Graph each function over a two-period interval. See Example 4....

Question

Graph each function over a two-period interval. See Example 4.

y = -3 + 2 sin x

Accepted Answer

Hello. Today we will be graphing the following trigonometric function and we'll be graphing two periods of the function. So we are given Y is equal to negative two plus three sine X. Now, before we graph the trigonometric function, there are four pieces of information that we need to find. We need to find the functions amplitude, the period, the phase shift and the vertical shift. In order to get these pieces of information, we will be comparing the given trigonometric function to the general function Y is equal to a multiplied by sine of B multiplied by X minus C plus D. Using this function. Let's start by identifying the function's amplitude. Now, the amplitude is going to be the value that tells us the maximum and minimum height that the sine function will reach the amplitude is defined as the absolute value of A. And this is where A is the leading coefficient in front of the sine function. Now let's quickly rewrite the given function by reordering the terms we can rewrite the function as Y is equal to three sine X minus two. The leading coefficient in front of the sine function is going to be positive three. That means the amplitude will be defined as the absolute value of three, which will simplify to be three. What this means is that the current function will have a maximum value of three and a minimum value of negative three. Now let's find the functions period. The period for sine function is defined as two pi divided by the absolute value of B. This is where B is the coefficient in front of the X minus C term. Now, in this case here, we only have a coefficient of uh of one in front of the X. This means that the value B will equal to one, meaning the period will be defined as two pi divided by the absolute value of one which will simplify to give us two pi the standard period for sine function is two pi. So there is no change to the period of the function. Next, we will need to find the face shift of the given function. Now you can think of the phase shift as a horizontal shift of the given graph. The phase shift is defined by the C term in the X minus C quantity. But in this case here we are not given ac term. So the C term will equal to zero. Therefore, the phase shift is zero. This means we do not have a horizontal shift of the given function. Finally, we will need to identify the vertical shift of the given function Now, the vertical shift is defined by the D term which we are adding or subtracting any extra values to the given trigonometric function. Currently, we are subtracting two from the function meaning we have a vertical shift of negative two. This means that we are shifting the entirety of the graph down by two units. Now that we have the standard or the four pieces of information that we need, we can start graphing the trigonometric function. Let's start by marking the traditional values for the sine function. Now for one period of the sine function, the sine function will start at the origin zero comma zero. And the first period will end at two pi the maximum value for the standard graph will be located at pi divided by two comma one. The minimum value is located at three pi divided by two comma negative one and we have a zero at pi. Now let's take a look at our amplitude since our amplitude is three, that means that the standard minimum and maximum values are going to be located at positive three and negative three. So our maximum value will be located at pi divided by two comma three. And our minimum value will be located at three pi divided by two comma negative three. Next, since we have no change to the period, we are not stretching the first period of the given function, we don't have a phase shift. So we're not moving any of the points, any units to the left or right. But we do have a vertical shift of negative two. What this means is that every point in the first period of the function will shift down to units. So let's go ahead and apply that change. So for the first period of the function, the sine function will start at zero. Comma negative two have a maximum value at pi divided by two comma one have a have the next point at pi comma negative two, the minimum value at three pi divided by two comma negative five and end at two pi comma negative two. This is going to be the shape of the first period of the function. Now we can extend this the next period of the function by graphing a similar shape to the first period. So the next point in the second period will be a five pi divided by two comma one. The following point at three pi comma negative two. The minimum value of the second period will be located at seven pi divided by two comma negative five. And the final period will end at four pi comma negative two. And if we extend the graph, this will be the second period of the given sine function. Finally this will be the final image of the given trigonometric function. And this will be the answer to the problem. So I hope this video helps you in understanding how to graph this trigonometric function and I'll go ahead and see you all in the next video.

Struggling with Trigonometry?

Graph each function over a two-period interval. See Example 4.
y = -3 + 2 sin x

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Graphs of the Sine and Cosine Functions practice set

Struggling with Trigonometry?

Graph each function over a two-period interval. See Example 4.y = -3 + 2 sin x

Watch next

Graphs of the Sine and Cosine Functions practice set

Graph each function over a two-period interval. See Example 4.
y = -3 + 2 sin x