a. Find the amplitude.
b. Find the period.
c. Find the vertical shift.
d. Find the horizontal shift.
e. Find the number of cycles between 0 and 2π.
f. Find the equation of the graph.
You must show all of your work to receive credit.
Answers
Answers:
a) Amplitude = 2b) Period = pic) Vertical shift = -2, which means it has been shifted down 2 units.d) Horizontal shift = 3pi/8, this shifting is to the right.e) There is one cycle between 0 and 2pi.f) The equation of the graph is y = 2*sin(2(x-3pi/8))-2========================================================
Explanations:
Part (a)
The highest point is when y = 0 and the lowest point is when y = -4. The vertical distance between the peak and valley is 4 units, which cuts in half to 2. This is the amplitude. It's the vertical distance from the center to either the peak or valley.
Note: Amplitude is always positive as it measures a distance.
---------------------
Part (b)
For x > 0, the first valley or lowest point occurs between 0 and pi/4. It appears to be the midpoint of the two values. So that would be (0+pi/4)/2 = pi/8.
The next valley occurs between pi and 5pi/4. Compute the midpoint to get (pi+5pi/4)/2 = (4pi/4+5pi/4)/2 = (9pi/4)/2 = 9pi/8
So we have the graph go from one valley x = pi/8 to the next valley over x = 9pi/8. This is a distance of pi units because 9pi/8-pi/8 = 8pi/8 = pi
The graph repeats itself every pi units, so the period is pi.
---------------------
Part (c)
The midline is normally through y = 0, aka the x axis. However, the graph shows the midline is through y = -2. This means the graph has been shifted down 2 units.
---------------------
Part (d)
This will depend on whether you use sine or cosine. This is entirely because cosine is a phase-shifted version of sine, and vice versa. I'll go with sine.
The parent sine function y = sin(x) goes through the origin (0,0)
However, as part (c) mentioned, we shifted the graph 2 units down. So we have y = sin(x)-2. But plugging x = 0 into this leads to the point (0,-2)
This doesn't match what the graph says. The graph shows the point (3pi/8, -2) on the red curve. The x coordinate 3pi/8 is the midpoint of pi/4 and pi/2
This must mean we need to shift the sine graph 3pi/8 units to the right.
---------------------
Part (e)
Start at the lowest point when x = pi/8. If you start the cycle here, then it ends when x = 9pi/8. See part (b).
So far we've completed one cycle. If we start at x = 9pi/8, then the next valley or lowest point is slightly beyond or to the right of x = 2pi. This means we run out of room and we haven't completed a full cycle.
Overall, one full cycle is between 0 and 2pi.
---------------------
Part (f)
Again I'm going to use sine instead of cosine. Refer back to part (d).
The general sine curve equation is
y = A*sin(B(x-C))+D
where
|A| = amplitudeB handles the period, specifically T = 2pi/B where T is the period. We can solve for B to get B = 2pi/TC = horizontal phase shiftD = vertical shift, and ties together with the midline equationIn this case, we found that
|A| = 2T = pi leads to B = 2pi/T = 2pi/pi = 2C = 3pi/8D = -2So,
y = A*sin(B(x-C))+D
will update to
y = 2*sin(2(x-3pi/8))-2
which is one way to express the equation of the red curve. Optionally you can distribute the 2 through to (x-3pi/8).
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__________________________________________________________
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__________________________________________________________
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__________________________________________________________
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