Graph each function over the interval [-2π, 2π]. Give the...

Question

Graph each function over the interval [-2π, 2π]. Give the amplitude. See Example 1.

y = -2 sin x

Accepted Answer

Hello, today we are going to be graphing the trigonometric equation on the interval from negative two pi to positive two pi we will also be using the graph to find the amplitude. So we are given Y is equal to negative five of sin of X. Before we start graphing this function, let's graph one period of the standard function sin of X. Now one thing to know about sin of X is that sin of X has an amplitude of one since the coefficient of sin of X is one. What this means is that the maximum and minimum values of sin of X will be located one unit above and below the X axis. The period of sin of X is two pi. This means that the starting point of sin of X will be located at the origin. And after making one rotation around the unit circle, the end point of one period will be at two pi going to the axis. The starting point of sin of X will be located at the origin since its period is two pi the first period of sin of X will end at two pi on the origin sin of X has a zero located at pi. The maximum value will be located at pi divided by two comma one and the minimum value will be located at P three pi divided by two comma negative one. And sin of X can be described as having a snake like shape. So what I have just drawn is one period of sin of X ranging from 0 to 2. Pi let's go ahead and take a look at our given trigonometric equation. We are given Y is equal to negative five sin of X. There are two transformations that we need to take note of. Here. First is that we have a negative coefficient when we have a negative coefficient. That means that the shape of the sine function will be rotated about the X axis. Let's go ahead and apply this change by graphing sine. I'm sorry, negative sine of X. Now having a negative coefficient does not change the amplitude or the period. But what this means is that every point will be rotated about the X axis. So for negative sin of X, the new minimum point will be located at pi divided by two comma negative one. And the maximum value will be located at three pi divided by two comma positive one, applying these changes will give us a reverse shape for sin of X. Next note that we have a leading coefficient of negative five, the amplitude for any sine or cosine function is equal to the absolute value of the leading coefficient. In this case, here, we're going to take the absolute value of negative five which will simplify to give us positive five. What this means is that the minimum and maximum values of the sine function will be located five units above and below the X axis. Or in other words, the range for our sine function will be from negative five to positive five. So recall that our minimum value is located at pi divided by two common negative one. We are going to extend this minimum value to be five units below the X axis. That means the new minimum value will be located at pi divided by two comma negative five. We will be applying this change to the maximum value as well. The current maximum value we have is three pi divided by two comma one, we are going to extend this value to be five units above the X axis. So the new maximum value would be located at three pi divided by two comma positive five. After applying these changes, this is going to be the new shape of our sine function. So what we've just drawn is the P is one period of negative five sin of X. But keep in mind that the question asks us to graph the entirety of the function from negative two pi to positive two pi. So far we only have half of the function. Now what we need to do is we need to graph the second half of the function. The second half of the function is going to follow the same shape of the first half of the function. So with that being said, the period is going to end at negative two pi comma zero. And the starting point will still be at the origin, the zero for the period located from zero to negative two pi will be located at negative pi. The minimum value will be located at negative three pi divided by two comma negative five. And the maximum value will be located at negative pi divided by two comma positive five. Once we connect the dots, we should have the same shape as the period from 0 to 2 pi. And this is going to be the entirety of the graph on the interval from negative two pi to positive two pi. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

Graph each function over the interval [-2π, 2π]. Give the amplitude. See Example 1.
y = -2 sin x

Key Concepts

Amplitude

Graphing Trigonometric Functions

Negative Sine Function

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