How do you find the period and amplitude for #y=1/4 cos((2x)/3)#?

1 Answer
Gió
Jun 28, 2016

I found:
#"Amplitude"=1/4#
#"period"=3pi#

Explanation:

The amplitude will be the number in front of your #cos#, i.e., #1/4#; this tells you that your function oscillate between #1/4# and #-1/4#.
The period is a bit more tricky; you use the number in front of the #x# of the argument of #cos#, i.e., #2/3#; let us call it #n#, so we have:
#"period"=(2pi)/n=(2pi)/(2/3)=3pi# this means that your function makes a complete oscillation in #3pi# radians.

Graphically:
graph{(1/4)cos(2x/3) [-10, 10, -5, 5]}

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