Answer:
![\textsf{b)} \quad f(x)=5 \sin\left((\pi)/(8)x\right)+3](https://img.qammunity.org/2023/formulas/mathematics/college/howp3npa0sg7k7ldmu9oubrbl97n60ybri.png)
Explanation:
Standard form of a sine function
![f(x)=\text{A} \sin\left[\text{B}(x+\text{C})\right]+\text{D}](https://img.qammunity.org/2023/formulas/mathematics/college/eek6c78pi3ze9v7rdc8xa85ia936iudt3d.png)
where:
- A = amplitude (height from the mid-line to the peak).
- 2π/B = period (horizontal distance between consecutive peaks).
- C = phase shift (horizontal shift - positive is to the left).
- D = vertical shift.
Given:
- Period = 16 units
- Mid-line: y = 3
- Maximum: y = 8
If the maximum is y = 8 and the mid-line is y = 3 then the amplitude is:
![\implies A=8-3=5](https://img.qammunity.org/2023/formulas/mathematics/college/qvmnj1cdyu2g2towcddtt04grekkyd97zp.png)
If the period is 16 units then:
![(2\pi)/(B)=16 \implies B=(2\pi)/(16)=(\pi)/(8)](https://img.qammunity.org/2023/formulas/mathematics/college/m6fozbztssyfuhonbittrvi3utvsd788df.png)
There appears to be no horizontal shift, so C = 0.
If the mid-line is y = 3 then the curve has been shifted 3 units up (vertically). Therefore, D = 3.
Substitute these values into the standard formula:
![\implies f(x)=5 \sin\left[(\pi)/(8)(x+0)\right]+3](https://img.qammunity.org/2023/formulas/mathematics/college/ey5bjmniewo17bb4hu20kaf17i6cvgor7f.png)
![\implies f(x)=5 \sin\left((\pi)/(8)x\right)+3](https://img.qammunity.org/2023/formulas/mathematics/college/5z7q56ld56ahfsedcl6btfsuafy6dxr128.png)