
How do you find period, amplitude, phase shift and midline of $f(x) = - 4\sin (2x + \pi ) - 5$?
Answer
495.3k+ views
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 $y = - 5$
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 $y = - 5$
Recently Updated Pages
NCERT Solutions For Class 12 Maths Application Of Integrals Exercise 8.1

NCERT Solutions For Class 12 Maths Miscellaneous Exercise - Application Of Integrals

JEE Main Maths Mock Test 2025: FREE Online Mock Test Series

NCERT Solutions For Class 12 Maths Vector Algebra Exercise 10.4

NCERT Solutions For Class 12 Maths Differential Equations Exercise 9.1

NEET Physics Mock Test 2025: Free Practice & Solutions

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE
