Q9c (1 Viewer)

[CecilyMare]

_pizza lol your comment is so convoluted D:

Stationary point at f(1) means that x=0 at f'(1) but it doesn't mean there can't be an inflexion point

 

[someth1ng]

Not pointed, but I doubt they'd deduct marks for it.

It's a possibility but I'm pretty sure I'd still lose marks...but it's still technically a minimum right? xD

 

[someth1ng]

_pizza lol your comment is so convoluted D:

Stationary point at f(1) means that x=0 at f'(1) but it doesn't mean there can't be an inflexion point

No, there can't be an inflexion point there. The inflexion point doesn't work.

 

[ThandraCityRail]

No, there can't be an inflexion point there. The inflexion point doesn't work.

a curve isn't differentiable at sharp points

so yes your initial curve is wrong

it should be more curvey like how someone posted earlier

bl

 

[_pizza]

_pizza lol your comment is so convoluted D:

English exam was last week, I don't need to be coherent any more! But yeah, what I was trying to say can be summed up like so:

as f(x) ---> 8
f'(x) -----> 0
So f'(x) wouldn't cross the x axis back into positive measure, only approach it.

And I didn't mean there couldn't be stat points, there quite clearly is one as you say when x = 0. Just where someth1ng drew pointy bit that implied stationary point, it wasn't, only inflexion.

There we go, hopefully its not still in relative Japanese..

 

[Gillywilly]

there is no point of inflexion at x = 1
it's just a x-intercept, minimum turning point at x = 3 and approaches x = 0
like the one shown above by the nice fellow

 

[mdunn94]

Who got this...

View attachment 23396

It's a shitty picture - I know.

Yep got that exactly wasnt to sure i dropped nuts in the exam

 

[jamesfirst]

I drew y = -2 as an asymtope ....


dang. And I made the graph go above the x - axis


can I still get 2/3 if I showed that it passed the x axis at x=1 and it was a stationary point in x=3 ?

 

[someth1ng]

can I still get 2/3 if I showed that it passed the x axis at x=1 and it was a stationary point in x=3 ?

Probably.

 

[jamesfirst]

Probably.

I think the marks are rewarded as:


- showing the correct x intercept

- showing the correct stationary point

- graphing it correctly

3/3

 

[_pizza]

there is no point of inflexion at x = 1
it's just a x-intercept, minimum turning point at x = 3 and approaches x = 0
like the one shown above by the nice fellow

Yeah exactly. I don't know why I wrote it as a stationary point, that's not what I was thinking while writing it and its not what I put in the exam, I did exactly what the correct diagram shows... LOL wtf that's so weird.

Talking is hard...

 

[Gillywilly]

me thinks probably 1 for x-intercept
1 for min. turning point
and 1 for showing curve approach the x-axis

 

[OmmU]

Yes but not as V shaped.

 

[damo_g]

Like this?

I screwed this one up, made a super stupid mistake. But how many marks do you think i'd get for drawing the curve above, except i started it from below the axis, and the asymptote was above the axis. It still cut at 1, had stat point at 3 (although max), and asymptote.

 

[jamesfirst]

me thinks probably 1 for x-intercept
1 for min. turning point
and 1 for showing curve approach the x-axis

turning point = stationary point ??? LOL that's how I always remembered it.


meh, at least I drew it in I'll get 2/3