1 Answer

Ethan Goldblum · Answer 1 · 2022-12-12T02:20:03+0000

Answer:

$\displaystyle 2$

Explanation:

$\displaystyle \boxed{y = -2cos\:((\pi)/(3)x - (2)/(3)\pi) + (1)/(2)} \\ \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow (1)/(2) \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{2} \hookrightarrow ((2)/(3)\pi)/((\pi)/(3)) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{6} \hookrightarrow (2)/((\pi)/(3))\pi \\ Amplitude \hookrightarrow 2$

OR

$\displaystyle \boxed{y = -2sin\:((\pi)/(3)x - (\pi)/(6)) + (1)/(2)} \\ \\ y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow (1)/(2) \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{(1)/(2)} \hookrightarrow ((\pi)/(6))/((\pi)/(3)) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{6} \hookrightarrow (2)/((\pi)/(3))\pi \\ Amplitude \hookrightarrow 2$

As you can see, the right-centred photograph displays the trigonometric graph of
$\displaystyle y = -2cos\:(\pi)/(3)x + (1)/(2),$ in which you need to replase "sine" with "cosine", then figure out the appropriate C-term that will horisontally shift and map the graph onto the sine graph [photograph farthest to 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 be very careful with your calculations. So, between the two photographs, we can tell that the cosine graph [centre right photograph] is shifted
$\displaystyle 2\:units$ to the left, which means that in order to match the sine graph [photograph farthest to the left], we need to shift the graph FORWARD
$\displaystyle 2\:units,$ which means the C-term will be positive, and by perfourming your calculations, you get
$\displaystyle \boxed{2} = ((2)/(3)\pi)/((\pi)/(3)).$ So, the cosine graph of the sine graph, accourding to the horisontal shift, is
$\displaystyle y = -2cos\:((\pi)/(3)x - (2)/(3)\pi) + (1)/(2).$ Intermediate, we have two sine graphs to deal with [centre left photograph and the photograph farthest to the left]. The left-centred photograph displays the trigonometric graph of
$\displaystyle y = -2sin\:(\pi)/(3)x + (1)/(2).$ So, just like before, we need to figure out the appropriate C-term that will horisontally shift the graph back into position ["position", meaning shifting this sine graph back in plase (matching the sine graph in the photograph on the left)]. So, between the two photographs, we can tell that the left-centred sine graph is shifted
$\displaystyle (1)/(2)\:unit$ to the left, which means that in order to match the leftward sine graph, we need to shift the graph FORWARD
$\displaystyle (1)/(2)\:unit,$ which means the C-term will be positive, and by perfourming your calculations, you get
$\displaystyle \boxed{(1)/(2)} = ((\pi)/(6))/((\pi)/(3)).$ So, when shifted back into original position, the sine graph is
$\displaystyle y = -2sin\:((\pi)/(3)x - (\pi)/(6)) + (1)/(2).$ 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 WILL hit
$\displaystyle [-2(1)/(2), (1)/(2)],$ from there to
$\displaystyle [3(1)/(2), (1)/(2)],$ they are obviously
$\displaystyle 6\:units$ apart, telling you that the period of the graph is
$\displaystyle 6.$ Now, the amplitude is obvious to figure out because it is the A-term, and to be certain, 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
$\displaystyle y = (1)/(2),$ in which each crest is extended two units beyond the midline, hence, your amplitude. Now... the sine graph in the photograph farthest to the right is the OPPOCITE of the sine graph in the photograph farthest to the left, and the reason for this is because of the negative inserted in front of the amplitude value. Whenever you insert a negative in front of the amplitude value of any trigonometric equation, the whole graph reflects over the midline. Keep this in mind moving forward. Now, with all that being said, no matter how far the graph shifts vertically, the midline will ALWAYS follow.