Question

A sine function had an amplitude of 3, period of 6pi, horizontal shift of 3pi/2, & vertical shift of -1.

Answer 1

Answer: $\bold{y=\dfrac{1}{2}}$

Step-by-step explanation:

f(x) = A sin (Bx - C) + D

• amplitude = |A|
• period =$\dfrac{2\pi}{B}$
• phase shift =$\dfrac{C}{B}$
• 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 $6\pi=\dfrac{2\pi}{B}\quad \rightarrow \quad B=\dfrac{2\pi}{6\pi}\quad \rightarrow \quad B=\dfrac{1}{3}$

C

$\text{phase shift is given as}\ \dfrac{3\pi}{2}\ \text{so}\ \dfrac{3\pi}{2}=\dfrac{C}{\frac{1}{3}}\quad \rightarrow\quad \dfrac{(\frac{1}{3})3\pi}{2}=C\quad \rightarrow\quad \dfrac{\pi}{2}=C$

D

vertical shift of -1 is given so -1 = D

Now, substitute the values of A, B, C, and D into the formula (above):

$f(x) = 3\ sin \bigg(\dfrac{1}{3}x - \dfrac{\pi}{2}\bigg) - 1$

Next, solve when x = 2π

$f(2\pi) = 3\ sin \bigg(\dfrac{1}{3}(2\pi) - \dfrac{\pi}{2}\bigg) - 1$

        $= 3\ sin \bigg(\dfrac{2\pi}{3} - \dfrac{\pi}{2}\bigg) - 1$

        $= 3\ sin \bigg(\dfrac{4\pi}{6} - \dfrac{3\pi}{6}\bigg) - 1$

        $= 3\ sin \bigg(\dfrac{\pi}{6}\bigg) - 1$

        $= 3\ \bigg(\dfrac{1}{2}\bigg) - 1$

        $=\dfrac{3}{2}-\dfrac{2}{2}$

        $=\dfrac{1}{2}$