curve sketching question ^^ (1 Viewer)

[ToO LaZy ^*]

heyas...
this is actually a 3U question, but i can't seem to find the topic 'curve sketching' under any of my maths excel books..arghhh!
so u's r my last hope.... (apart from teacher)

so here it is: sketch showing TPs, inflex, or asymptotes...

1. y = x / x^2+1

2. y = 1-x^2 / 1+x^2
and last one....

3. y = 2x+3 / x^2-4

thx guys and gals...ur help is greatly appreciated

 

[Mill]

What exactly is your problem?

Like can you find the TPs, inflexions, asymptotes?

Where do you get stuck?

Graphing programs are also useful to check your graph after you've drawn it and aren't sure.

 

[ToO LaZy ^*]

oh..i can find the TPs and asymp but not the inflexions..
n sumtimes i get the shape a bit wrong.

oh can u send me a site with that graphing program plz? =)

 

[Grey Council]

oh right

differentiate, then let that equal to zero. But i'm sure you know that.

Can you please post up your working (as much as you've done) for the first question?

lol, if you really just want a graph, please say so, but I think you'll learn more if you do it yourself.

 

[Mill]

There's a few...

If you can be bothered to learn how to use it, I would recommend MATLAB. It's a university maths program. It needs to be purchased though, or maybe you can get a hold of the 'shareware version'.

There's a few others around like GnuPlot and Graphmatica, but I haven't lots of experience with them and I'm not sure if they will do some equations (like the crazyyyyy ones... they will do your generic y = f(x) ) ones.

 

[ToO LaZy ^*]

wait a sec...i'll do the Q again..
let me find the problem then i'll get back to u

 

[evilc]

MATLAB is good! I used it last year at uni

 

[ToO LaZy ^*]

okok..found it..
q2
i found the y-interc...ok, got that bit =)
now im trying to do the first derivative..
and i end up with:

-4x / (1+x^2)^2
and so...to find the TPs...u need to make it equal to 0 right??
and any number ^2 cannot = 0 (denominator)

this is where im stuck....=(

 

[ToO LaZy ^*]

ohk..thx for the programs..=)

 

[Grey Council]

why not? i don't see why any number squared can't equal to zero.

so the critical points are as follows:
zeros: -1, 1
turning points: 0, 1

 

[ToO LaZy ^*]

oh. sorry...left out the positive constant..
edit: x^2+2 cannot equal 0

ps..i don't have a scanner

 

[ToO LaZy ^*]

but i'll be getting one soon hopefully

 

[Grey Council]

get the derivative, times both sides by:
(1+x^2)^2
it doesn't matter that that cannot be zero. For a turning point the gradient = 0, ie the derivative = 0 when turning point.
So
-4x / (1+x^2)^2 = 0
solve, you get x = 0

 

[ToO LaZy ^*]

yes...then sub x=0 into y equation and u get (0,1) as TP...
then to find inflex. do the second derivative...which turns out to be:

-4(1+x^2)^2+16x^2(1+x^2) / (1+x^2)^4

....i wouldve shown the whole working but it'll take too long and i dun hv a scanner..

then sub x=0 into equation and u get 4....which is a minimum...
but the answer has a WAYYY diff answer

 

[Grey Council]

wha?? im so confused.
sorry, i kinda taught myself graphs, not very good at this.

I don't think there IS an inflexion. mmmm
asymptote is:
y=-1

the curve is like a bell. goes from -1 (less than) y (less than or equal to) 1 ____know what i mean? kinda like -1 < y < 1, but the second one is less than or equal to.

 

[Grey Council]

umm, we ARE doing the second one, aren't we?

 

[ToO LaZy ^*]

yeah..there is an inflex..
i've got the whole graph and points in front of me...
i just don't know how 2 get the inflex...weird.

 

[ToO LaZy ^*]

yep
Q2

 

[ToO LaZy ^*]

u must of gotten confused from the other question i posted in the '3u thread'

 

[Grey Council]

the heck, i have a bell shaped curve thing.
its an even function.
never goes above 1, or below -1
its flat flat flat till it reaches x = -5, then it starts to rise. Goes to x=0, y=1 then mirrors its movement on the other side of the axis