How do you find the amplitude and period of #y=2sin.5x#?

1 Answer
Gió
Sep 12, 2015

Amplitude is #2# and period is #4pi#

Explanation:

The amplitude #A# is represented by the number in front of the #sin# function so:
#A=2#
The period is given as #(2pi)/k# where #k# is the number in front of #x# in the argument of #sin#:
so period#=(2pi)/0.5=4pi~~12.5# that represents the interval needed to complete one oscillation.
Graphically:
graph{2*sin(0.5x) [-16.02, 16.02, -8.01, 8.01]}

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