
What is the amplitude, period, phase shift and vertical displacement of \[y = - 2\,\cos \,2\,\left( {x + 4} \right) - 1\]?
Answer
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Hint: The cosine function has a domain whole real line or the set of real numbers and whose range is bounded between the closed interval \[ - 1\] to\[1\].
The sinusoidal function is of the form: \[y = a\,\cos \,b\,\left( {x + c} \right) + d\] where,
Amplitude: \[|a|\]
Period: \[\dfrac{{2\pi }}{{\left| b \right|}}\]
Phase shift: \[c\]
Vertical displacement: \[d\]
Complete step-by-step solution:
In the given problem, we have \[y = - 2\,\cos \,2\,\left( {x + 4} \right) - 1\]
Here, \[a = - 2,\,\,b = 2,\,\,c = 4\,\,{\text{and}}\,\,d = - 1\]
According to the definition of sinusoidal function:
Amplitude: \[|a| = | - 2|\]
Since, absolute values are always positive.
Therefore, the amplitude is \[|a| = 2\].
Period: \[\dfrac{{2\pi }}{{\left| b \right|}} = \dfrac{{2\pi }}{{\left| 2 \right|}}\]
Simplifying the above expression we get
\[\dfrac{{2\pi }}{{\left| b \right|}} = \dfrac{{2\pi }}{2}\]
By cancelling the both like terms in the numerator and the denominator in the above equation we get,
\[\dfrac{{2\pi }}{{\left| b \right|}} = \pi \]
Phase shift: \[c = 4\], this tells us that the graph of \[y = - 2\,\cos \,2\,\left( {x + 4} \right) - 1\] shift to the left \[4\]units.
Vertical displacement: \[d = - 1\], this tells us that the graph of \[y = - 2\,\cos \,2\,\left( {x + 4} \right) - 1\] shifts $1$ units to the downwards.
Note: The phase shifts in the sinusoidal functions having the left and right shift accordingly when the values become positive and negative respectively. The vertical displacement shifts its position of the sinusoidal graphs to vertically upwards or vertically downwards accordingly as the value becomes negative and positive respectively. The cosine function has the special characters in which it gives the outputs as periodic functions with the periods \[2\pi \] and it is even function by satisfying the function \[f\left( { - x} \right) = f\left( x \right)\] and thus it becomes an even function because \[\cos \left( { - x} \right) = \cos \left( x \right)\] and hence its satisfies the properties of even function. Therefore, the cosine function has a different new character which gives the functional values to be prior in such a way that it becomes positive in several manners when compared to other trigonometric functions.
The sinusoidal function is of the form: \[y = a\,\cos \,b\,\left( {x + c} \right) + d\] where,
Amplitude: \[|a|\]
Period: \[\dfrac{{2\pi }}{{\left| b \right|}}\]
Phase shift: \[c\]
Vertical displacement: \[d\]
Complete step-by-step solution:
In the given problem, we have \[y = - 2\,\cos \,2\,\left( {x + 4} \right) - 1\]
Here, \[a = - 2,\,\,b = 2,\,\,c = 4\,\,{\text{and}}\,\,d = - 1\]
According to the definition of sinusoidal function:
Amplitude: \[|a| = | - 2|\]
Since, absolute values are always positive.
Therefore, the amplitude is \[|a| = 2\].
Period: \[\dfrac{{2\pi }}{{\left| b \right|}} = \dfrac{{2\pi }}{{\left| 2 \right|}}\]
Simplifying the above expression we get
\[\dfrac{{2\pi }}{{\left| b \right|}} = \dfrac{{2\pi }}{2}\]
By cancelling the both like terms in the numerator and the denominator in the above equation we get,
\[\dfrac{{2\pi }}{{\left| b \right|}} = \pi \]
Phase shift: \[c = 4\], this tells us that the graph of \[y = - 2\,\cos \,2\,\left( {x + 4} \right) - 1\] shift to the left \[4\]units.
Vertical displacement: \[d = - 1\], this tells us that the graph of \[y = - 2\,\cos \,2\,\left( {x + 4} \right) - 1\] shifts $1$ units to the downwards.
Note: The phase shifts in the sinusoidal functions having the left and right shift accordingly when the values become positive and negative respectively. The vertical displacement shifts its position of the sinusoidal graphs to vertically upwards or vertically downwards accordingly as the value becomes negative and positive respectively. The cosine function has the special characters in which it gives the outputs as periodic functions with the periods \[2\pi \] and it is even function by satisfying the function \[f\left( { - x} \right) = f\left( x \right)\] and thus it becomes an even function because \[\cos \left( { - x} \right) = \cos \left( x \right)\] and hence its satisfies the properties of even function. Therefore, the cosine function has a different new character which gives the functional values to be prior in such a way that it becomes positive in several manners when compared to other trigonometric functions.
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