Curve sketching (1 Viewer)

[vice lord]

Hey wen sktching the absolute value of sin(x) for example is it sharp pts at intercepts or tp.. same q. fir f(x)^2

 

[mitchy_boy]

I just sketched y=|sin(x)| using graphmatica and the sharp points seem to be at the x axis. I hope that's answering the question.

 

[annabackwards]

Yeah, the reflected turning points are sharp for absolute value graphs

 

[mitchy_boy]

Yeh download graphmatica vice lord, just google graphmatica,it really helps you understand things when they're a little iffy.

 

[vice lord]

Yeh ill download graphamatica now thx alot for ur help...Btw how bout the sin(x)^2 curve?

 

[ninetypercent]

yeah I lost a mark because my y = l cosx l graph wasn't sharp in the x intercepts

 

[zeleboy]

Yeah, the reflected turning points are sharp for absolute value graphs

"I'm using the FF addon Leechblock to block BoS 22-7 until 4th of Nov, so I won't be on much. Study hard"

yeah...

 

[annabackwards]

"I'm using the FF addon Leechblock to block BoS 22-7 until 4th of Nov, so I won't be on much. Study hard"

yeah...

I am, 22-7 aka 22 hours 7 days a week.

So i'm on from 1200-1300 and 1800-1900 only

 

[untouchablecuz]

y=(sinx)²=(f(x))²

dy/dx=2f'(x)f(x)

thus the x coordinates of y=f(x) become x coordinates of turning points for y=(f(x))²

also, the x coordinates of the original turning points of y=f(x) remain to be the x coordinates of the turning points of y=(f(x))²

 

[The Nomad]

Hey wen sktching the absolute value of sin(x) for example is it sharp pts at intercepts or tp.. same q. fir f(x)^2

Go study and stop posting on BOS!

No really, I'm going to punch you when I see you