Answer: [tex]\bold{y=\dfrac{1}{2}}[/tex]
Step-by-step explanation:
f(x) = A sin (Bx - C) + D
- amplitude = |A|
- period =[tex]\dfrac{2\pi}{B}[/tex]
- phase shift =[tex]\dfrac{C}{B}[/tex]
- vertical shift = D
A
amplitude of 3 is given so 3 = |A| → A = ± 3, since it is stated that this is a positive function, then A = 3
B
period of 6π is given so [tex]6\pi=\dfrac{2\pi}{B}\quad \rightarrow \quad B=\dfrac{2\pi}{6\pi}\quad \rightarrow \quad B=\dfrac{1}{3}[/tex]
C
[tex]\text{phase shift is given as}\ \dfrac{3\pi}{2}\ \text{so}\ \dfrac{3\pi}{2}=\dfrac{C}{\frac{1}{3}}\quad \rightarrow\quad \dfrac{(\frac{1}{3})3\pi}{2}=C\quad \rightarrow\quad \dfrac{\pi}{2}=C[/tex]
D
vertical shift of -1 is given so -1 = D
Now, substitute the values of A, B, C, and D into the formula (above):
[tex]f(x) = 3\ sin \bigg(\dfrac{1}{3}x - \dfrac{\pi}{2}\bigg) - 1[/tex]
Next, solve when x = 2π
[tex]f(2\pi) = 3\ sin \bigg(\dfrac{1}{3}(2\pi) - \dfrac{\pi}{2}\bigg) - 1[/tex]
[tex]= 3\ sin \bigg(\dfrac{2\pi}{3} - \dfrac{\pi}{2}\bigg) - 1[/tex]
[tex]= 3\ sin \bigg(\dfrac{4\pi}{6} - \dfrac{3\pi}{6}\bigg) - 1[/tex]
[tex]= 3\ sin \bigg(\dfrac{\pi}{6}\bigg) - 1[/tex]
[tex]= 3\ \bigg(\dfrac{1}{2}\bigg) - 1[/tex]
[tex]=\dfrac{3}{2}-\dfrac{2}{2}[/tex]
[tex]=\dfrac{1}{2}[/tex]
