determine the amplitude, the period, and the phase shift of the function. y = 8 sin(1/4 x - π/2) write the…

Explanation:

Step1: Identify the form parameters

For the function $y = 8\sin(\frac{1}{4}x-\frac{\pi}{2})$, we can rewrite it in the form $y = A\sin\left[B\left(x - \frac{C}{B}\right)\right]+D$. Here $A = 8$, $B=\frac{1}{4}$, $C=\frac{\pi}{2}$, $D = 0$.

Step2: Calculate the amplitude

The amplitude of the sine - function $y = A\sin\left[B\left(x - \frac{C}{B}\right)\right]+D$ is given by $|A|$. So, the amplitude $|A|=|8| = 8$.

Step3: Calculate the period

The period of the sine - function $y = A\sin\left[B\left(x - \frac{C}{B}\right)\right]+D$ is $T=\frac{2\pi}{|B|}$. Substituting $B=\frac{1}{4}$, we get $T=\frac{2\pi}{\left|\frac{1}{4}\right|}=8\pi$.

Step4: Calculate the phase - shift

The phase - shift of the sine - function $y = A\sin\left[B\left(x - \frac{C}{B}\right)\right]+D$ is $\frac{C}{B}$. Substituting $B = \frac{1}{4}$ and $C=\frac{\pi}{2}$, we have $\frac{C}{B}=\frac{\frac{\pi}{2}}{\frac{1}{4}} = 2\pi$.

Step5: Rewrite the function in the required form

$y=8\sin\left[\frac{1}{4}\left(x - 2\pi\right)\right]+0$

Answer:

Amplitude: 8; Period: $8\pi$; Phase - shift: $2\pi$; $y = 8\sin\left[\frac{1}{4}\left(x - 2\pi\right)\right]$