Question

What trigonometric function represents the graph? (6 points)

Answer 1

Answer:

Option A is correct.i.e, $f(x)=4\,sin(x-\frac{\pi}{2})$

Step-by-step explanation:

We are given with a graph .

A points from graph which satisfies it = $(\frac{\pi}{2},0)\:,\:(\pi,4)\:,\:(\frac{3\pi}{2},0)\:,\:(2\pi,0)$

We check which equation satisfy points.

Option A:

$f(x)=4\,sin(x-\frac{\pi}{2})$

Let, $y=4\,sin(x-\frac{\pi}{2})$

we have $(\frac{\pi}{2},0)$

LHS = y = 0

$RHS=4\,sin(x-\frac{\pi}{2})=4\,sin(\frac{\pi}{2}-\frac{\pi}{2})=4\,sin\,0=4\times0=0$

LHS = RHS

Thus, This Option is correct.

Option B:

$f(x)=4\,cos(x-\frac{\pi}{2})$

Let, $y=4\,cos(x-\frac{\pi}{2})$

we have $(\frac{\pi}{2},0)$

LHS = y = 0

$RHS=4\,cos(x-\frac{\pi}{2})=4\,cos(\frac{\pi}{2}-\frac{\pi}{2})=4\,cos\,0=4\times1=4$

LHS ≠ RHS

Thus, This Option is not correct.

Option C:

$f(x)=4\,sin(x-\frac{\pi}{2})+1$

Let, $y=4\,sin(x-\frac{\pi}{2})+1$

we have $(\frac{\pi}{2},0)$

LHS = y = 0

$RHS=4\,sin(x-\frac{\pi}{2})+1=4\,sin(\frac{\pi}{2}-\frac{\pi}{2})+1=4\,sin\,0+1=4\times0+1=1$

LHS ≠ RHS

Thus, This Option is not correct.

Option D:

$f(x)=4\,cos(x-\frac{\pi}{2})+1$

Let, $y=4\,cos(x-\frac{\pi}{2})+1$

we have $(\frac{\pi}{2},0)$

LHS = y = 0

$RHS=4\,cos(x-\frac{\pi}{2})+1=4\,cos(\frac{\pi}{2}-\frac{\pi}{2})+1=4\,cos\,0+1=4\times1+1=5$

LHS ≠ RHS

Thus, This Option is not correct.

Therefore, Option A is correct.i.e, $f(x)=4\,sin(x-\frac{\pi}{2})$

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)