Step 1:
Write the function.
[tex]f(t)\text{ = 3cos(}\frac{1}{2}(t-\pi))\text{ + 2}[/tex]Step 2
Let's find the amplitude.
Amplitude definition
[tex]\begin{gathered} \mathrm{For\: f}\mleft(\mathrm{x}\mright)\mathrm{=A\cdot}\cos \mleft(\mathrm{Bx-C}\mright)\mathrm{+D\: the\: amplitude\: is\: |A|} \\ \text{From f(t) = 3cos(}\frac{1}{2}(t-\pi))\text{ + 2} \\ \text{The amplitude = |A| = |3| = 3} \\ \text{Amplitude = 3} \end{gathered}[/tex]Midline:
From the graph of the function
[tex]\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{ = }\frac{6}{2}\text{ = 3} \end{gathered}[/tex]Period
[tex]\begin{gathered} \text{Period = }\frac{2\pi}{|b|} \\ We\text{ have the standard form} \\ acos(bx\text{ + c) }\pm d \\ \text{Now rewrite the function f(t) = 3cos(}\frac{1}{2}(t-\pi))\text{ + 2} \\ \text{f(t) = 3cos}(\frac{1}{2}t-\frac{1}{2}\pi)\text{ + 2} \\ \text{Therefore, b = }\frac{1}{2} \\ \text{Hence,} \\ \text{Period = }\frac{2\pi}{\frac{1}{2}} \\ \text{Period = 2}\pi\times\text{ 2} \\ \text{Period = 4}\pi \end{gathered}[/tex]Horizontal shift
[tex]\begin{gathered} \text{Phase shift or horizontal shift = }\frac{c}{b} \\ \text{b = }\frac{1}{2}\text{ , c = }\frac{\pi}{2} \\ \text{Horizontal shift }=\text{ }\frac{\frac{\pi}{2}}{\frac{1}{2}} \\ \text{Horizontal shift = }\pi \end{gathered}[/tex]Graph
