Question

Use the graph below to answer the question that follows:
What trigonometric function represents the graph?

f(x) = −3 sin(x − pi over 2)
f(x) = −3 cos(x − pi over 2)
f(x) = 3 cos(x − pi over 2)
f(x) = 3 sin(x − pi over 2)

Answer 1

Answer:

(C)

Step-by-step explanation:

In order to get the graph of the trigonometric function, we must compare the given function with asin(bx-c)+d and acos(bx-c)+d.

Thus, if we take f(x)= -3sin(x-$\frac{\pi }{2}$

Comparing this equation with asin(bx-c)+d, we have,

Amplitude (a)=$-3$, b=$1$,c=$\frac{\pi }{2}$ and d=$0$

Period=\frac{2x}{\left | b \right |} =$2\pi$

Phase shift =$\frac{c}{b}$ =$\frac{\pi }{2}$ ($\frac{\pi }{2}$ towards right)

Vertical shift= $0$

Thus, on the basis of these measures, we have the graph.

For f(x)= $-3cos(x-\frac{\pi }{2} )$, comparing with acos(bx-c)+d, we have:

Amplitude= $-3$,

b=$1$,

c=$\frac{\pi }{2}$

d=$0$

Period= $2\pi$

Phase shift= $\frac{\pi }{2}$ ($\frac{\pi }{2}$ toards right)

With these we can draw the graph.

Similarly we can find the variables like above and if we compare all the graphs, we  get the result that the graph of $3cos(x-\frac{\pi }{2} )$= f(x) matches the given graph.

Hence option C is correct.

Answer 2

Answer:

Option

c

Step-by-step explanation:

Given is a graph which has maximum at 3 and minimum at -3.

X intercepts are 0, pi, 2pi and 3pi

Period = 2pi

We know that sin curve has x intercepts as multiples of pi but when there is a phase shift of pi/2 we see that cos curve is the most appropriate

Since amptlitude is 3, and middle line is x axis,

we have y = 3 * trig function with coefficient 1 for x

Hence appropriate function is

f(x) = 3cos (x-pi/2)

Option C would be correct.