oblique assymptotes (1 Viewer)

[maths lover]

well im doing some work on oblique asymptotes and wondering if these have any:

• (X^2+4)/(X-1)
• ((x^2-1)+5)/(x-1)
• (x+1)+5/(x-1) this one has one i think y=x+1

 

[ohexploitable]

They all have oblique asymptotes.

 

[SpiralFlex]

If the degree of the power of the numerator is 1 above the degree of the denominator. Also, yes they all have oblique asymptotes, you can also find oblique asymptotes by division for the first two.

 

[maths lover]

If the degree of the power of the numerator is 1 above the degree of the denominator. Also, yes they all have oblique asymptotes, you can also find oblique asymptotes by division.

what do you mean by division

 

[SpiralFlex]

what do you mean by division

I guess you haven't learnt polynomial division yet, don't worry about it.

 

[Deep Blue]

Say you have a function,

f(x) = g(x) + { c / h(x) } where f(x) is your function, g(x) is your oblique asymptote as h(x) goes to infinity and c is the arbitrary constant. In your above examples you need to use the division transformation first. That is the case for a linear oblique asymptote. I don't think you do anything more difficult than that in 3 unit, unless of course you have posted in the wrong section.

 

[maths lover]

yes i have learnt polynomials

 

[xV1P3R]

Long divide the top by the bottom to get a whole polynomial number and a fraction. Your non-fraction is then your oblique asymptote.

 

[Deep Blue]

yes i have learnt polynomials

Did what I say make sense?

 

[hscishard]

Are those three questions? Because that's the way you find the oblique asymptotes without long division

 

[deterministic]

Just divide each term in the numerator and denominator by the highest power of x in the denominator, cancel out any term that goes to 0 as x goes to infinity (ie. anything with powers of x in the denominator after cancellation). Anything left over will be your asymptote. If there is x left over, then it is oblique.

Eg.

hence x-1 is your oblique asymptote

 

[cutemouse]

If y=ax+b is an oblique asymptote to the curve y=f(x) then

 

[cutemouse]

Just divide each term in the numerator and denominator by the highest power of x in the denominator, cancel out any term that goes to 0 as x goes to infinity (ie. anything with powers of x in the denominator after cancellation). Anything left over will be your asymptote. If there is x left over, then it is oblique.

Eg.

hence x-1 is your oblique asymptote

What about , which has an oblique asymptote y=x+1?

 

[maths lover]

What about , which has an oblique asymptote y=x+1?

thinking the same thing.