y= -2cos(pi/6) + 8
whats the maximum and minimum?
Assuming you meant
y= -2cos(pi/6 x) + 8
then the center line is at y=8
and the amplitude is 2
so, what do you think?
min is 6 and max is 10?
To find the maximum and minimum values of the given equation y = -2cos(pi/6) + 8, we need to understand the basic properties of the cosine function.
The cosine function oscillates between its maximum value of 1 and its minimum value of -1 over a full period of 2π (or 360 degrees). The standard form of the cosine function is y = A*cos(Bx + C), where A represents the amplitude, B represents the period, and C represents the phase shift.
In this case, the equation y = -2cos(pi/6) + 8 can be written as y = -2cos((1/6)π) + 8. Comparing this with the standard form, we can see that A = -2 (amplitude), B = 1 (period = 2π/B = 2π/1 = 2π), and C = 0 (no phase shift).
Since the period of the cosine function is 2π, the maximum and minimum values occur at the boundary points of the period. In this case, the first boundary point is at x = 0 (or 2πk, where k is an integer). Plugging in x = 0 into the equation, we get y = -2cos(0) + 8 = -2(1) + 8 = 6. Thus, the maximum value of y is 6.
The second boundary point is at x = 2π (or 2π(k + 1/2), where k is an integer). Plugging in x = 2π into the equation, we get y = -2cos((1/6)π) + 8 = -2(-1) + 8 = 10. Thus, the minimum value of y is 10.
Therefore, the maximum value is 6 and the minimum value is 10.