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NO LINKS!! Please help me with this problem Part 2e​

NO LINKS!! Please help me with this problem Part 2e​-example-1
User Rcruz
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Answer:


\textsf{b)} \quad f(x)=5 \sin\left((\pi)/(8)x\right)+3

Explanation:

Standard form of a sine function


f(x)=\text{A} \sin\left[\text{B}(x+\text{C})\right]+\text{D}

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

If the period is 16 units then:


(2\pi)/(B)=16 \implies B=(2\pi)/(16)=(\pi)/(8)

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


\implies f(x)=5 \sin\left((\pi)/(8)x\right)+3

NO LINKS!! Please help me with this problem Part 2e​-example-1
User Sheikh Hasib
by
8.6k points

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