Help plz it’s due today
1 answer:
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Problem 1
See the attached image. Specifically see figure 1.
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Problem 2
See the attached image. Specifically see figure 2.
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Problem 3
Answer: bottom right corner
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f(x) = 4*sin(2pi*x)
f(x) = a*sin(b*x)
a = 4 is the amplitude
b = 2pi
T = 2pi/b = 2pi/2pi = 1 is the period
The graph that has a period of 1 and amplitude 4 is the bottom row of choices. We can rule out the graph on the left because sine starts off increasing as you move away from the origin and go from left to right.
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Problem 4
Answer: f(x) = -10cos(2pi*x/3)
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T = period = 3
b = 2pi/T = 2pi/3
amplitude = (max - min)/2
amplitude = (20-0)/2
amplitude = 10
y = 10 is the midline since d = (20+0)/2 = 10
f(x) = a*cos(bx)+d
f(x) = -10*cos(2pi*x/3)+10
Note: the value of 'a' is negative because of the reflection over the x axis
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Problem 5
Answer: Choice D
f(x) = 9*sin(6pi*x)+6
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min = -3
max = 15
midline: d = (max + min)/2 = (15-3)/2 = 6
amplitude: a = (max - min)/2 = (15+3)/2 = 9
f = frequency = 3
T = period = 1/f = 1/3
b = 2pi/T = 2pi/(1/3) = 6pi
f(x) = a*sin(bx)+d
f(x) = 9*sin(6pi*x)+6
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See the attached image. Specifically see figure 1.
==============================================
Problem 2
See the attached image. Specifically see figure 2.
==============================================
Problem 3
Answer: bottom right corner
------------------------
f(x) = 4*sin(2pi*x)
f(x) = a*sin(b*x)
a = 4 is the amplitude
b = 2pi
T = 2pi/b = 2pi/2pi = 1 is the period
The graph that has a period of 1 and amplitude 4 is the bottom row of choices. We can rule out the graph on the left because sine starts off increasing as you move away from the origin and go from left to right.
==============================================
Problem 4
Answer: f(x) = -10cos(2pi*x/3)
------------------------
T = period = 3
b = 2pi/T = 2pi/3
amplitude = (max - min)/2
amplitude = (20-0)/2
amplitude = 10
y = 10 is the midline since d = (20+0)/2 = 10
f(x) = a*cos(bx)+d
f(x) = -10*cos(2pi*x/3)+10
Note: the value of 'a' is negative because of the reflection over the x axis
==============================================
Problem 5
Answer: Choice D
f(x) = 9*sin(6pi*x)+6
------------------------
min = -3
max = 15
midline: d = (max + min)/2 = (15-3)/2 = 6
amplitude: a = (max - min)/2 = (15+3)/2 = 9
f = frequency = 3
T = period = 1/f = 1/3
b = 2pi/T = 2pi/(1/3) = 6pi
f(x) = a*sin(bx)+d
f(x) = 9*sin(6pi*x)+6
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