+0  
 
0
52
5
avatar+1690 

having some trouble on this one as well,

 

thank you in advance!

 Mar 4, 2020
 #1
avatar+1690 
+1

i think the middle line would be at (0,0.25)?

 Mar 4, 2020
 #2
avatar+109499 
+2

The  midline  is  the   same  as  with  the  "regular"  sine curve....we  would have to have a plus or minus added to  the function to affect the  normal midline of  y  =  0 

 

The only  thing tricky about  this one  is that  it  is  a  "compressed" sine  curve wth  a  "phase shift"

 

Note that  we  can write  this  in  the following  manner

 

f(x)  =  sin (2pi ) ( x  - 1/4)

 

This  says  that  there  are  2pi  periods  in  2pi  rads....so   .....every period = 1 (rad)

 

Also.....the  regualr  sine curve  is   shifted  1/4  rads  to the right

 

Look at the regular  sine  curve  and this function on the same graph :

 

https://www.desmos.com/calculator/itkg2m4xrj

 

Note  that  the point   (.25,0)  "starts" the  shifted curve  and the point (1.25, 0)  completes  the period......

 

Also note that  the  curve  is shifted (1/4) rad to the right  of the regular curve

 

 

 

cool cool cool

 Mar 4, 2020
 #3
avatar+1690 
+1

ops i head said the midline opposite, I meant to write (0.25,0) , my mistake.

 

I have though attached what you have explained and created it, is this correct?(:

jjennylove  Mar 4, 2020
 #4
avatar+22043 
+3

Looks OK    ,  jjl

ElectricPavlov  Mar 5, 2020
 #5
avatar+1690 
0

thanks!

jjennylove  Mar 5, 2020

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