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Hmm probably.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.
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.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![]()