MHB Graph of $y=\sin{x}-2$ on the domain $[0,2\pi]$

[karush]

Graph $y=\sin{x}-2$ on the domain $[0,2\pi]$
This is a sample math problem in preparation for the entrance exam for the USAF Academy
Even not asked I thot also the Period, Amplitude, PS and list some observations that should be know to graph without an app

1. we know that sin(0)=0 so sin(x) goes thru origin

$Y_{sin}=A\sin\left[\omega\left(x-\dfrac{\phi}{\omega} \right) \right]+B
\implies A\sin\left(\omega x-\phi \right)+B$
A=Amplitude B=Vertical Shift
T=Period= $\quad\dfrac{2\pi}{\omega}$
PS=Phase Shift $\quad\dfrac{\phi}{\omega}$
ok this get ? at times
and,,,,,

 

[skeeter]

$y = \sin{x} - 2$

just shift $y=\sin{x}$ down 2 units … why are you making it more complicated than necessary?

 

[karush]

$y = \sin{x} - 2$

just shift $y=\sin{x}$ down 2 units … why are you making it more complicated than necessary?

well I know this is a very simple one but I get confused on PS and T
A and VS are easy

$A\sin\left(\omega x-\phi \right)+B\implies (1)\sin\left((1) x-(0) \right)+(-2)$

$T=\dfrac{2\pi}{1}=2\pi$ $PS=\quad\dfrac{0}{1}=0$