Question

What trigonometric function represents the graph? (6 points)

A. f(x) = 4 sin(x − pi over 2 )

B. f(x) = 4 cos(x − pi over 2 )

C. f(x) = 4 sin(x − pi over 2 ) + 1

D. f(x) = 4 cos(x − pi over 2 ) + 1

Answer 1

Option A is correct.i.e,
`f(x)=4\,sin(x-(\pi)/(2))`

Explanation:

We are given with a graph .

A points from graph which satisfies it =
`((\pi)/(2),0)\:,\:(\pi,4)\:,\:((3\pi)/(2),0)\:,\:(2\pi,0)`

We check which equation satisfy points.

Option A:

`f(x)=4\,sin(x-(\pi)/(2))`

Let,
`y=4\,sin(x-(\pi)/(2))`

we have
`((\pi)/(2),0)`

LHS = y = 0

`RHS=4\,sin(x-(\pi)/(2))=4\,sin((\pi)/(2)-(\pi)/(2))=4\,sin\,0=4*0=0`

LHS = RHS

Thus, This Option is correct.

Option B:

`f(x)=4\,cos(x-(\pi)/(2))`

Let,
`y=4\,cos(x-(\pi)/(2))`

we have
`((\pi)/(2),0)`

LHS = y = 0

`RHS=4\,cos(x-(\pi)/(2))=4\,cos((\pi)/(2)-(\pi)/(2))=4\,cos\,0=4*1=4`

LHS ≠ RHS

Thus, This Option is not correct.

Option C:

`f(x)=4\,sin(x-(\pi)/(2))+1`

Let,
`y=4\,sin(x-(\pi)/(2))+1`

we have
`((\pi)/(2),0)`

LHS = y = 0

`RHS=4\,sin(x-(\pi)/(2))+1=4\,sin((\pi)/(2)-(\pi)/(2))+1=4\,sin\,0+1=4*0+1=1`

LHS ≠ RHS

Thus, This Option is not correct.

Option D:

`f(x)=4\,cos(x-(\pi)/(2))+1`

Let,
`y=4\,cos(x-(\pi)/(2))+1`

we have
`((\pi)/(2),0)`

LHS = y = 0

`RHS=4\,cos(x-(\pi)/(2))+1=4\,cos((\pi)/(2)-(\pi)/(2))+1=4\,cos\,0+1=4*1+1=5`

LHS ≠ RHS

Thus, This Option is not correct.

Therefore, Option A is correct.i.e,
`f(x)=4\,sin(x-(\pi)/(2))`

Answer 2

The correct option is: A. f(x) = 4sin(x − π/2), because:

1. When you evaluate x=π/2 in the function f(x) = 4sin(x − π/2), you obtain:

f(π/2) = 4sin(π/2− π/2)
f(π/2) = 4sin(0)
f(π/2) = 4(0)
f(π/2) = 0 (As you can see in the graphic)

2. If you evaluate x=π in the same function, then you have:

f(π) = 4sin(π− π/2)
f(π) = 4sin(π/2)
f(π) = 4 (As it is shown in the graphic)