#Bx+C vs Bx-C (Phase Shifts in Sinusoidal Waves)
keen spruceBOT
surreal ledge
From my personal experience, i found this form best fit my style of calculation:
Asin(B(x-C))+D
cuz using this and the rules of transform of graphs, it will be much easier to see what's happening on the graph
faint brook
and if C happens to be something like + pi/2 over here it'd be
Asin(B(x-(+pi/2))+D which simplifies to Asin(Bx-(pi/2)*B)+D?
so would that mean it'll shift to the left then? (since **minus **pi/2 * B)
if C were to be -pi/2 it'd be like
Asin(B(x-(-pi/2))+D which simplifies to Asin(Bx+(pi/2)*B)+D?
and that'd mean it'll shift to the right side? since (**plus **pi/2 * B)?
just checking if my understanding is clear, this would make it like a gazillion times easier
surreal ledge
reading
faint brook
eitherways it definitely works, just making sure im not learning through rote memorization
surreal ledge
my attempt:
let's use this as an example,
compare 5sin(πx/4)+3
and 5sin(πx/4-π/8)+3 = 5sin(π/4(x-0.5))+3
the -"+0.5" means it shift to positive side by 0.5 unit
so for Asin(B(x-C))+D
A: vertical
B: horizontal
C: horizontal shift right
D: horizontal shift up
faint brook
oh right, the minus from 4"-"0.5 comes from this form : Asin(B(x-C))+D
faint brook
oh all good, this form will help ALOT
you have no idea, I couldnt find a standard form anywhere
surreal ledge
so for Asin(B(x-C))+D
A: vertical
B: horizontal
C: horizontal shift right
D: h...
that is, with this from Shift from C, D would be more intuitive
for further reference:
https://www.mathsisfun.com/sets/function-transformations.html
faint brook
Thanks alot
surreal ledge
surreal ledge
been struggling for like 2 days, ill try to close the question now
you can type .close
faint brook
.close