1. enter the minimum value for thhe function shown in the graph.
2. (picture)
3. the graph of f(x)=cos(x) is transformed to a new function, g(x) , by reflecting it over the x-axis and shifting it 2 units down.
what is the equation of the new function g(x) ?
enter your answer in the box.
4. enter the amplitude of the function f(x) .
f(x)= −2sin(3x)−1
5. what is the equation of the midline for the function f(x) ?
f(x)=1/2sin(x)+6
enter your answer in the box.
6.what is the frequency of the function f(x)?
f(x)=−2sin(x/4)+3
express the answer in fraction form.
enter your answer in the box.

Problem 1)

-5.5

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Look at one of the lowest points (there are infinitely many). The y coordinate of each of these valley points is y = -5.5. This is the smallest that y can get, therefore its the function min. 

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Problem 2)

See the attached image figure 2.

Amplitude = 3 ---> a = 3
Period is 4pi so T = 4pi
b = 2pi/T = 2pi/(4pi) = 1/2 = 0.5
midline is y = 2 telling us that d = 2
There is no phase shift, so c = 0

y = a*sin(bx-c)+d
y = 3*sin(0.5x-0)+2
y = 3*sin(0.5x)+2
y = 3*sin(x/2)+2
is the function

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Problem 3)

-cos(x)-2

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g(x) = -f(x)-2
g(x) = -cos(x)-2

The negative out front the f(x) reflects it over the x axis
The minus two at the end shifts everything 2 units down

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Problem 4)

The amplitude is 2. This is the absolute value of the number out front of the sine, which is -2, so |-2| = 2.

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Problem 5)

The midline equation is y = 6

The equation is in the form a*sin(bx-c)+d where d = 6 leads us to the midline.

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Problem 6)

8pi

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Focus on the argument of sine
x/4 = (1/4)x = bx
b = 1/4
T = period = 2pi/b = 2pi/(1/4) = 8pi

11/36 is .305 repeating