Question

30 points to whoever can answer this please!

Complete the general form of the equation of a sinusoidal function having an amplitude of 6, a period of 2pi/3, and a phase shift to the left 1 unit.

Answer 1

A common sinusoidal function is:


`y = asin(bx)`

so,always amplitude is a, so a=6

and we know period is:


`p = ( |b| )/(2\pi)`

so


`( |b| )/(2\pi) = (2\pi)/(3)`


`b = + or - \frac{4 {\pi}^(2) }{3}`

and we have 2 equations that while we can consider which is appropriate ,to have its graph!


`6 \sin( \frac{4 {\pi}^(2) }{3}x )`
and when you want to shift it 1 unit left we have:

`6 \sin( \frac{4 {\pi}^(2) }{3}x +1)`

Answer 2

y=6sin3(x+1)

Explanation: