SOLUTION: The period of the following function is u and the phase shift is v units to the left.
f(x) = 10 cos (6x + π/2) - 2
What is the value of u+v
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-> SOLUTION: The period of the following function is u and the phase shift is v units to the left.
f(x) = 10 cos (6x + π/2) - 2
What is the value of u+v
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Question 1193295: The period of the following function is u and the phase shift is v units to the left.
f(x) = 10 cos (6x + π/2) - 2
What is the value of u+v Found 2 solutions by greenestamps, Theo:Answer by greenestamps(13215) (Show Source):
You can put this solution on YOUR website! you have:
f(x) = 10 * cos(6x + pi/2) - 2
the period is u and the phase shift is v.
general form of the equation is:
y = a * cos(b * (x - c)) + d
a is the amplitude
b is the frequency
c is the horizontal shift
d is the vertical shift
change your equation to conform to the general form.
f(x) = 10 * cos(6x + pi/2) - 2 becomes:
f(x) = 10 * cos(6 * (x + pi/12)) - 2
the frequency is 6 and the horizontal shift is pi/12.
the period is equal to 2pi / frequency = 2pi/6 = pi/3
this makes the period equal to pi/3.
that makes u = pi/3 radians.
the shift to the left is equal to pi/12 radians.
the first graph is without the horizontal shift to the left.
you can see that there are 6 full cycles of the cosine function in the period between 0 and 2pi.
0 and 2pi would normally hold 1 full cycle of the cosine function.
the normal period for 1 full cycle of the cosine function is 2pi.
the period for the 6 full cycles that fit into a period of 2pi is 2pi/6 = pi/3.
6 * pi/3 = 6pi/3 = 2pi.
i showed you the graph without the horizontal shift so you can compare it to what it looks like with the horizontal shift.
the second graph is with the horizontal shift to the left.
there are two equations that make the same graph.
they are y = 10 * cos(6x + pi/2) - 2 and y = 10 * cos(6 * (x + pi/12)) - 2
the shift to the left is pi/12 radians.
you can see the horizontal shift to the left because the cosine function starts at y = 8.
without the shift, the cosine function would be equal to 8 at x = 0.
with the shift, the cosine function is equal to 8 at x = -pi/12.
at x = 0, the value of the cosine function is equal to -2.
this is because, at x = 0, the cosine function becomes:
y = 10 * cos(6 * (0 + pi/12)) - 2 which becomes:
y = 10 * cos(6pi/12) - 2 which becomes:
y = 10 * cos(pi/2) - 2 which is equal to -2.
what this says is that each new cycle of the cosine function starts at f(x) = -2 rather than f(x) = 8 becuase of the horizontal shift to the left.
