How do you graph #1/2sin(x-pi)#?

1 Answer
Gió
Apr 18, 2015

Observing the equation of your function you can deduce a lot of things.
Considering: #y=1/2sin(1*x-pi)# you have a sine curve with:

1]
The #1# in front of #x# (indicated as #k#) allows you to evaluate the length #lambda# of your sine curve; #k=(2pi)/lambda# or #1=(2pi)/lambda# and #lambda=2pi#;

2]
The #1/2# is the Amplitude of your sine curve (maximum height);

3]
#-pi# in the argument of #sin# means that your sine is "moved" starting at a value that is #sin(-pi)# when #x=0#; consequently it will start not "going up" as the normal sine functions but down:

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