑦 equals sin π‘₯ has a relative minima at three πœ‹ over two and negative one. On the graph of the function we’ve been given, this point does appear to have been translated two units up. It has coordinates three πœ‹ over two and one. So we can indeed assume that the graph of 𝑦 equals sin of π‘₯ is translated two units up or by the vector zero, two.

So let’s think about how we use function notation to describe that. Well, we know that, suppose we have a function 𝑓 of π‘₯, we can map that onto the function 𝑓 of π‘₯ plus π‘Ž by a single translation π‘Ž units up or by the vector zero, π‘Ž. So let’s define our original graph 𝑦 equals sin of π‘₯ to actually be 𝑓 of π‘₯ equals sin of π‘₯. We know that to translate this graph two units up, we need the function 𝑓 of π‘₯ plus two. Well, of course, if the function 𝑓 of π‘₯ is sin π‘₯, adding two to the entire function just gives us sin π‘₯ plus two. And so the graph of the function given in this question has equation 𝑦 equals sin of π‘₯ plus two.

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