Nombre de usuarioPeerlessCucumber
Preguntas 14
Respuestas 48


Equation: y = -2 cos 4 x + 3


Desmos Graph:



- Questions -


1) Midline: y = 3                    yes


2) Topline: 5                           y=5   It is a line!


3) Bottom Line: 1                      y=1


4) Wavelength: π/2                          2pi/4 = pi/2      GOOD


5) Quarter Of The Wavelength: π/4     X     one quarter of pi/2   IS NOT pi/4           \(\frac{1}{4}*\frac{\pi}{2}=\frac{\pi}{8}\)


6) What would be sensible points for the x-axis? π/4, 3π/4, 5π/4, 7π/4, 9π/4 (?)

pi/8, 2pi/8=pi/4, 3pi/8, 4pi/8=pi/2   etc         


7) Will it start on the midline, above or under, why? ​It will start below the midline because it is a cos graph.  

NO   Sin graphs start in the middle because sin0=0

+cos0=1    -cos0=-1      This equation has the minus sign so it will start (cross the y axis) at the bottom.


8) Will is start going down or going up, why?​ It has begun at its lowest point as a cos graph and will begin upwards. 

Up because it is is already at the bottom. 


9) y-intercept: 1     YES    (this is the bottom!)


10) Plot 4 (or 5) points in advance just by knowing what the properties of the graph will be.

 I mean maxima, minima, and points on the midline: (π/8, 3), (3π/8, 3), (5π/8, 3), (7π/8, 3), (9π/8, 3) [Or] (π/4, 5), (3π/4, 5) [Etc]      (0,1)


11) Graph it from x = 0 to x = Wavelength: 


12) What is the first maxima? (π/4, 5)


13) What is the first minima? (π/2, 1)


14) What are points on the midline? (π/8, 3), (3π/8, 3), (5π/8, 3), (7π/8, 3), (9π/8, 3) (?)   

Good but I also wanted the ones at the top and bottom (0,-1), pi/4,5), (pi/2,-1)

Look at the graph.


Hopefully these are all correct!     You are getting there :)

You need to do a freehand sketch as you answer the questions. They will complement each other and make it easier.


Here is a graph with a better step on the x axis. I have made the x step equal to one quarter of a wave.

I have only added the shaded green so that you can easily see one wavelength.

See how it starts at the bottom.

when x=0   y=-2cos(4*0)+3 = -2*1+3 = 1   (which is at the bottom of the wave)



10 feb. 2020
   y = 7 sin 3 x -6  
  Midline    y = -6                  correct
  Topline    1 (?)                   correct
  Bottom line    -13 (?)                  correct
  Wavelength    2π/3                    correct
  Quarter of the Wavelength    π/3 (?)                  wrong              \(\frac{2\pi}{3}\div4 = \frac{\pi}{6} \)  

  What would be sensible points for the x-axis?  

  As in where it intersects?        every quarter

like, ... 0, pi/6, 2pi/6 3pi/6, 4pi/6, 5pi/6  ..

these fractions should be simplified though

  Will it start on the midline, above or under, why?    It will start on the midline because it is a sine graph.   correct
  Will is start going down or going up, why?    It will start upwards because of the positive number in front.  correct 
  y-intercept  -6     correct

  Plot 4 (or 5) Points in Advance 

  (I mean maxima, minima, and points on the midline.)  

  (4π/3, 6) & (3π/2, 1) & (5π/2, -15) &        (8π/3, -6)

(0,-6) (2pi/3,-6)(pi/3,-6)(pi/6,1)(3pi/6,-13)

But it is easier if you do it straight onto the rough graph.

  Graph it from x=0 to x= wavelength

  I'm not sure what this means.  

I mean graph it by hand for one whole wavelength, starting at the y axis.

  What is the first maxima?  (π/6, 1)          yes
  What is the first minima?  (π/2, -13)         yes
  What points are on the midline?  (4π/3, 6) & (8π/3, -6)       NO you can fix this



Sorry that I couldn't answer all of these fully or properly I assume, still getting the hang of it.


I just improved the scale of the graph

Also if you hit the region circle on the side you will see one full wavelength. 

You can see also how it is divided nicely into 4 quarters.

8 feb. 2020

Melody:  I have put my response in the middle of your answers.  


First Question

(please include the question so I don't have to go look for it)

Response to first answer.

y = 4-3 sin (x/4)

The first lowest point should be (2pi, 1) and the first highest point should be (6pi, 7) At least, I think this is correct. 

YES, that is perfect, I hope you worked it out with a rough sketch BEFORE you used Desmos.     I like your graph  wink





Second Question  



  y = 4 cos (2x) -3

you did this one perfectly

  y = f cos (gx) - t  

y = f cos (gx) - t  

melody's correction

  Midline    y = - 3  y = - tyes 
  Amplitude    4   fyes  
  y-intercept    1  (-3+4=1)  g     x-t+f

it is easier

to get this

as you hand

sketch the graph

  Is the y-intercept at the top,

  middle, or bottom?  

  Top  Topyes 
  Wavelength     π (2pi/2=pi)  2π (?)   x2pπ/g 
  A Maximum Point  (π, 1)  (2π, 0) (?)(0,-t+f), (2π/g, -t+f)

don't worry

about these last 2,

they were too

hard anyway.

  A Minimum Point  (π/2, -7)  (π, -2)  (?)(π/g,-t-f) 


I'm not sure how to fill in wavelength, the maximum point, and the minimum point. If you input this into the desmos graph, it gives a rough example, but considering it can be moved due to all these just being variables, I'm just answering with what the baseline graph would look like.

Good, Once you get the wavelength consistently correct that will work well.

anyway those last two questions, max and min with letters was a bit hard, don't get hung up on those.


Please correct me, I think I'm starting to make sense of it more consistently somewhat though. 



Note: I do relatively well usually once I make sense of the concept, suddenly something will just click and I understand it. I avoid over-practicing until I feel I have a general sense of what is wrong and right with the material, so that I don't form bad habits or get myself confused. This has been a really great experience having someone work with me one-on-one. I appreciate your patience and willingness a great deal, even if I might be a bit slow. 

6 feb. 2020

Are we assuming 'a' and 'k' are being multiplied by one? I'm sorry, I've gotten a little lost now. 



Melody's response.


a and k are just constants.  They are numbers, you just don't know what that number is.

You have to treat them just like you would treat a number.


You should now kow that y=6 is a vertical line where the y values of all the points are 6.



Say I said 

graph y=k  where k=6     

that would be exactly the same graph.

If I said

graph y=k  where k=2

then I would have to graph y=2


If i said 

y=3sinx you should be able to tell me all about that graph.  You would know that the amplitude was 3

If I said

y=asinx you should be able to tell me all about that graph too.  You would know that the amplitude was a


Can you get you head around what I want now?

5 feb. 2020
  y = sin x y = sin (2x)
Wavelength π
Midline y = 0 y = 0
Amplitude 1 1
y-intercept 0 0


Desmos graph for both equations.


All qualities of both graphs appear to be the same with the exception of wavelength, and so I assume the (2x) affects how the wavelength appears? 

4 feb. 2020


Question 1-


Desmos Graph:  (From checking afterward.)


a) Midline: y = -1

b) y-intercept: -1

c) Wavelength: π

d) Amplitude: 1

e) Does it start going up or down?: It begins moving upwards. 

f) What is the equation?: y = 1 sin x -1



Question 2-


Desmos Graph:  (From checking afterward.)


a) Midline: y = 4

b) y-intercept: 4

c) Wavelength: π

d) Amplitude: 1

e) Does it start going up or down?: It begins moving downwards. 

f) What is the equation?: y = -1 sin x + 4


2 feb. 2020

Equation: y = sin x

a) Midline: y = 0

b) y-intercept: 0

c) Wavelength: 

d) Amplitude: 1




Equation: y = -sin


a) Midline: y = 0

b) y-intercept: 0

c) Wavelength: 

d) Amplitude: -1 (?)      No, the amplitude is always positive. It stays +1


         What I see from both of these graphs is they share similar properties almost entirely, except for the second graph being flipped. With that said, does that mean the amplitude will be considered -1? Or will it remain a positive always? Either way, they're incredibly similar graphs, but one appears to be flipped because of the negative. 


EXCELLENT.   If you times by minus -1 the graph flips over, (It reflects across the y axis)





Equation: y = 3 sin x


a) Midline: y = 0

b) y-intercept: 0

c) Wavelength: 

d) Amplitude: 3




Equation: y = -3 sin x


a) Midline: y = 0

b) y-intercept: 0

c) Wavelength:

d) Amplitude: -3 (?)      +3


         Similarly, comparing these two graphs, the only noticeable difference is that the figure appears to be flipped because of the negative. Once again, I'm not sure if amplitude should ever be written in negatives, and if not, then the answer should be just the definite form of what is already written, I assume. I hope these are correct. 


That is all great, you are learning really well.

1 feb. 2020

y= 4(sinx)-3       (I added the equation so it is with the answers - Melody)


a) Midline: y = -3

b) y-intercept: -3

c) Wavelength: 2π

d) Amplitude: 4?


This is the graph from afterward.




I am here now so I willl add my response here.


Your answer is spot on.

You seem hesitant about amplitude:

Yes it is 4 and you can see that on the formula.

To get it from the graph find the highest y value and subtract the lowest y value and then halve it.  (+1--7)/2 = 8/2 = 4


get the highest y value and subtract the midline y value.     In this case     +1 -  - 3 = 1+3=4


Make sure you underst this from the graph.



New question:    Get this one right and I will move onto a slightly new idea.  


what will be

a) midline

b) y intercept

c) wavelength

d) amplitude



Peerless Cucumber: *Wave*




a) Midline: y = 6

b) y-intercept: 6

c) Wavelength: 2π

d) Amplitude: 8

Here is the graph!

31 ene. 2020


a) Midline: y = 5

b) y-intercept: 5

c) Wavelength: 2π

d) Amplitude: 1


These all are somewhat similar to y = sin x, however, when it becomes +5, it merely moves up 5 on the graph, and the logic seems consistent when replaced by a 2, 3, or any other number, for example. Whereas by changing the number it is multiplied by effects the amplitude?   

30 ene. 2020


a) Midline: y = 0

b) y-intercept: 0

c) Wavelength: 2π

d) Amplitude: 10


think​ these are correct. I'm getting a little more confident in them. laugh

30 ene. 2020