Curve sketching (1 Viewer)

[fullonoob]

Sketch y = x(3+rootx) and y = x(3-rootx)

I don't know how to get the oblique asymptote y= 3x
and how you get point (4,4). Explain please with full working out.

Cambridge book sucks

 

[Cazic]

http://community.boredofstudies.org/14/mathematics-extension-2/232196/graphing.html

No idea wtf oblique asymptote is

 

[Lukybear]

I dont think oblique would be 3x. You sure?

y=3x + x^(3/2)
as x >> inf y inf as well?

 

[Lukybear]

Actually, your getting confused between oblique asymptotes and horizontal tangent.

As the curve approaches zeor, dy/dx >>>3. Therefore a horizontal tangent of y=3x exist.

Oblique isotopes are something altogether. It occurs with large x, and y approaching an asymptote that is not the vertcies.

I.e.

y=x + 1/x

as x>>> inf+ y>>>x+ etc... therefore y=x is an oblique asymptote.

 

[fullonoob]

Actually, your getting confused between oblique asymptotes and horizontal tangent.

As the curve approaches zeor, dy/dx >>>3. Therefore a horizontal tangent of y=3x exist.

Oblique isotopes are something altogether. It occurs with large x, and y approaching an asymptote that is not the vertcies.

I.e.

y=x + 1/x

as x>>> inf+ y>>>x+ etc... therefore y=x is an oblique asymptote.

Hmm probably.
dy/dx= (6rootx + 3x) / 2 root x right?
now if split the fraction up
3 + (3x/ 2 rootx) = dy/dx
when x->0+
dy/dx = 3
but how do you get y = 3x and how do you know if its a horizontal
Btw i got the point (4,4) now, don't need to explain that.
But the y = 3x asymptote would help

 

[Lukybear]

Hmm probably.
dy/dx= (6rootx + 3x) / 2 root x right?
now if split the fraction up
3 + (3x/ 2 rootx) = dy/dx
when x->0+
dy/dx = 3
but how do you get y = 3x and how do you know if its a horizontal
Btw i got the point (4,4) now, don't need to explain that.
But the y = 3x asymptote would help

there is no y=3x asymptote. Ive got confused. There is no horizontal tangent. Just a tangent as x>>0 of y=3x. Basically, at the origin, the curve melts into the tangent. If you graph this, the curve actually goes away from the y=3x.

 

[fullonoob]

How do you know it goes away from y= 3x
-> i'm not sure where y= 3x comes from
After you lim the dy/dx you get = 3
But how does it become y=3x
Thats why im confused

 

[Lukybear]

it dosent. At all

forget about the y=3x, its meaningless.

To find the curvature of the graph, you take 1st and 2nd derivate. Itll get you x>0, and x>0, hence it concaves upwards. Look at the link the Cazic posted.