1 Answer

Mutexkid · Answer 1 · 2021-03-13T02:11:46+0000

Step-by-step Step-by-step explanation:

$\displaystyle \boxed{y = -cos\:(2x - (\pi)/(2)) + 1} \\ \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 1 \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{(\pi)/(4)} \hookrightarrow ((\pi)/(2))/(2) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{\pi} \hookrightarrow (2)/(2)\pi \\ Amplitude \hookrightarrow 1$

OR

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

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 there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the centre photograph displays the trigonometric graph of
$\displaystyle y = -cos\:2x + 1,$ 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 [centre photograph] is shifted
$\displaystyle (\pi)/(4)\:unit$ to the left, which means that in order to match the sine graph [photograph on the left], we need to shift the graph FORWARD
$\displaystyle (\pi)/(4)\:unit,$ which means the C-term will be positive, and by perfourming your calculations, you will arrive at
$\displaystyle \boxed{(\pi)/(4)} = ((\pi)/(2))/(2).$ So, the cosine graph of the sine graph, accourding to the horisontal shift, is
$\displaystyle y = -cos\:(2x - (\pi)/(2)) + 1.$ 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
$\displaystyle [-\pi, 1],$ from there to
$\displaystyle [-2\pi, 1],$ they are obviously
$\displaystyle \pi\:units$ apart, telling you that the period of the graph is
$\displaystyle \pi.$ 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
$\displaystyle y = 1,$ in which each crest is extended one unit beyond the midline, hence, your amplitude. Now, there is one more piese of information you should know -- 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.