Horizontal Asymptotes and Slant Asymptotes of Rational Functions

This algebra video tutorial explains how to identify the horizontal asymptotes and slant asymptotes of rational functions by comparing the degree of the numerator with the degree of the denominator of the rational expression. The equation of the slant asymptote can be determine using long division if the degree of the numerator exceeds the degree of the denominator by exactly 1. This algebra video tutorial contains plenty of examples and practice problems.
Fundamental Theorem of Algebra:
• Fundamental Theorem of...
Rational Expressions - Basic Intro:
• Rational Expressions -...
Simplifying Rational Expressions:
• Simplifying Rational E...
Multiplying Rational Expressions:
• Multiplying Rational E...
Dividing Rational Expressions:
• Dividing Rational Expr...
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Adding & Subtracting Rational Expressions:
• Adding and Subtracting...
Rational Expressions - Unlike Denominators:
• Adding and Subtracting...
Simplifying Complex Rational Expressions:
• Simplifying Complex Ra...
How To Solve Rational Equations:
• Solving Rational Equat...
Rational Equations - Extraneous Solutions:
• Extraneous Solutions o...
_____________________________________
Horizontal and Slant Asymptotes:
• Horizontal Asymptotes ...
Finding Rational Functions Given 2 Points:
• How To Find a Rational...
Rational Functions - X and Y Intercepts:
• How To Find The X and ...
Graphing Advanced Rational Functions:
• Graphing Advanced Rati...
Rational Inequalities:
• Rational Inequalities
________________________________________
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very late reply but there is a quicker way called synthetic division, when you're dividing by those trinomials with a binomial it makes it so much easier.

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[@nejcocepek8357]

in the first case of slant asymptote the asymptote does not exist since the remainder of long division is zero and if we cancel out the terms we see that we get linear function with equation y = x + 2 with a missing point at x = -3 which lies on the asymptote and that contradicts with the definition of an asymptote

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exactly i was confused on why he said it was slant

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[@taiedy]

Sorry, how can there be a slant asymptote if the equation at 5:20 factors cleanly into [x+2] without a remainder? Shouldn't there be a slant asymptote only if it leaves a remainder?

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[@sonjahenderson1390]

Im confused about if these rules ALWAYS apply. For instance take (5x + 1) / (x^2 - x - 1) and graph it in the graphing calculator. There appears to be a horizontal asymptote at y = 0. However there is a vertical like that strikes through y = 0 in between 2 other lines in the function. Does the horizontal asymptote still apply?? Do these set of rules apply no matter what? If anyone could help that would be wonderful!! Thank you

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I'm talking about: x^2+5x+6/X+3, but algebraic long division is probably something to get marks in the exam, rather than factoring, both ways are not that hard, after you explained them.

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is there a way to rationally calculate these without knowing any tricks like "If the degree is the same devide the coefficients".
I want to know how to look at a function and see in my head where the asymptotes would be

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By the way, synthetic division can be used to find the asymptote, although it might be easier to use polynomial long division in certain situations.

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Is synthetic devision the same as long division?

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[@jamesforteza5142]

Y=(x²-3x+4)/(x²-6x+8)
Y=1 is the horizontal asymptote but when you graph it it is only applicable on the right side of the graph but when you extend the line y=1 to the left it will intersect the graph of the given rational equation.

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its easy to remember if u think like this...
Think the rational expression as a building
if its top heavy just by one it will collapse but not desintagrate resulting in a oblique line
if its top heavy by more than one it will collapse and desintagrate, theres no line!
if its bottom heavy well the building its on the floor as it should be, y=0
if top and bottom are equally heavy, then divide those bad boys

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[@mr.9754]

The other oblique asymptote is good, but I think it's best to write out the full answer (kind of like a mixed number). The rational function, after long division, equals 2x + (-16x+6)/(xx+4).
Then analyze as x approaches infinity....(-16x+6)/(xx+4) will approach zero....ie leaving y=2x as the oblique asymptote.

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