Question

Write a sine function that has a midline of 5, an amplitude of 3 and a period of 5/4

Answer 1

`\displaystyle y = 3sin\:1(3)/(5)\pi{x} + 5`

Explanation:

`\displaystyle \boxed{y = 3cos\:(1(3)/(5)\pi{x} - (\pi)/(2)) + 5} \\ \\ 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 5 \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{(5)/(16)} \hookrightarrow ((\pi)/(2))/(1(3)/(5)\pi) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{1(1)/(4)} \hookrightarrow (2)/(1(3)/(5)\pi)\pi \\ Amplitude \hookrightarrow 3`

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 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{1(1)/(4)} \hookrightarrow (2)/(1(3)/(5)\pi)\pi \\ Amplitude \hookrightarrow 3`

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

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