Consider the function
f(x)=-5cos((2/3)x-(pi/6))
a). Is there a reflection? Does the graph start at a maximum, a minimum, or an x-intercept?
For Amplitude I got 5
For Period I got 3pi
And for Phase shift I got 3pi/12 or => pi/4
yes, amplitude is five and period is 3 pi
when x = 0:
y = 5 cos (-pi/6) = 5 (.866)
that is not a max or a min or and x intercept
However there is always a reflection about the points where the argument of the cos function is zero because cos -z = cos z
that is when
(4/6)x = pi/6
or x = pi/4 which is 45 degrees
of course every time that argument increases or decreases by 2 pi (which is x increases or decreases by 3 pi as you said) you have another max and symmetry
Yes, (pi/6)/(2/3) = pi/4 = phase shift
see:
http://www.mathsisfun.com/algebra/amplitude-period-frequency-phase-shift.html
To determine if there is a reflection and where the graph starts, we need to examine the coefficient in front of the cosine function.
The given function is f(x) = -5cos((2/3)x - (pi/6)), where the coefficient in front of the cosine function is -5.
a) Reflection:
If the coefficient in front of the cosine function is negative, there is a reflection about the x-axis. In this case, since the coefficient is -5, there is a reflection.
b) Start of the graph:
To determine where the graph starts, we need to consider the phase shift. The phase shift is given by (pi/6) divided by the coefficient of x, which is 2/3.
Phase shift = (pi/6) / (2/3) = (pi/6) * (3/2) = pi/4
Therefore, the graph starts at a minimum value, corresponding to the phase shift of pi/4.
In summary,
a) There is a reflection about the x-axis.
b) The graph starts at a minimum value.
You correctly calculated the amplitude as 5, the period as 3pi, and the phase shift as pi/4.