Can anyone do this? (1 Viewer)

[FD3S-R]

question : Sketch the following showing all asymptotes, intercepts and important features, y = root [(x-1)(x-2)] / (x-3)]


thanks

 

[Jase]

with or without calculus?

well im gonna go with the unsafe way and just use ordinates.

Well theres an asymptote at x=3. the graph is only above x-axis.
the graphy is only defined for 1>x>2 (greater than or equal to actaully) and x>3
so bassically its this this funny bulge between 1 and 2. asymptote at x=3, and a funny concave up curve in teh x>3

 

[FD3S-R]

the question didnt indicate no calculus so we can use it i guess, its got to do with ordinates

my move was to take it as (x-1)(x-2)/(x-3), sketch that first and then change the graph as a root was in place, kinda looked very weird , guys if u can, have a go at this question and post results

thanks

 

[withoutaface]

From graphmatica...

 

[mojako]

the question didnt indicate no calculus so we can use it i guess, its got to do with ordinates

my move was to take it as (x-1)(x-2)/(x-3), sketch that first and then change the graph as a root was in place, kinda looked very weird , guys if u can, have a go at this question and post results

thanks

that shud be a correct method...

 

[BillyMak]

From graphmatica...

That is the graph of y² = (x-1)(x-2)/(x-3), not y = [(x-1)(x-2)/(x-3)]^1/2 right?

 

[BillyMak]

And is it the graph of y = [(x-1)(x-2)/(x-3)]^1/2 or y = [(x-1)(x-2)]^1/2/(x-3) that the question asks?

 

[mojako]

And is it the graph of y = [(x-1)(x-2)/(x-3)]^1/2 or y = [(x-1)(x-2)]^1/2/(x-3) that the question asks?

read the first post
EDIT: sorry.. I didn't notice the ambiguity there.

 

[mouiseravioli]

read the first post
EDIT: sorry.. I didn't notice the ambiguity there.

ur a geek i bet u get bashed at school every day

 

[mojako]

ur a geek i bet u get bashed at school every day

I feel compelled to say thank you to the forum search feature.
EDIT: if that makes sense... probably not since my brain is severely damaged due to excessive bashing at my school
but, if u look at my profile carefully.. I'm in a school of dark magic... don't try me.. I can easily turn u into a frog... I've mastered all the secret spells of Hogwart's.. I have dementors lying around here.. ready to come to ur house... jk

 

[Archman]

That makes perfect sense.
there is this interesting trend with people on 7 posts.

 

[BillyMak]

Unfortunately I've managed to run into a few other posts by that guy. I don't think he'll be around for too much longer

 

[FD3S-R]

bah can we not flame, i just want to find an answer or method of doing this question

its [{(x-1)(x-2)}/{x-3}]^1/2

i drew the graph looks weird but i can only find 2 stat pts, im missing 2 for some reason

help thx

 

[mojako]

That is the graph of y² = (x-1)(x-2)/(x-3), not y = [(x-1)(x-2)/(x-3)]^1/2 right?

now that I really look into this Q and this thread,
the first one in that quote is y=+- the second one.
justin's pic looks like [(x-1)(x-2)/(x-3)]/2

ok, now answering the Q.
numerator = (x-1)(x-2) = (x^2-3x+2)
then divide by (x-3) with long division,
(x^2-3x+2)/(x-3) = x + 2/(x-3)
this of course is a recognisable graph.
make a rough sketch of it.
then take the square root, and sqrt is only defined when the thing inside the sqrt is >=0.
for stat pts, u can just find the stat pts for y=x + 2/(x-3)
the x-coordinates will be the same for the sqrt
and yes the graph looks weird and funny.. like Billy Mark said.
and there are only 2 stat pts.. how many do u want

now I have the graph which I drew by inspection(tm) ^^

ok?(tm)

(tm) inspection(tm) and ok?(tm) are trademarks of third parties

 

[FD3S-R]

this is the answer, yeh it was the same as my answer so im happy

it has a oblique asymptote and x=3 asymptote

i just graphed (x-1)(x-2)/(x-3), then eliminated anything below x axis and reflected rest about x axis (property of rooting)

 

[mojako]

this is the answer, yeh it was the same as my answer so im happy

it has a oblique asymptote and x=3 asymptote

i just graphed (x-1)(x-2)/(x-3), then eliminated anything below x axis and reflected rest about x axis (property of rooting)

its just the thing above y=axis though
EDIT: above x-axis

 

[Archman]

well i think that's the graph for y^2 = stuff
not y = sqrt(stuff)
the sqrt sign by convention takes the positive value.

 

[SeDaTeD]

That looks like an elliptic curve rotated a bit, if you ignore one branch. Is that what it's called, an elliptic curve?

 

[Jase]

so i was right? yay..

 

[SeDaTeD]

Yeh, I get ya. I just meant that it LOOKS like one.
And the modular stuff is kinda beyond me.