Question

The question is the following image...

Answer 1

i have y=5cos(4x+???)

Answer 2

amp = 5 which is the max. 2pi/4 is pi/2 which is the period for the equation... i know that

Answer 3

notice the period of the function, is \(\pi\) meaning \(\bf y=5sin({\color{red}{ \square }}x)\qquad period\implies \cfrac{2\pi}{{\color{red}{ \square }}}=\pi\implies \cfrac{2\pi}{\pi}={\color{red}{ \square }}\iff 2\)

Answer 4

ohhh the period is actually pi. okay i did have that mistaken

Answer 5

notice in the question though it requests that the function is named as a cosine function

Answer 6

right... recall that putting a negative sign in front of the sine or cosine, flips it upside-down

Answer 7

hmmm.... lemme.. ahemm recheck that =)

Answer 8

-pi/2?

Answer 9

is that what we add on the end there?

Answer 10

I gather you'd need to push the cosine function by a phase shift of \(\cfrac{3\pi}{2}\)

Answer 11

i think the function needs to be y=5cos(2x+-pi/2)

Answer 12

i graphed it and used the table to plug in the five critical x values for a single period and the numbers matched up

Answer 13

thank you for your help on getting that period ironed out

Answer 14

\(\bf y=5cos\left({\color{blue}{ 2}}x{\color{red}{ +3\pi}}\right)\qquad \textit{phase shift}\implies \cfrac{{\color{red}{ 3\pi}}}{{\color{blue}{ 2}}}\)

Answer 15

you are not correct. in your equation when x=0, y=-5, which is incorrect

Answer 16

hmmm heheh, was thinking about it... that

Answer 17

i do agree that that is how you calculate phase shift