Need help with retarded Implicit Diffrentiation Graph... Thanks (1 Viewer)

[blackops23]

Hi guys, I've just come across a graph which I'm curious about and I'd like to know why it possesses an incredibly strange property - the actual graph itself is its "asymptotes" - or maybe not.

Here's the graph: x^2 - 3xy + y^2 = 0 ----- (1)

Now here's what I did in an attempt to sketch it:
------------------------------------

y' = (3y - 2x)/(2y - 3x)

2y-3x --> 0, y' --> infinity (i.e. vertical tangent)
y=3x/2 , sub in ---(1)
Therefore vertical tangent at x=0, (0,0

3y-2x=0, y' = 0, (turning points)
y=2x/3, sub in ---(1)
Therefore turning point at (0,0)

But you can't have a vertical tangent and a turning point at one distinct point.
Then I realised that when you sub in x=0, y=0, in y', y'=0/0, therefore there's an indeterminable gradient at (0,0)

So I went on to find the asymptotes:

x^2 - 3xy + y^2 = 0
[divide by x^2]

1 - (3y/x) + (y^2/x^2) = 0

Completing the square:

1- (2.25) + (y/x - 1.5)^2 = 0
therefore:

(y/x - 1.5)^2 = 1.25 = 5/4
(y/x) - 1.5 = root(5)/2

therefore asymptotes are:

y= [(3 + sqrt(5))/ 2]x which is approximately y=2.5(x), i.e the GRADIENT of the line is = [3 + sqrt(5)]/2 = approx. 2.6
and

y= [(3 - sqrt(5))/2]x which is approximately y=x, i.e .e the GRADIENT of the line is = [3 - sqrt(5)]/2 = approx.

--------------------------------------------------

So now I had the asymptotes...But nothing else...

So I gave up and sketched it on my graphing software, and the graph was two straight lines intersecting at (0,0), but the strange thing was that these lines were the asymptotes I had found...

But then that means the lines I found aren't really asymptotes. So I want to know, what on earth did I do wrong, and what would have YOU DONE to sketch this graph??


Thanks guys. =)

 

[Bored Of Fail 2]

wheres jm01 when you need him

 

[blackops23]

wheres jm01 when you need him

haha lol, besides you got any idea on this one?

 

[jyu]

x^2 - 3xy + y^2 = 0
[divide by x^2]

1 - (3y/x) + (y^2/x^2) = 0

Completing the square:

1- (2.25) + (y/x - 1.5)^2 = 0
therefore:

(y/x - 1.5)^2 = 1.25 = 5/4
(y/x) - 1.5 = root(5)/4

therefore asymptotes are:

y= [(6 + sqrt(5))/ 4]x which is approximately y=2x, i.e the GRADIENT of the line is = [6 + sqrt(5)]/4 = approx. 2.05
and

y= [(6 - sqrt(5))/4]x which is approximately y=x, i.e .e the GRADIENT of the line is = [6 - sqrt(5)]/4 = approx. 0.95

I think



So the graph is two lines.

(0,0) is the intersecting point because it satisfies the original equation

.

 

[jyu]

 

[Bored Of Fail 2]

seems unnessesary to give a piece wise definition lol

 

[ohexploitable]

seems unnessesary to give a piece wise definition lol

Yah, was about to point this out too...

 

[jyu]

It is necessary because dy/dx is indeterminable at (0,0).

 

[Bored Of Fail 2]

lol whatever, I dont remember that rule

 

[NewiJapper]

x = 3.14159

 

[Bored Of Fail 2]

x = 3.14159

haahahhhahahahahahahaah.......................no

 

[blackops23]

I think



So the graph is two lines.

(0,0) is the intersecting point because it satisfies the original equation

.

So these two lines aren't the asymptotes - well obviously. But I thought the method I followed was the standard procedure for finding the asymptotes, so what the hell did I actually end up finding? All this time I was trying to sketch the graph in a way so it'd approach the asymptotes...

So say, in an exam, how would you have known that the graph itself would be those 2 lines, and the fact that they aren't asymptotes??

 

[jyu]

Try to sketch these
(x+y)(x-y)=0
(x+y)(x+2y)=0
(x-y)(x^2+y)=0

 

[Rezen]

Solve the equation as a quadratic in terms of x and it becomes obvious the graph is those two lines.

 

[blackops23]

So does this graph display this strange property because it is:

x^2 - 3xy + y^2 = 0

i.e, because the graph is equal to 0???

 

[jyu]

If the rhs = 0 and the the lhs can be factorised.

E.g. (x+y)(x-y)(y-x^2+3)=0

 

[blackops23]

aHAH! So my "method" for finding the asymptotes wasn't really finding them at all... Since the RHS is zero, I guess the only thing I managed to do was simply rearrange the original equation to obtain the two lines... I see now, thanks guys.