determine the equation of a sine function with\n• minimum (π/2, -6)\n• maximum (5π/6, 4)\nselect all that…

Explanation:

Step1: Find the amplitude

The amplitude $A$ is half the difference between the maximum and minimum values. $A=\frac{4 - (-6)}{2}=\frac{10}{2} = 5$. The function can be of the form $y = A\sin(B(x - C))+D$. Since the function goes from a minimum to a maximum, and the standard - sine function goes from 0 to a maximum first, the sine function is reflected about the $x$ - axis, so $A=- 5$.

Step2: Find the vertical shift

The vertical shift $D$ is the average of the maximum and minimum values. $D=\frac{4+( - 6)}{2}=\frac{-2}{2}=-1$.

Step3: Find the period

The period $T$ can be found from the $x$ - values of the maximum and minimum. The distance between a minimum and a maximum is $\frac{T}{2}$. Let $x_1=\frac{\pi}{2}$ and $x_2 = \frac{5\pi}{6}$. Then $\frac{T}{2}=\frac{5\pi}{6}-\frac{\pi}{2}=\frac{5\pi - 3\pi}{6}=\frac{\pi}{3}$, so $T=\frac{2\pi}{3}$. Using the formula $T=\frac{2\pi}{B}$, we get $\frac{2\pi}{B}=\frac{2\pi}{3}$, so $B = 3$.

Step4: Find the phase - shift

We know the general form of the sine function is $y=-5\sin(3(x - C))-1$. We can use the minimum point $(\frac{\pi}{2},-6)$. Substituting $x = \frac{\pi}{2}$ and $y=-6$ into $y=-5\sin(3(x - C))-1$ gives: $-6=-5\sin(3(\frac{\pi}{2}-C))-1$. First, simplify the equation: $-6 + 1=-5\sin(3(\frac{\pi}{2}-C))$, so $-5=-5\sin(3(\frac{\pi}{2}-C))$, then $\sin(3(\frac{\pi}{2}-C)) = 1$. We know that $\sin\theta=1$ when $\theta=\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}$. So $3(\frac{\pi}{2}-C)=\frac{\pi}{2}+2k\pi$. Solve for $C$: $\frac{3\pi}{2}-3C=\frac{\pi}{2}+2k\pi$, $-3C=\frac{\pi}{2}-\frac{3\pi}{2}+2k\pi$, $-3C=-\pi + 2k\pi$, $C=\frac{\pi}{3}-\frac{2k\pi}{3}$. When $k = 0$, $C=\frac{\pi}{3}$. The equation of the sine function is $y=-5\sin(3(x-\frac{\pi}{3}))-1$.

Answer:

B. $y=-5\sin3(x - \frac{\pi}{3})-1$