Write a sine function that has an amplitude of 5, a midline of y = 3 and a period of 3π/7

Answer 1

The sin function is of the format y=a sin (b(x+c)) + d

a is the amplitude

the period = 2π/b

c is the phase shift

d is the vertical shift

For your problem,

a=5

c=0 as there is no phase shift

3π/7 = 2π/b

b=2π•7/3π

b=14/3

d is the midline value = 3

Placing those values back into the equation yields:

y=a sin (b(x+c)) + d

y=5 sin (14x/7) + 3

Answer 2

y = 5sin(14x/3) + 3