Question

Given f(t)=3cos(1/2(t-pi))+2 , what’s the amplitude, midline, period, horizontal shift and graph look like

Answer 1

Step 1:

Write the function.

`f(t)\text{ = 3cos(}(1)/(2)(t-\pi))\text{ + 2}`

Step 2

Let's find the amplitude.

Amplitude definition

`\begin{gathered} \mathrm{For\: f}\mleft(\mathrm{x}\mright)\mathrm{=A\cdot}\cos \mleft(\mathrm{Bx-C}\mright)\mathrmA \\ \text{From f(t) = 3cos(}(1)/(2)(t-\pi))\text{ + 2} \\ \text = 3 \\ \text{Amplitude = 3} \end{gathered}`

Midline:

From the graph of the function

`\begin{gathered} \text{Midline = }\frac{maximum\text{ - minimum}}{2} \\ \text{maximum = 5, minimum = -1} \\ \text{Midline = }\frac{5\text{ - (-1)}}{2} \\ \text{Midline = }\frac{5\text{ + 1}}{2}\text{ = }(6)/(2)\text{ = 3} \end{gathered}`

Period

`\begin{gathered} \text{Period = }(2\pi)/(|b|) \\ We\text{ have the standard form} \\ acos(bx\text{ + c) }\pm d \\ \text{Now rewrite the function f(t) = 3cos(}(1)/(2)(t-\pi))\text{ + 2} \\ \text{f(t) = 3cos}((1)/(2)t-(1)/(2)\pi)\text{ + 2} \\ \text{Therefore, b = }(1)/(2) \\ \text{Hence,} \\ \text{Period = }(2\pi)/((1)/(2)) \\ \text{Period = 2}\pi*\text{ 2} \\ \text{Period = 4}\pi \end{gathered}`

Horizontal shift

`\begin{gathered} \text{Phase shift or horizontal shift = }(c)/(b) \\ \text{b = }(1)/(2)\text{ , c = }(\pi)/(2) \\ \text{Horizontal shift }=\text{ }((\pi)/(2))/((1)/(2)) \\ \text{Horizontal shift = }\pi \end{gathered}`

Graph