Trig

[sayonara]

For the equation , how would you work out the turning points? The horizontal translation part is confusing me :S.

[QuantumJG]

ok so finding the turning points of this function is just finding out when dy/dx = 0. For simplicity let's just look at y = cosec(2x) and then we can look at the linear transformations.

y = 1/sin(2x)

y' = 1/sin^2(2x) [sin(2x)x0 - 1x2cos(2x)]

= 2cos(2x)/sin^2(2x)

so let y' = 0

=> 2cos(2x)/sin^2(2x) = 0

=> you have cos(2x) = 0...1

and cosec(2x) = 0 ...2, but cosec(2x) =/= 0 for any R

thus 2 is redundant and 1 is the only useable formula.

so when does cos(2x) = 0

well x = .5arcos(0) = -npi/4, ..., -5pi/4, -3pi/4, -pi/4, pi/4, 3pi/4, 5pi/4, ..., npi/4

x = {npi/4| where n is an odd integer}

now if we now have:

2x + pi/2 thats the same as just shifting the graph pi/4 units to the left.

so with this we just get our set of values for x and subtract pi/4 from them

so, x = {npi/4 - pi/4| where n is an odd integer} = {pi/4(n-1)|where n is an odd integer} = {kpi/4| where k is an even integer}

[sayonara]

:S, is there a faster way then differentiating it? Currently, I'm just finding the x value between asymptotes.

[Flaming_Arrow]

:S, is there a faster way then differentiating it? Currently, I'm just finding the x value between asymptotes.

to find the asymptotes solve sin( 2x + pi/4) = 0

[sayonara]

:S, is there a faster way then differentiating it? Currently, I'm just finding the x value between asymptotes.

to find the asymptotes solve sin( 2x + pi/4) = 0

I know how to solve the asymptotes, but is there a different method besides differentiating to find TP?

[TrueTears]

Yes there is.

For

Asymptotes occur at where

Turning points occur at where

Apply transformations when appropriate.

EDIT: This is the way I always use to work out TP and asymptote for and , but be weary for (and obviously ) as the vertical transformations is not included in the x axis intercept general formula.

[/0]

For the equation , how would you work out the turning points? The horizontal translation part is confusing me :S.

Putting it in this form can help:



In this form you apply dilations then translations:

This is a dilation by 1/2 from the y-axis followed by a translation of in the negative x-axis direction.

[TrueTears]

For the equation , how would you work out the turning points? The horizontal translation part is confusing me :S.

Putting it in this form can help:



In this form you apply dilations then translations:

This is a dilation by 1/2 from the y-axis followed by a translation of in the negative x-axis direction.

And remember for cosec vertical transformations, dilations won't affect the general result.

[sayonara]

Thanks everyone!

[sayonara]

Another question, for , how would you solve for x-intercepts?

[/0]

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