Horizontal Asymptotes and Slant Asymptotes of Rational Functions

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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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MR. Organic Chemistry tutor ,thank you for a well explained and solid analysis of Horizontal and Slant Asymptotes in precalculus.

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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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[@noeldiaz1354]

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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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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@@daniaalsadi7208 you end up getting the same answer in both but the way you get that answer is little different.

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imo synthetic division is more convenient to use

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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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its amazing how you didnt stretch this into an 80 minute lesson for absolutely no reason!

[@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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8:28 but that one has an asymptote where it's a cubic function

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*Why do bad teachers get paid?:* tenure this is when school sign contracts with teacher pretty much guarantees them a job for life no matter what. That's why when you get a bad teacher most of the time they get away with it. Even if they are so bad and the school decides to fire them the school has to go through a complicated process that can last at least a year or longer. Chances of a teacher getting fired is 1 in 2000 compared to a doctor which is 1 in 57. The American public school system sucks. We need more teachers like this dude.

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[@jstagirlinthisworld]

So, there is a major maths error in this video. Starting at 5:00 We are given the equation: (x² + 5x + 6) y = --------------------- (x + 3) which, when properly factored, is: (x + 2)(x + 3) y = ----------------------- (x + 3) So, even though the numerator is 1 degree larger than the denominator, it also shares a common factor with the denominator of (x + 3). Before dealing with asymptotes, we have to remove any *holes* that may be present in the graph, and this matching (x + 3) factor tells us that there is actually a HOLE at x = -3, not a slant asymptote. Removing this hole from our equation gives us our actual graph which is now reduced to just the graph of: y = (x +2) with y's domain restricted to all real numbers *except* x = -3 that is, our function is simply: y = x + 2 Domain: (-∞, -3) ∪ (-3, ∞) Plug this into a graphing calculator if you want to double check - the graph is just a straight line that is undefined at x = -3 You can't have an oblique asymptote if your graph itself is just a line

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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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4:00 - isn't this just equivalent to x+2, a linear function with no asymptotes?

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[@bobu5213]

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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[@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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[@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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THANK YOU 💚👀✨ I didn't understand how to find Slant asymptote, but I'll understand I watched this video twice, and now I better understand how to find slant asymptote :з)

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The third example at the end.. why is it bottom heavy? Wouldn't they both have a degree of 1? 10:05

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May you please help me with a video on how to work with vertical and horizontal asymptote that involves square roots

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Didn’t need long division for the first slant asymptote since you could factor the numerator which cancels with the denominator, leaving x+2.

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