Given: The sine graph shown in the image
To Determine: The amplitude, the period and the sine function
Solution
The general sine function is as shown below
From the given graph,
The amplitude is
![\begin{gathered} A=2 \\ Period=270^0-90^0=180^0 \\ Period=(2\pi)/(B) \\ \pi=180^0 \\ 180^0=(2*180^0)/(B) \\ B*180^0=360^0 \\ B=(360^0)/(180^0) \\ B=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vgq3urrxcnzuwa1dje20g51l27nm8irx73.png)
![\begin{gathered} C=phase(shift)=-45^0 \\ D=-3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/t91md0nbu6yo4rfalgc5uegkskt0ahplj9.png)
So, the function of the graph would be
![\begin{gathered} y=Asin(B(x+C))+D \\ A=2 \\ B=2 \\ C=-45^0 \\ D=-3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/magl1gdfqcyv715ybaqlze9zzcc1hph08g.png)
![\begin{gathered} f(x)=2sin(2(x-45^0)-3 \\ f(x)=2sin(2x-90^0)-3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yl57ibyoshzls95zyf0u5i9h8ng5rj4o2n.png)
![\begin{gathered} OR \\ f(x)=2sin(2x-(\pi)/(2))-3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/486xibfm5z4p7gxufmh0gsqpmpt8fbvrpr.png)
Hence, the sine function of the given graph is
y = 2sin(2x - 90⁰) - 3
OR
y = 2sin(2x - π/2) - 3