How do you find the amplitude, period, and shift for #y= -sin (4/3)x#?

1 Answer
Dec 16, 2015

amplitude#=1#
period#=270^@#
phase shift#=#none
vertical shift#=#none

Explanation:

The general equation for a sine graph is:

#y=color(red)asin(color(orange)k(x-color(green)d)+color(blue)c#

where:
#color(red)a=|"amplitude"|#
#color(orange)k=360^@/("period")#
#color(green)d=#phase shift
#color(blue)c=#vertical shift

In this case, since the value of #a# is #1#, the amplitude is #1#. It would not be #-1# because the amplitude is always an absolute value, or always positive.

The period is found by substituting the #k# value, #4/3#, into the equation, period#=360^@/k#:

period#=360^@/k#
#=360^@/(4/3)#
#=360^@-:4/3#
#=360^@*3/4#
#=color(red)cancelcolor(black)(360^@)^(90^@)*3/color(red)cancelcolor(black)4#
#=270^@#

The phase shift is #0# in this case since there is no indication after the #x# variable in the equation.

The vertical shift is #0# in this case since there is no indication at the end of the equation.