# How do you find the amplitude, period, phase shift, and vertical shift for y = 5 - 3 sin (2theta - pi)?

Aug 8, 2015

Normally you may have learned an equation similar to this one:

$y = A \sin \left(n \theta + \phi\right) + k$

(or replace with whatever variables you're using.)

Basically:
$A$ is the amplitude, and $A = 3$ (recall that the amplitude goes from the apparent axis [EX: if shifted up 2 units, then the axis is $y = 2$] to the highest point in the y range)
$\phi$ is the phase shift, and $\phi = - \pi$, meaning $\pi$ to the right
$k$ is the vertical shift, and $k = + 5$, meaning $5$ upwards

The period can be found by realizing that the period of $\sin \theta$ is $2 \pi$. Knowing that, deriving the period is fairly simple. Since doubling the value of $n$ halves the period:

$n = \frac{2 \pi}{T} \implies T = \frac{2 \pi}{n}$

Testing this, from the above information, you can infer that the period is $\pi$. So:

$\pi = \frac{2 \pi}{n} \implies n \pi = 2 \pi$
$n = 2$

and $n$ is indeed $2$ in the equation.

Overall, this equation is $\sin \theta$ flipped over the x-axis, compressed to a period of $\pi$, stretched in the vertical directions to an amplitude of $3$, then shifted right $\pi$ units and up $5$ units.

$y = 5 - 3 \sin \left(2 \theta - \pi\right)$:
graph{5-3sin(2x - pi) [-3.2, 3.2, -0.5, 8.2]}