Question

How do you find the amplitude and period for $y=cos\left( \dfrac{1}{2}x \right)-2$ ?

Answer 1

Hint: Now the given function can be written in the form $acos\left( b \left( x-c \right)+d \right)$ hence we get the values of a and b. Now the amplitude of such functions is equal to a and the period of the function is calculated by the formula $\dfrac{2\pi}{\left| b \right|}$ . Hence we have the amplitude and period of the function.

Complete step by step solution:
Now we consider the given function $y=cos\left( \dfrac{1}{2}x \right)-2$ .
We want to find the amplitude and period of the function.
Let us first understand the meaning of this term.
Amplitude of the function means the maximum value a function can achieve is measured from the axis.
Hence to the amplitude of the function is the maximum height of the function.
Similarly the period of the function is nothing but the distance between two crests or two troughs. Basically period is nothing but the distance after which the function starts to repeat it’s value.
Now for any trigonometric function of the form $y=acos\left( b \left( x-c \right)+d \right)$ the amplitude is equal to a and the period of the function is given by $\dfrac{2\pi}{\left| b \right|}$ .
Now consider the given function.
Here we can write the function as $y=cos\left( \dfrac{1}{2} \left( x-0 \right) \right) -2$ hence we get a = 1 and b = $\dfrac{1}{2}$
Hence the amplitude of the function is 1 and the period of the function is $\dfrac{2\pi}{\dfrac{1}{2}}=4\pi$ .

Hence the amplitude of the given function is 1 and the period of the given function is $4\pi$

Note: Now note that we have a constant in the function. The constant tells us the vertical shift of the function. Hence form functions of the form $acos \left (bx-c \right) +d $ the vertical shift of the function is given equal to d and is the shift of the function from original function in vertical direction.