Slidey
But pieces of what?
- Joined
- Jun 12, 2004
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- HSC
- 2005
Um... for y=(x^3+x^2)/x^2, the actualy curve IS y=x+1, that's not the asymptote. No asymptotes exist, however, a discontinuity at x=0 does.mojako said:oblique asymptote is a slanted line
to find out if its oblique or horiz:
u find the limiting value of y as x approaches infinity
if the limiting value is a constant (number) then the asymptote is horizontal
if the limiting value is infinity or negative infinity then it's oblique (or not oblique, but not horizontal either)
to find the oblique asymptote:
->
you can divide, for example:
y = (x3+x2) / x2
y = x3/x2 + x2/x2
y = x + 1
so asymptote is y=x+1
->
take limit, for example:
y = x + 1/x
as x-> infinity, the 1/x part approaches zero
so as x-> infinity, y=x
then asymptote is y=x
quote=blackfriday : "oblique asymptotes usually occur when the highest power in the numerator is higher than the highest power in the denominator."
higher by 1
So tell me the asymptote for y=(x^3+1)/x, besides a vertical at x=0. Does it have any others? Is perhaps y=x^2 an asymptote? I have a good feeling it is.
And what do you call an oblique asymptote which the curve actually cuts?