Find out how to determine the equation of the oblique asymptotes algebraically.?

Answer 1

okay

i'm guessing you have a rational function

in that case, it probably looks something like y = (4x^2 + 2)/(x + 1)

it's not important that the numerator has an x^2 term and the denominator has an x term. the important thing for oblique asymptotes is that the numerator's highest degree is ONE MORE than the denominator's highest. for example, if the highest on top is an x^6 term, then the highest degree on the bottom must be an x^5

so knowing how you recognize an oblique asymptote, you find it by long division

**divide the numerator by the denominator, and ignore any remainders**

the resulting expression is your oblique asymptote

Answer 2

The slope of the curve approaches the asymptote as x approaches infinity.

Let's try an example. A hyperbola is a well-known conic section with oblique asymptotes. There is a completely algebraic (and geometric) way to determine the asymptotes, but since I'm using this as an example, let's use the more generalized method.

Suppose we have the standard form of the hyperbola,

(x^2)/a^2 - (y^2)/b^2 = 1

Solve for y:

(y^2)/b^2 = (x^2)/a^2 - 1 = (x^2 - a^2)/a^2

y^2 = (b/a)^2 (x^2 - a^2)

y = +/- (b/a) sqrt(x^2 - a^2)

Now we want to know the slope of this curve as x gets very large. (That's the slope of the asymptote.)

The easiest way to do this is to take the derivative (calculus), and that's what I'm going to do. If you insist on algebra only, that can also be done (the "delta process"), but I won't do that unless you add a note to your question.

So now we take the derivative of y:

y = +/- (b/a) sqrt(x^2 - a^2)

dy/dx = +/- (b/a)(1/2)(2x) / sqrt(x^2 - a^2)

dy/dx = +/- (b/a) [x / sqrt(x^2 - a^2)]

Now let x get very large -- something like 100,000,000. Then the expression inside the brackets [ ] reduces to x/x = 1, and

dy/dx = +/- b/a for large x.

These two lines, one with positive slope and one with negative, and both crossing at the origin, are the oblique asymptotes of the hyperbola.

And as I said, if you want a completely algebraic solution (no calculus), add a clarification to your question.

Answer 3

f(x)= 2x(x-2)/(x+a million) * Your function isn't defined for x=-a million ; f(x) is going to infinity (resp. -infinity) at the same time as x is going to -a million (yet >-a million) (resp. to -a million (yet<-a million)) and also you've a vertical asymptote at x=-a million * you've now to study what occurs at the same time as x is going to infinity ( ? or - ?): are you able to discover an indirect asymptote y=ax+b ? you ought to study in case you'll detect a and b inclusive of Lim (x-->inf.) (f(x)-g(x)) = 0 ? (a million) f(x)-g(x) = (2x(x-2) - (x+a million)(ax+b) ) /(x+a million) = (2x^2 -4x -ax^2 -bx - ax -b) / (x+a million)= ((2-a)x^2 -x(4+a+b) -b) / (x+a million) to satisfy the identity (a million) you want 2-a=0 and four+a+b=0 then a=2 and b=-4-2=-6 The equation of your indirect asymptote is y=2x-6

Answer 4

Given a real function f, try to find numbers a and b such that

.. lim .. f(x) - (ax + b) = 0
x-> +-inf

then y = ax + b is oblique asymptote.