Question

Given f(t) = -2sin(1/2t - pi/2) - 1, find the amplitude, midline, period, horizontal shift and sketch a graph with labeled axis’s

Answer 1

Given:

`f(t)=-2\sin ((1)/(2)t-(\pi)/(2))-1`

To get the wave parameters, we compare it to the wave equation.

`\begin{gathered} y=A\sin (wt\pm kx) \\ y=A\sin (2\pi ft-(2\pi x)/(\lambda)) \end{gathered}`

Comparing both equations, the following can be deduced;

`\begin{gathered} y=A\sin (2\pi ft-(2\pi x)/(\lambda)) \\ f(t)=-2\sin ((1)/(2)t-(\pi)/(2))-1 \\ \\ \text{Amplitude, A = 2} \\ \text{Midline is at y = -1} \end{gathered}`

To get the period,

`\begin{gathered} 2\pi ft=(1)/(2)t \\ f=(t)/(2(2\pi t)) \\ f=(1)/(4\pi) \\ \\ R\text{ecall the formula relating frequency and period,} \\ T=(1)/(f) \\ T=(1)/((1)/(4\pi)) \\ T=4\pi \\ \\ \text{The period is 4}\pi \end{gathered}`

The horizontal shift is;

`(\pi)/(2)`

Hence,

`\begin{gathered} A=2\ldots\ldots\ldots\ldots.\ldots\ldots..\ldots\text{Amplitude} \\ y=-1\ldots\ldots\ldots.\ldots..\ldots\ldots.midl\text{ine} \\ T=4\pi\ldots\ldots\ldots\ldots\ldots\ldots........period \\ Shift=(\pi)/(2)\ldots.\ldots\ldots\ldots\ldots..horizontal\text{ shift} \end{gathered}`

The graph is as shown below;