What is the amplitude and period of #y=5/3sin(-2/3x)#?

Question

1853 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-28T18:28:38

#Amplitude=5/3#
#Period=3pi#

Consider the form #asin(bx-c)+d#

The amplitude is #|a|#

and the period is #{2pi)/|b|#

We can see from your problem that
#a=5/3# and #b=-2/3#

So for amplitude:

#Amplitude=|5/3|# #---># #Amplitude=5/3#

and for period:

#Period= (2pi)/|-2/3|# #---># #Period= (2pi)/(2/3)#

Consider this as a multiplication for better understanding...

#Period= (2pi)/1-:2/3# #---># #Period= (2pi)/1*3/2#

#Period= (6pi)/2# #---># #Period=3pi#