Question

Find the amplitude, period, phase shift, and midline of this function. $y = \sin( \frac{\pi}{6}x + \pi) - 3$

Answer 1

Standard form of a function is

$a\sin (bx+c)\pm d$$y=\sin (\frac{\pi}{6}x+\pi)-3$

So by comparing the given equation to the standard form, we get

a=1, b=pi/6, c=pi, d=3

amplitude can be written as mod a or

$\lvert a\rvert=\lvert1\rvert=1$

So the amplitude is 1

Now for the period, we use generalization as

$\frac{2\pi}{\lvert b\rvert}\text{radians}$$=\frac{2\pi}{\lvert\frac{\pi}{6}\rvert}=\frac{2\pi}{\frac{\pi}{6}}=2\times6\times\frac{\pi}{\pi}=12$

So the period is 12

Here, c is the phase shift, so c=pi

Hence phase shift is pi

Midline is the line that runs between maximum and minimum values

Since, the amplitude is 1 and the phase shift is -pi, the minimum and maximum values are

$\min =\text{ 1-}\pi\text{ }=1-\pi_{}$$\max =1+\pi$

Midline will be the centre of region (1+pi, 1-pi)

So the midline will be

$\frac{1+\pi-(1-\pi)}{2}=\frac{2\pi}{2}=\pi$