Question

#8 help with algebra 2 question. That’s the only picture I have. I tried writing it out.

Answer 1

Given a cosine function graph;

The general cosine function is

`y=A\cos(Bx-C)+D`

Where

`\begin{gathered} A\text{ is the amplitude} \\ Period=(2\pi)/(B) \\ C\text{ is the phase shift} \\ D\text{ is the vertical shift} \end{gathered}`

From the graph,

The midline is y = 1

The amplitude, A, is

`\begin{gathered} A=4-1=3 \\ A=3 \end{gathered}`

The amplitude, A is 3

Where,

`\begin{gathered} Period=12 \\ Period=(2\pi)/(B) \\ 12=(2\pi)/(B) \\ Crossmultiply \\ 12B=2\pi \\ Duvide\text{ both sides by 12} \\ (12B)/(12)=(2\pi)/(12) \\ B=(\pi)/(6) \end{gathered}`

The phase shift, C = 0, and the vertical, D, is 1

Thus, the equation of the graph is

`\begin{gathered} y=A\cos(Bx-C)+D \\ Where \\ A=3 \\ B=(\pi)/(6) \\ C=0 \\ D=1 \\ y=3\cos((\pi)/(6)x)+1 \end{gathered}`

The graph is shown below

Hence, the equation is

`y=3\cos((\pi)/(6)x)+1`