How do you find oblique asymptotes? (1 Viewer)

[kaz1]

How do you find oblique asymptotes? In the function y= x + 1/x the oblique asymptote is y=x. I found this asymptote by by putting x close to infinity. Is there a better way of doing it? The correct way? I find my way too noobish and I think it might not work all the time.

 

[u-borat]

uh well, lets take a harder example.

like y=(2x+4)/x

if you separate the terms....

y= 2+4/x

now you know that y=2 is an oblique asymptote.

the reason for this is because 4/x CAN NEVER EQUAL ZERO.

so in ur case, 1/x can never equal zero, hence y=x is an asymptoteeee

 

[kaz1]

uh well, lets take a harder example.

like y=(2x+4)/x

if you separate the terms....

y= 2+4/x

now you know that y=2 is an oblique asymptote.

the reason for this is because 4/x CAN NEVER EQUAL ZERO.

so in ur case, 1/x can never equal zero, hence y=x is an asymptoteeee

Thanks. I understand now.

 

[tacogym27101990]

uh well, lets take a harder example.

like y=(2x+4)/x

if you separate the terms....

y= 2+4/x

now you know that y=2 is an oblique asymptote.

the reason for this is because 4/x CAN NEVER EQUAL ZERO.

so in ur case, 1/x can never equal zero, hence y=x is an asymptoteeee

isn't because as x >>> inifinity, 1/x becomes negligible

 

[tommykins]

Using tacogym's analogy, heres what he's saying -

y= x + 1/x

If we graph y = x and y = 1/x, and add them on, as x -> infinity, the graph approaches y = x

You can tell it approaches above the line y= x because it's an addition of 1/x, but if it was a negative, it would approach from below the assymptote.

In 2unit/3unit - general rule is just dividing all terms by the HIGHEST degree in the denominator.

ie. 3x+5/x^2+x highest degree is x^2, division gives.
(3/x+5/x)/(1+1/x) - as x -> infinity
3/x -> 0
5/x -> 0
1/x -> 0

so you have 0/1 = 0

thus the line y = 0 is an asymptote.

 

[u-borat]

thanks tommmmmmmy.

why am i here i want maths out of my life forever.