How do you find the period for #y=4sin(2x)+1#?

1 Answer
Aug 22, 2015

#y = 4 sin(2x)+1# has a period of #pi#

Explanation:

Starting from a base equation:
#color(white)("XXXX") y_theta = sin(theta) = color(red)(1)*sin(color(blue)(theta))+color(green)(0)#
#color(white)("XXXXXXXX")#has an amplitude of #color(red)(1)#
#color(white)("XXXXXXXX")#is centered on #y = color(green)(0)#
#color(white)("XXXXXXXX")#and has a period of #2pi#
To say that #y_theta# has a period of #2pi#
#color(white)("XXXX")#means that #y_theta# has a repeating pattern for #(0+k) <= theta <= (2pi+k)# for any constant #k#

#color(white)("XXXX") y = color(red)(4) sin(color(blue)(2x)) + color(green)(1)#
#color(white)("XXXXXXXX")#has an amplitude of #color(red)(4)#
#color(white)("XXXXXXXX")#is centered on #y = color(green)(1)#
#color(white)("XXXXXXXX")#and has a repeating pattern for
#color(white)("XXXXXXXXXXX")(0+k) <= 2x <= (2pi+k)#
#color(white)("XXXXXXXXXXX")#or
#color(white)("XXXXXXXXXXX")(0+hatk) <= x <= (pi+hatk)# for
constant #hatk#
#color(white)("XXXXXXX")# That is, it has a period of #pi#