Question 1193295
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.


<img src = "http://theo.x10hosting.com/2022/041521.jpg" >


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.


<img src = "http://theo.x10hosting.com/2022/041522.jpg" >


let me know if you have any questions.
theo