contestada

Write an equation for a cosine function with an amplitude of 3, a period of 2, a phase shift of -2, and a vertical displacement of 5.


y=3cos⁡2π(x+2)+5


y=3cos⁡2π(x−5)+2


y=5cos⁡3π(x−2)+2


y=3cos⁡2π(x+2) / 2+5

Respuesta :

luisejr77

Answer:

Last option

[tex]y = 3cos(\pi(x+2)) + 5[/tex]

Step-by-step explanation:

The general cosine function has the following form  

[tex]y = Acos(b(x-\phi)) + k[/tex]

Where A is the amplitude: half the vertical distance between the highest peak and the lowest peak of the wave.  

[tex]\frac{2\pi}{b}[/tex] is the period: time it takes the wave to complete a cycle.

k is the vertical displacement.

[tex]\phi[/tex] is the shift phase

In this problem :

[tex]A = 3[/tex]

[tex]\frac{2\pi}{b}=2\\\\ b=\frac{2\pi}{2}\\\\ b=\pi[/tex]

[tex]\phi =-2\\\\k = 5[/tex]

So  The function is:

[tex]y = 3cos(\pi(x+2)) + 5[/tex]

Otras preguntas