Answer:
i) amplitude is 4/2 =2
ii) period of the graph = 3.14 ( radians), (or 180°)
iii) the frequency of the given function is f = 2Hz ( hertz).
iv) The midline for the function is y = -3.
v) The maximum value of the problem of the function is -1.
vi) the minimum value of the function is -5.
Step-by-step explanation:
i) the Peak to Peak is = -1 - (-5) = 4
therefore amplitude is 4/2 =2
ii) The period of the graph = 3.14
iii) one cycle is over 2π radians
one cycle of the given function is over π radians. Therefore the given
function will have 2 cycles over 2π radians. Therefore the frequency of the given function is f = 2 hertz.
iv) The midline for the function is y = -3.
v) The maximum value of the problem of the function is -1
vi) the minimum value of the function is -5.
