Question

Which graph below has a midline of about -1, an amplitude of 2, and a period of 3pi

Answer 1

the last one

Explanation:

The two on the right bounce up and down around -1. Two up and two down from -1. But only the one on the bottom right takes 3 pi to come back to the same place it started.

Answer 2

Explanation:

`\displaystyle y = 2cos\:((2)/(3)x - (\pi)/(2)) - 1 \\ y = 2sin\:(2)/(3)x - 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{(3)/(4)\pi} \hookrightarrow ((\pi)/(2))/((2)/(3)) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{3\pi} \hookrightarrow (2)/((2)/(3))\pi \\ Amplitude \hookrightarrow 2`

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{3\pi} \hookrightarrow (2)/((2)/(3))\pi \\ Amplitude \hookrightarrow 2`

You will need the above information to help you interpret the graph. So, first off, you MUST figure the period out by using wavelengths. So, looking at where the graph hits
`\displaystyle [-4(1)/(2)\pi, -1],` from there to
`\displaystyle [-1(1)/(2)\pi, -1],` they are obviously
`\displaystyle 3\pi\:units` apart, telling you that the period of the graph is
`\displaystyle 3\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 two units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

*If you wanted to know the equation(s) of the graph, look above. If you have any questions on how the equation(s) came about, do not hesitate to ask.

I am delighted to assist you at any time.