Write the equation of the trigonometric graph

Answer:

y = sin(4(x+π/8)) + 1

Step-by-step explanation:

For a trigonometric equation of form

y = Asin(B(x+C)) + D,

the amplitude is A, the period is 2π/B, the phase shift is C, and the vertical shift is D (shifts are relative to sin(x) = y)

First, the amplitude is the distance from the center to a top/bottom point (also known as a peak/trough respectively). The center of the function given is at y=1, and the top is at y=2, Therefore, 2-1= 1 is our amplitude.

Next, the period is the distance between one peak to the next closest peak, or any matching point to the next matching point. One peak of this function is at x=0 and another is at x= π/2, so the period is (π/2 - 0) = π/2. The period is equal to 2π/B, so

2π/B  = π/2

multiply both sides by b to remove a denominator

2π = π/2 * B

divide both sides by π

2 = 1/2 * B

multiply both sides by 2 to isolate b

4 = B

After that, the phase shift is the horizontal shift from sin(x). In the base function sin(x), one center is at x=0. However, on the graph, the closest centers to x=0 are at x=± π/8. Therefore, π/8 is the phase shift.

Finally, the vertical shift is how far the function is shifted vertically from sin(x). In sin(x), the centers are at y=0. In the function given, the centers are at y=1, symbolizing a vertical shift of 1.

Our function is therefore

y = Asin(B(x+C)) + D

A = 1

B = 4

C = π/8

D = 1

y = sin(4(x+π/8)) + 1

Answer(s):

[tex]\displaystyle y = sin\:(4x + \frac{\pi}{2}) + 1 \\ y = -cos\:(4x \pm \pi) + 1 \\ y = cos\:4x + 1[/tex]

Explanation:

[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{\pi}{8}} \hookrightarrow \frac{-\frac{\pi}{2}}{4} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\frac{\pi}{2}} \hookrightarrow \frac{2}{4}\pi \\ Amplitude \hookrightarrow 1[/tex]

OR

[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 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\frac{\pi}{2}} \hookrightarrow \frac{2}{4}\pi \\ Amplitude \hookrightarrow 1[/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 cosine graph, if you plan on writing your equation as a function of sine, 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 = sin\:4x + 1,[/tex] in which you need to replase “cosine” with “sine”, then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosine 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 sine graph [photograph on the right] is shifted [tex]\displaystyle \frac{\pi}{8}\:unit[/tex] to the right, which means that in order to match the cosine graph [photograph on the left], we need to shift the graph BACK [tex]\displaystyle \frac{\pi}{8}\:unit,[/tex] which means the C-term will be negative, and perfourming your calculations, you will arrive at [tex]\displaystyle \boxed{-\frac{\pi}{8}} = \frac{-\frac{\pi}{2}}{4}.[/tex] So, the sine graph of the cosine graph, accourding to the horisontal shift, is [tex]\displaystyle y = sin\:(4x + \frac{\pi}{2}) + 1.[/tex] Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph hits [tex]\displaystyle [0, 2],[/tex] from there to [tex]\displaystyle [\frac{\pi}{2}, 2],[/tex] they are obviously [tex]\displaystyle \frac{\pi}{2}\:unit[/tex] apart, telling you that the period of the graph is [tex]\displaystyle \frac{\pi}{2}.[/tex] 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 one unit 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.

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Math IIIB Performance Task: Piecewise-Defined Functions
Question Set 2

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