# How do you find period, amplitude, phase shift and midline of f(x) = −4 sin(2x + π) − 5?

Jun 10, 2018

#### Explanation:

we have standard form
$a \sin \left(b x + c\right) \pm d$
$| a | \text{ is amplitude," (2pi)/|b|" is period," " c is phase shift (or horizontal shift), d is vertical shift}$
comparing the equation with standard form
$a = - 4 , b = 2 , c = \pi , d = - 5$

midline is the line that runs between the maximum and minimum value($i . e$ amplitudes)
since the new amplitude is 4 and graph is shifted 5 units in negative $y - \text{axis}$ ($d = - 5$)

therefore maximum value is $4 - 5 = - 1$ and minimum value is $- 4 - 5 = - 9$ midline will be centre of the region$\left(- 1 , - 9\right) \text{ " i.e" } - 5$

phase change =$\pi$ units right(c is positive)
period is $\frac{2 \pi}{2} = \pi$

it will be more clear from graph
graph{-4sin(2x+pi)-5 [-16.02, 16.01, -8.01, 8.01]}