what is the function of cos that has an amplitude of 4, a period of 2pi and has a point at (0,2)?

Question

_{I need answers with detailed steps}

Answer 1

y = Acos(Bx+C)

 

A = amplitude = 4

 

Period = 2π/B = 2π   So, B = 1

 

So far, we have y = 4cos(x+C)

 

We need to determine C so that the graph goes through the point (0, 2).

 

2 = 4cosC   So, cosC = 1/2

 

                             C = π/3

 

The equation is y = 4cos(x + π/3)

 

 

 

 

Answer 2

For trig functions the following form is used:

 

f(x) = A*cos(Bx - C) + D

 

 

A = Amplitude

 

2pi/B = period

**note: it's only 2pi/B because the period of sin and cos is 2pi. If we were dealing with tan it'd be pi/B, since the period of tan is just pi.

 

C = phase shift (along x-axis)

 

D = vertical shift (along y-axis)

 

Since your amplitude is 4, A = 4.

Since your period is 2pi:

 

2pi/B = 2pi. B = 1

 

You have a point at (0,2), which implies your phase shift is 0.

Using the same logic above, you have a vertical shift of 2.

 

Putting all your information together:

 

f(x) = 4*cos(1*x - 0) + 2

 

or rather,

 

f(x) = 4*cos(x) + 2