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In this video I go over a quick example on Slant Asymptote Lines and this time look again at the special case, which is a rational function in which the numerator has a degree just one higher than the denominator. The example I cover is determining the slant asymptote line of the function f(x) = x3/(x2+1). Since it is a rational function, i.e. a division of two polynomials, and with the numerator a degree of 3, which is one more than the numerator’s degree of 2, we can simply use polynomial long division because as illustrated in my earlier video, the quotient is the asymptote line, and in this case is y = x. Nonetheless, I prove that the quotient is in fact the linear asymptote by applying the definition of a slanted asymptote, which is the limit as x approaches infinite of the difference between f(x) and its linear or slanted asymptote must approach zero, which it in fact does in this case. Graphing the two functions, we can clearly see that f(x) approaches y = x on both the positive and negative infinity scales. This is a very good video into illustrating how to solve for slanted asymptotes for rational functions, so make sure to watch this video!
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Example
Find the slant asymptote line of the function f(x) = x3/(x2 + 1).
Solution
Recall that a Slant Asymptote is defined as:

Recall that a rational function is defined as being the division of two polynomials on the domain where the denominator is not equal to 0.

And since f(x) is a rational function with the numerator one degree higher than the denominator, then as shown in my earlier video we can use polynomial long division to determine its slant asymptote, which is the quotient!


Thus as x approaches infinity, f(x) - x approaches 0 which is the same as f(x) approaching the line y = x, thus y = x is a slant asymptote line.
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Retrieved: 13 November 2017
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Notice how even as x approaches negative infinity, the same asymptote line holds true.

Keep up the great work ;)
Happy to see the same quality!
Thanks! Quality is the name of the game at MES.