Question

How do you find the amplitude, period, phase shift for $y=-\sin \left( 5x+3 \right)$?

Answer 1

Compare the given function with the general form of the sine function given as: $y=a\sin \left( b\left( x+c \right) \right)+d$. Here, ‘a’ will be the amplitude of the given function. Now, to find the period of the function, use the formula: $T'=\dfrac{T}{\left| b \right|}$, where T is the period of the function $\sin x$. Finally, determine the phase shift of $y=-\sin \left( 5x+3 \right)$ by comparing it with $y=a\sin \left( b\left( x+c \right) \right)+d$ and finding the value of ‘c’.

Complete step by step solution:
Here, we have been provided with the function $y=-\sin \left( 5x+3 \right)$ and we are asked to find its amplitude, period and phase shift.
Now, the general form of the sine function is given as $y=a\sin \left( b\left( x+c \right) \right)+d$, where we have the vertical compression or stretch (a), also known as the amplitude, horizontal compression or stretch (b), horizontal shift (c), also known as the phase shift, and vertical shift (d) of the function. So, on converting the given function into the general form and comparing the values, we have,
$\Rightarrow y=-1\sin \left( 5\left( x+\dfrac{3}{5} \right) \right)+0$
$\Rightarrow a=-1,b=5,c=\dfrac{3}{5}$ and $d=0$.
(i) Amplitude: - Amplitude of a function is defined as the minimum and maximum value that a function can approach. Clearly, we have the value of ‘a’ equal to -1 which is negative but we have to take its mod to get the amplitude. Therefore, the amplitude of the given function is 1 unit.
(ii) Time period: - Time period of a function is defined as the interval of x after which the value of the function starts repeating itself. It is also simply called the period and the function is known as periodic function. The period of the function $y=a\sin \left( b\left( x+c \right) \right)+d$ is given as $T'=\dfrac{T}{\left| b \right|}$, where T is the period of $\sin x$.
Now, we know that the period of $\sin x$ is $T=2\pi$, so we have,

& \Rightarrow T'=\dfrac{2\pi }{\left| 5 \right|} \\\ & \Rightarrow T'=\dfrac{2\pi }{5} \\\ \end{aligned}$$ Therefore, the period of the provided function is $$\dfrac{2\pi }{5}$$. (iii) Phase shift: - Phase shift of a function is defined as how far a function is shifted horizontally from the usual position. Now, for sine function given as $y=a\sin \left( b\left( x+c \right) \right)+d$ the phase shift is given by the term ‘c’: - 1\. If ‘c’ is negative then phase shift is $\left| c \right|$ units to the right. 2\. If ‘c’ is positive then phase shift is $\left| c \right|$ units to the left. In the above question we have $$-\sin \left( 5\left( x+\dfrac{3}{5} \right) \right)$$, so on comparing we have $c=\dfrac{3}{5}$ which is positive. So, it is the $${{2}^{nd}}$$ condition. Hence, the phase shift is $\dfrac{3}{5}$ units to the left. **Note:** One may note that we have another type of shift known as vertical shift which is given as ‘d’ units according to the general form. In the above question it was 0 so there was no vertical shift. Note that phase shift has no effect on time period. Time period changes only when the coefficient of x is changed in the function. You must remember the formula $$T'=\dfrac{T}{\left| b \right|}$$ as it will be used in chapter integration.