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Hey guys, forgot how to do this. All I remember is that the problem is given in factor form and product form and you have to choose the easier one to work with, but I'm totally drawing a blank on what to do. I know that a horizontal asymptote will be of the form y = x, but thats about it. Any help much appreciated.
Let R(x) = (4x^2)(2x - 1)/7(5x + 3)(x + 2)^2 = (8x^3 - 4x^2)/(35x^2 - 119X^2 + 56x + 84)
Find an equation of the horizontal asymptote of R(x).
There is none. :eek:Quote:
Originally Posted by leviathanwave
The horizontal asymptote is determined by the degree in the numerator and denominator, and (in case you forgot) the degree of a polynomial is the highest exponent of that polynomial.
In this problem, the degree of the numerator is 3 (from 8x^3) and in degree of the denominator is 2 (from 35x^2). Since the degree in the numerator is greater than the degree of the denominator, we don't get a horizontal asymptote. [NOTE: the only time we get a horizontal asymptote is when the degrees are the same or if the degree of the denominator is greater.]
In this case, all we get is a slant asymptote. To find out what that asymptote is, use polynomial division. (If you have never heard of a "slant asymptote" or "polynomial division" then forget I mentioned this and just know there is no horizontal asymptote.)
Hello, leviathanwave!
There seems to be typos in the problem . . .
For a horizontal asymptote, take the limit of R(x) as x→∞Quote:
R(x) .= .[4x²(2x - 1)] / [7(5x - 3)(x + 2)²] .= .(8x³ - 4x²)/(35x³ - 119x² + 56x - 84)
Find an equation of the horizontal asymptote of R(x).
Divide top and bottom by x³ and take the limit:
. . . . . . . . . . . 8 - 4/x . . . . . . . . . . . 8
. lim .-------------------------------- . = . ---
x→∞ .35 - 119/x + 56/x² - 84/x³ . . . .35
Therefore, the horizontal asymptote is: .y .= .8/35