Question

Write a cosine function that has an amplitude of 5, a midline of 4 and a period of 1.

Answer 1

`\displaystyle y = 5cos\:2\pi{x} + 4`

Explanation:

`\displaystyle \boxed{y = 5sin\:(2\pi{x} + (\pi)/(2)) + 4} \\ \\ 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 4 \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{-(1)/(4)} \hookrightarrow (-(\pi)/(2))/(2\pi) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{1} \hookrightarrow (2)/(2\pi)\pi \\ Amplitude \hookrightarrow 5`

OR

`\displaystyle 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 4 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{1} \hookrightarrow (2)/(2\pi)\pi \\ Amplitude \hookrightarrow 5`

From the above information, you now should have an ideya of how to interpret trigonometric equations like these.

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