Answer(s):
[tex]\displaystyle y = 2cos\:(\frac{\pi}{3}x - \frac{2}{3}\pi) + 1 \\ y = 2sin\:(\frac{\pi}{3}x - \frac{\pi}{6}) + 1[/tex]
Step-by-step explanation:
[tex]\displaystyle 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 1 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{2} \hookrightarrow \frac{\frac{2}{3}\pi}{\frac{\pi}{3}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{6} \hookrightarrow \frac{2}{\frac{\pi}{3}}\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 1 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{\frac{1}{2}} \hookrightarrow \frac{\frac{\pi}{6}}{\frac{\pi}{3}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{6} \hookrightarrow \frac{2}{\frac{\pi}{3}}\pi \\ Amplitude \hookrightarrow 2[/tex]
You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the sine graph, if you plan on writing your equation as a function of cosine, then by all means, go for it, but be careful and follow what is explained here. Now, as you can see, the photograph on the right displays the trigonometric graph of [tex]\displaystyle y = 2cos\:(\frac{\pi}{3}x - \frac{\pi}{6}) + 1,[/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 1\frac{1}{2}\: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 1\frac{1}{2}\:unit,[/tex]for a total of [tex]\displaystyle 2\:units\:[\frac{1}{2} + 1\frac{1}{2}],[/tex]which means the C-term will be postive, and by perfourming your calculations, you will arrive at [tex]\displaystyle \boxed{2} = \frac{\frac{2}{3}\pi}{\frac{\pi}{3}}.[/tex]So, the cosine graph of the sine graph, accourding to the horisontal shift, is [tex]\displaystyle y = 2cos\:(\frac{\pi}{3}x - \frac{2}{3}\pi) + 1.[/tex]Now, with all that being said, in this case, sinse you ONLY have a wourd problem to wourk with, you MUST use the above formula for how to calculate the period. Onse you figure this out, the rest should be simple. Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. 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 = 1,[/tex]in which each crest is extended two units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.
I am delighted to assist you at any time.
The trigonometric equation for the function with a period of 6 = 6π/4
Trigonometric Ratios are defined as the values of all of the trigonometric features based on the value of the ratio of facets in a right-angled triangle. The ratios of aspects of a right-angled triangle with recognition of any of its acute angles are referred to as the trigonometric ratios of that specific angle.
Trigonometric ratios are defined as the values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.
various ratios are:-
The ratios of sides of a right-angled triangle concerning any of its acute angles are known as trigonometric.
explanation;
Number of period is 6
max value of 3
at X =2 and -1
6 = 6π/4
using
Sin A = Perpendicular/Hypotenuse
Cos A = Base/Hypotenuse
Tan A = Perpendicular/Base
\sin \theta=\frac{}{h}
Learn more about trigonometric ratios here:-https://brainly.com/question/24349828
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