[Question] Graphs (1 Viewer)

[richz]

About oblique asymptotes, how do u know when a graph has an oblique asymptote and when a graph doesnt have one??

 

[Slidey]

Perform polynomial division. It will have an oblique asymptote if there are any linear factors. Exempli gratia:

y=(x^2+1)/x

Divide by x: y=x+1/x

Oblique asymptote at y=x.

I believe you must also test whether it truly is an asymptote by letting x -> + infinity, and ensuring y-> + infinity, but I'm not sure on that one - maybe inspection is enough.

As an extension, it is interesting to note that for, say the curve y=(x^3+1)/x, y=x^2+1/x, the curve DOES have asymptotic behaviour at y=x^2, but for that matter it also has asymptotic behaviour at y=1/x...

 

[richz]

so you mean, when you do long division and the equation is linear that is the asymtote

 

[richz]

eg. x^2/x-1 it says it has an oblique asymptote at y=x+1

 

[Slidey]

y=x^2/x-1, to the top: subtract one and then add one to maintain equality:
y=([x^2-1] + 1)/(x-1)
y=(x^2-1)/(x-1) + 1/(x-1)
y=(x+1) + 1/(x-1)
.'. oblique asymptote at y=x+1

Does that help?

 

[richz]

ok yes i undestand that, a graph has an oblique asymptote when long division is done and the equation is linear

 

[richz]

wtf am i saying dw abt that

 

[richz]

just dont understand what u mean by linear factors

 

[Slidey]

Oh, sorry!

"linear factor" isn't the best term to use. However, what I mean is anything in the form: ax+b

Linear means straight. In this case it means x to the power of 1.

The 'linear factor' of y=(x+1)+1/(x-1) is x+1

See here, for further information: http://www.boredofstudies.org/community/showthread.php?t=60249

 

[richz]

yes thnx i understand now if u do long division and there is no 'linear factor' the asymyptote is horizontal

 

[Slidey]

For the purposes of your 4unit graphing, yes.

 

[~ ReNcH ~]

For the purposes of your 4unit graphing, yes.

Implicit in that statement is that there must be further "rules" concerning curve sketching beyond the MX2 syllabus...do you think you could elaborate on some of them? (purely out of curiosity).

 

[Slidey]

Implicit in that statement is that there must be further "rules" concerning curve sketching beyond the MX2 syllabus...do you think you could elaborate on some of them? (purely out of curiosity).

It was in reference to this comment: "yes thnx i understand now if u do long division and there is no 'linear factor' the asymyptote is horizontal"

Which is not entirely correct. It is correct for probably any graph we'll come accross in 4unit, but suppose that the asymptote is neither linear nor constant - it won't be oblique nor horizontal (see y=(x^3+1)/x).

Other than that, I think that there would certainly be rules and concepts beyond 4unit, but I am not of much help there, sorry.

 

[~ ReNcH ~]

It was in reference to this comment: "yes thnx i understand now if u do long division and there is no 'linear factor' the asymyptote is horizontal"

Which is not entirely correct. It is correct for probably any graph we'll come accross in 4unit, but suppose that the asymptote is neither linear nor constant - it won't be oblique nor horizontal (see y=(x^3+1)/x).

Other than that, I think that there would certainly be rules and concepts beyond 4unit, but I am not of much help there, sorry.

I was going to ask about a case such as y=(x^3+1)/x. In that case would the asymptote necessarily be the curve y=x^2?

 

[Slidey]

In that case, both the curves 1/x and x^2 would be asymptotic on the curve y=(x^3+1)/x. That is:
For large |x|, y=(x^3+1)/x behaves as a parabola, for small |x| it behaves as a hyperbola.

Try graphing it and you'll see what I mean.

These are just observations, though.