SOLUTION: Finding the number of cycles in 2 pi, the horizontal length of each of the 4 quadrants of each cycle =, vertical shift, horizontal shift of each cycle, and determining a reflection

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Question 1135312: Finding the number of cycles in 2 pi, the horizontal length of each of the 4 quadrants of each cycle =, vertical shift, horizontal shift of each cycle, and determining a reflection over the x axis.

f(theta)=4sin(3theta+theta/4)-10
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
this graph is a little different.

i use x for theta and y for f(x), so the formula in my language is:

y = 4 * sin(3x + x/4) - 10

it's the same formula except it's easier to graph this way using the graphing calculator at desmos.com.

within the parentheses, 3x + x/4 is equivalent to 12x/4 + x/4; which is equal to 13x/4.

that translates to the frequency being 13/4 * x.

there is no horizontal shift.

it's just a frequency change.

if the frequency is 13/4, then the period of the sine function is equal to 2 * pi / (13/4) which is equal to 2 * pi * 4 / 13 which is equal to 8/13 * pi.

your sine function of y = 4 * sin(3x + x/4) - 10 translates to y = 4 * sin(13/4 * x) - 10.

the amplitude is 4.
the vertical displacement is -10.
the frequency is 13/4
the period is 8/13 * pi.

here's the graph from 0 radians to 8pi/13 radians.

here's the graph from 8pi/13 radians to 16pi/13 radians.

the amplitude is 4.
the center line is at y = -10
that means the high point of the graph is y = -10 + 4 which becomes y = -6.
that means the low point of the graph is y = -10 - 4 which becomes y = -14.
the period is 8pi/13 radians.
that means one cycle goes from 0 to 8pi/13 and from 8pi/13 to 16pi/13.