Transformation of a sin(x) function

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Transformation of a sin(x) function

Postby barnamah » Sun 03 April, 2011 17:28

Write the equation of the sine function where:

- the amplitude is 5
-the period is `6pi`
-the displacement is `pi/5` units to the left

the main equation to apply the above transformation is:

a is vertical stretch (amplitude)
k is horizontal compression
d is translation to the left (-) or right (+)
c is translation up (+) or down (-)

we have amplitude 5 so a=5,
to get k, we know that `(2pi)/k=period` so `(2pi)/k=6pi` solve for k :
`(2pi)/k=6pi` multiply both side by `k` and we have `k*(2pi)/k=6pi*k`

the k on left cancel out and we have `2pi=6pi*k`.
now we have `pi` on both side and it cancels out and we have:

Now divide both side by `6` and we have: `(2)/(6)=(6*k)/(6)`. Now 6 on the right cancel out each other and we have: `2/6=k` simplify, we get:
`k=1/3` and to get `1/(1/3)=3` so we use 3 for the period

to get d, we have `pi/5` to to the left meaning `-pi/5=x` set it =0 and have `0=x+pi/5`

Putting it together we have:
`y=5sin(3(x+pi/5))` and we have simplify the inner parenthesis values and have `y=5sin(3x+(3pi)/5)`

so our final transformed function is: `y=5sin(3x+(3pi)/5)`
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