slant asymptoes-i cant understand it (1 Viewer)

[ronaldinho]

How do i find slant or parabolic asymptotoes. I thought u just divide by the highest power of x

but some1 else said to use polynomials.. COuld some1 please explain it to me and why it works? and how come when u divide by the highest power of x u still get the same shape of the asymptote

thanks bye

 

[airie]

How do i find slant or parabolic asymptotoes. I thought u just divide by the highest power of x

but some1 else said to use polynomials.. COuld some1 please explain it to me and why it works? and how come when u divide by the highest power of x u still get the same shape of the asymptote

thanks bye

o.0

Do you mean the graph of a function y=u/v, where both u and v are functions of x, on a Cartesian plane? Like, y=(x³+2x²-5)/(x+1)?

If so, you NEVER simply divide the whole expression in x by "highest power of x", but you rearrange the expression. In the example above, the degree of the numerator is higher than the degree of the denominator, thus there exists a parabolic asymptote (as the difference of the degrees is 2). If, however, the degree of the numerator is less than that of the denominator, there exists a horizontal asymptote, and no slant/parabolic asymptote.

Basically,
- the vertical asymptote is the value to which x approaches when the value of y approaches +/- infinity;
- the horizontal/slant/parabolic/cubic etc... asymptote describes values to which y approaches as the value of x approaches +/- infinity.

So when you're trying to find an asymptote in the second case, usually you'd try to rearrange the expression of y, in terms of x, so that a fraction where x (or any other variable on the usual x-axis) appears in the denominator but not in the numerator (ie. the numerator is a constant) is one of the terms. This is because the value of the fraction would approach zero as x approaches +/- infinity, therefore as x approaches +/- infinity, the value of the whole expression (ie. the value of y) would approach the value of the rest of the terms in the expression.

If the degree of the numerator is greater than that of the denominator, say y=u/v (u and v both being functions of x with a positive power), there exists an asymtote of the second category, as explained above. Use long division (or any other method to your liking) to write u in the form vQ(x)+R(x), and therefore rearrange the expression on the right into the form Q(x)+R(x)/v where Q(x) and R(x) are functions of x (x has a non-negative power ie. not in the denominator). Therefore as x approaches +/- infinity the value of R(x)/v would approach zero, thus the whole expression y=Q(x)+R(x)/v would approach Q(x).

If the degree of the numerator is less than or equal to that of the denominator, usually there exists a horizontal asymptote. Divide BOTH the numerator and the denominator by the highest power of x in the denominator (since deg(denominator)>=deg(numerator) anyway), so that both the top and the bottom are additions/multiplications of constant terms and negative powers of x. Again, as x approaches +/- infinity, the value of negative powers of x would approach zero, so the top approaches zero (when degree of original numerator is less than that of denominator) or a constant (when they are equal), say c_n, while the bottom approaches a constant, say c_d. The value of y therefore approaches zero or c_n/c_d when x approaches +/- infinity.

 

[muttiah]

thanks for the thorough explanation. i understoond everything except the bit below.. i have to use that method in an example then refer back to this bit and then i will get it

p.s where are the smilie faces?

o.0

If the degree of the numerator is greater than that of the denominator, say y=u/v (u and v both being functions of x with a positive power), there exists an asymtote of the second category, as explained above. Use long division (or any other method to your liking) to write u in the form vQ(x)+R(x), and therefore rearrange the expression on the right into the form Q(x)+R(x)/v where Q(x) and R(x) are functions of x (x has a non-negative power ie. not in the denominator). Therefore as x approaches +/- infinity the value of R(x)/v would approach zero, thus the whole expression y=Q(x)+R(x)/v would approach Q(x).

 

[SoulSearcher]

Ok, let's have the example y = x²/(x+1)
Since the degree in the numerator is higher than the degree in the denominator, we can simplify it as so:
x²/(x+1)
= [(x²-1)+1)]/(x+1)
=[(x-1)(x+1) + 1]/(x+1)
= (x-1)(x+1)/(x+1) + 1/(x+1)
= x-1 + 1/(x+1)

As x approaches infinity, 1/(x+1) would approach 0, therefore the equation would approach the line x-1. Therefore x-1 is a slant asymptote.

 

[airie]

p.s where are the smilie faces?

Hey...x.x Well, I had "o.0" at the beginning... (Since when am I synonymous with smiles? <.<)

If the degree of the numerator is greater than that of the denominator, say y=u/v (u and v both being functions of x with a positive power), there exists an asymtote of the second category, as explained above. Use long division (or any other method to your liking) to write u in the form vQ(x)+R(x), and therefore rearrange the expression on the right into the form Q(x)+R(x)/v where Q(x) and R(x) are functions of x (x has a non-negative power ie. not in the denominator). Therefore as x approaches +/- infinity the value of R(x)/v would approach zero, thus the whole expression y=Q(x)+R(x)/v would approach Q(x).

Yeah, basically what I meant there was that you'd try to get an expression with non-negative powers of x, plus a term of fraction where x ONLY exists in the denominator. You would do this by writing the numerator as the denominator multiplied by a quotient polynomial (Q(x), with non-negative powers of x) plus a remainder polynomial, just like what you did in primary school with long division of numbers. Then the expression would become Q(x)+R(x)/v, since you carried out the division (which would not affect the value of the expression since it was in there before already), and the fraction R(x)/v would approach zero as x approaches +/- infinity, thus the whole expression would just approach the rest of the expression, as SoulSearcher has shown above with his example

 

[Riviet]

Note that sometimes your fraction will be hard to spot and manipulate by inspection, so you can always resort to the trusty long division.

 

[fishy89sg]

isnt is also called "oblique" asymptotes?

 

[Riviet]

isnt is also called "oblique" asymptotes?

Yes, that's what I'd normally refer to them as.