Jun 07, 9:22:26 AM
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Write a sine function that has an
amplitude of 2, a midline of 3 and a period of 3/2.

Answer:

[tex]\displaystyle y = 2sin\:1\frac{1}{3}\pi{x} + 3[/tex]

Step-by-step explanation:

[tex]\displaystyle \boxed{y = 2cos\:(1\frac{1}{3}\pi{x} - \frac{\pi}{2}) + 3} \\ \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 3 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{\frac{3}{8}} \hookrightarrow \frac{\frac{\pi}{2}}{1\frac{1}{3}\pi} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{1\frac{1}{2}} \hookrightarrow \frac{2}{1\frac{1}{3}\pi}\pi \\ Amplitude \hookrightarrow 2[/tex]

OR

[tex]\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 3 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{1\frac{1}{2}} \hookrightarrow \frac{2}{1\frac{1}{3}\pi}\pi \\ Amplitude \hookrightarrow 2[/tex]

You will need the above information to help you interpret the graph. First off, keep in mind that although the exercise told you to write the sine equation based on the speculations it gave you, if you plan on writing your equation as a function of cosine, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the photograph on the right displays the trigonometric graph of [tex]\displaystyle y = 2cos\:1\frac{1}{3}\pi{x} + 3,[/tex]in which you need to replase "sine" with "cosine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the sine graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the cosine graph [photograph on the right] is shifted [tex]\displaystyle \frac{3}{8}\:unit[/tex]to the left, which means that in order to match the sine graph [photograph on the left], we need to shift the graph FORWARD [tex]\displaystyle \frac{3}{8}\:unit,[/tex]which means the C-term will be positive, and by perfourming your calculations, you will arrive at [tex]\displaystyle \boxed{\frac{3}{8}} = \frac{\frac{\pi}{2}}{1\frac{1}{3}\pi}.[/tex]So, the cosine graph of the sine graph, accourding to the horisontal shift, is [tex]\displaystyle y = 2cos\:(1\frac{1}{3}\pi{x} - \frac{\pi}{2}) + 3.[/tex]Now, with all that being said, in this case, sinse you ONLY have the exercise to wourk with, take a look at the above information next to [tex]\displaystyle Wavelength\:[Period].[/tex]It displays the formula on how to define each wavelength of the graph. You just need to remember that the B-term has [tex]\displaystyle \pi[/tex]in it as well, meaning both of them strike each other out, leaving you with just a fraction. Now, the amplitude is obvious to figure out because it is the A-term, so this is self-explanatory. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at [tex]\displaystyle y = 3,[/tex]in which each crest is extended three units beyond the midline, hence, your amplitude. So, no matter what the vertical shift is, that will ALWAYS be the equation of the midline, and if viewed from a graph, no matter how far it shifts vertically, the midline will ALWAYS follow.

I am delighted to assist you at any time.

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