Question

Help please! Will medal and fan... Algebra 2 question.

Answer 1

@ganeshie8 Could you possibly help?

Answer 2

I Have my answer but I need someone to check it for me.

Answer 3

Amplitude is the distance between the x-axis and the maximum value.

Answer 4

Period is the distance between 2 repeated points.

Answer 5

Is that right?

Answer 6

I agree with number 1.

Answer 7

You mean the A part?

Answer 8

yeah.

Answer 9

I also agree with the values of theta at which maximums occur

Answer 10

I don't agree with minimum nor zeros

Answer 11

Okay how can I fix them

Answer 12

Every increment is pi/4, right?

Answer 13

Right

Answer 14

So the minimum occurs at the first, third, fifth, and seventh increment, which is the same as pi/4, 3pi/4, 5pi/4, 7pi/4

Answer 15

Oh okay! I see! So that's the minimum?

Answer 16

yeah

Answer 17

How can I get the zeros if my method was not correct? :/

Answer 18

notice that zeros pass the midpoint of each increment

Answer 19

assuming it's drawn to scale

Answer 20

Okay, what do I do with that information in order to apply it to the zeros?

Answer 21

The first x-intercept pass the midpoint of the first increment.
So, it would be (pi/4)/2=pi/8

Answer 22

Okay, how can I find the rest of the numerical values in relation to the zeros, after pi/8? What calculation do I use?

Answer 23

also notice that every x-intercept is apart by pi/4

Answer 24

So the next would be pi/12? :/

Answer 25

pi/8+pi/4=pi/8+2pi/8=3pi/8

Answer 26

What's the next one?

Answer 27

I don't know :(

Answer 28

The roots increase by pi/4 each time.
pi/8
pi/8+pi/4=3pi/8
pi/8+pi/4*2=5pi/8
pi/8+pi/4*3=7pi/8
...
pi/8+pi/4*7=15pi/8

Answer 29

So the next would be 9pi/8?

Answer 30

ok

Answer 31

So how would I format that into one of the zeros?

Answer 32

all of the zeros would be (2k+1)pi/8 where k=0,1,2,3,4,5,6,7

Answer 33

i have to go
sorry

Answer 34

ok thank you so much!