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Use the graph below to answer the question that follows:

What trigonometric function represents the graph? (6 points)
Select one:
a. f(x) = −3 sin(x − pi over 2 )
b. f(x) = −3 cos(x − pi over 2 )
c. f(x) = 3 cos(x − pi over 2 )
d. f(x) = 3 sin(x − pi over 2 )

Use the graph below to answer the question that follows What trigonometric function represents the graph 6 points Select one a fx 3 sinx pi over 2 b fx 3 cosx p class=

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caylus
Hello,

[tex]if\ x=0\ then \\ a)f(x)=-3*sin(-\dfrac{\pi}{2})=-3*(-1)=3\ and\ not \ 0\\ d)f(x)=3*sin(-\dfrac{\pi}{2})=3*(-1)=-3\ and\ not \ 0\\ \\\\ if\ x= \dfrac{\pi}{2} \ then\ \\ b)f(x)=-3*cos(0)=-3\ and\ not\ 3\\ [/tex]

So answer C

Answer:

The  trigonometric function which represents the graph is:

         c)   [tex]f(x)=-3\cos (x-\dfrac{\pi}{2})[/tex]

Step-by-step explanation:

From the graph we may observe that when x=π/2 we have the value of the function  f(x)= 3

Hence, we will check in which this point hold true:

a)

      [tex]f(x)=-3\sin (x-\dfrac{\pi}{2})[/tex]

Now, when x=π/2 we have:

   [tex]f(x)=-3\sin (\dfrac{\pi}{2}-\dfrac{\pi}{2})\\\\\\f(x)=-3\sin (0)\\\\\\f(x)=0\neq 3[/tex]

Hence, option: a is incorrect.

b)

[tex]f(x)=-3\cos (x-\dfrac{\pi}{2})[/tex]

Now, when x=π/2 we have:

   [tex]f(x)=-3\cos (\dfrac{\pi}{2}-\dfrac{\pi}{2})\\\\\\f(x)=-3\cos (0)\\\\\\f(x)=-3\neq 3[/tex]

Hence, option: b is incorrect.

d)

[tex]f(x)=3\sin (x-\dfrac{\pi}{2})[/tex]

Now, when x=π/2 we have:

   [tex]f(x)=3\sin (\dfrac{\pi}{2}-\dfrac{\pi}{2})\\\\\\f(x)=3\sin (0)\\\\\\f(x)=0\neq 3[/tex]

Hence, option: d is incorrect.

c)

   [tex]f(x)=-3\cos (x-\dfrac{\pi}{2})[/tex]

Now, when x=π/2 we have:

   [tex]f(x)=3\cos (\dfrac{\pi}{2}-\dfrac{\pi}{2})\\\\\\f(x)=3\cos (0)\\\\\\f(x)=3[/tex]

Similarly we will see that the value of this function matches all the points that on the graph.

        Hence, option: c is correct.

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