Question

How do you find period, amplitude, phase shift and midline of $f(x) = - 4\sin (2x + \pi ) - 5$?

Answer 1

Hint: We know the period, amplitude, phase shift for standard form. Compare our expression with the standard form to obtain these values. Midline can be found easily by taking the mean of maxima and minima.

Complete step by step solution:
Standard form of trigonometric curve,
$= a\sin (bx + c) + d$
In the standard form the coefficients indicate the following data:
$= \left| a \right|$ is amplitude,
$= \dfrac{{2\pi }}{{\left| b \right|}}$ is period,
$= c$ is phase shift,
$ \Rightarrow y = d$ is the midline.
Comparing the expression with trigonometric standard form of expression,
$ \Rightarrow a = - 4$
$ \Rightarrow b = 2$
$ \Rightarrow c = \pi $
$ \Rightarrow d = - 5$
Thus calculating the required parameters by substituting the information obtained from the question.
Period $ = \dfrac{{2\pi }}{2} = \pi $
Amplitude $ = 4$
Phase shift $ = \pi $ (positive sign indicates right)
Midline is $y = - 5$

Additional information:
Midline can be considered as the mean value of the curve. The graph oscillates about the midline with an amplitude equivalent to coefficient of trigonometric operation with a period obtained from the coefficient of the variable (here ‘x’).

Note:
Alternative method
The period of $\sin x$is $2\pi $and amplitude is 1.
Substituting $x$ by $\;2x$ the period will shrink to $\pi $.
Substituting $\;2x$ by $\;2x + \pi $ there will be phase shift of $\pi $.
Multiplying the expression by -4,
$= - 4\sin (2x + \pi )$
then the amplitude will increase to 4.
Subtracting -5 from the expression,
$= - 4\sin (2x + \pi ) - 5$
The midline will be the non-trigonometric constant part.
Hence,
Period $ = \pi $
Amplitude $ = 4$
Phase shift $ = \pi $
Midline is $y = - 5$