Consider the function f(x)= 3\sin (\frac{\pi}{3}(x−4))+5. State the am

Consider the function ${\displaystyle f\left(x\right)=3\sin \left(\frac{\pi }{3}(x-4)\right)+5}$.
State the amplitude A, period P, and midline. State the phase shift and vertical translation. In the full period [0, P], state the maximum and minimum y-values and their corresponding x-values.
Enter the exact answers.
Amplitude: ${\displaystyle A=}$
Period: ${\displaystyle P=}$
Midline: ${\displaystyle y=}$
The phase shift is:
a. up 4 units
b. 4 units to the right
c. 4 units to the left
d. 5 units to the left
The vertical translation is:
a. up 4 units
b. down 5 units
c. up 5 units
d. down 4 units
Hints for the maximum and minimum values of f(x):
The maximum value of ${\displaystyle y=\sin \left(x\right)}$ is ${\displaystyle y=1}$ and the corresponding x values are ${\displaystyle x=\frac{\pi }{2}}$ and multiples of ${\displaystyle 2\pi }$ less than and more than this x value. You may want to solve ${\displaystyle \frac{\pi }{3}(x-4)=\frac{\pi }{2}}$.
The minimum value of ${\displaystyle y=\sin \left(x\right)}$ is ${\displaystyle y=-1}$ and the corresponding x values are ${\displaystyle x=\frac{3\pi }{2}}$ and multiples of ${\displaystyle 2\pi }$ less than and more than this x value. You may want to solve ${\displaystyle \frac{\pi }{3}(x-4)=\frac{3\pi }{2}}$.
If you get a value for x that is less than 0, you could add multiples of P to get into the next cycles.
If you get a value for x that is more than P, you could subtract multiples of P to get into the previous cycles.
For x in the interval [0, P], the maximum y-value and corresponding x-value is at:
x=
y=
For x in the interval [0, P], the minimum y-value and corresponding x-value is at:
${\displaystyle x=}$
${\displaystyle y=}$