Finding Vertical and Horizontal Asymptotes of Rational Functions

This video steps through 6 different rational functions and finds the vertical and horizontal asymptotes of each. A graph of each is also supplied. On the graph, the function is a black solid line and its asymptotes are green dashed lines.

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[@sghand1218]

Thank you so much for showing this example of asymptotes. It made learning so much clearer. I am now looking to solve Oblique asymptotes, but cannot find your video for that? Do you have one? Also, I am so thankful you have the Vertex and other functions in your teaching videos. Your videos will get me through this college algebra!! Thank you again.

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[@hsmathsolutions]

I'm glad I could help!! :-)

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[@hsmathsolutions]

Glad it was helpful!

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[@ziyolamusic2066]

Yeah it's my summative exam tom and this is the Coverage for summative assessment. Gen. Math: asymptote problem solving involving rational equation(motion and number problem), exponential equation, exponential inequality, exponential function, logarithms and logarithmic properties. Pre cal: Systems of Non linear equation Series and sequence Sigma notation Fundamental identities

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[@joycexie8537]

4:48 I have a question, if shouldn't you factor out the numerator and the denominator first and cancel out (x+2), which would be a hole?

[@hsmathsolutions]

Thank you for your question. In that example, the numerator is prime and will not factor; thus no terms will cancel out. But you are right, if it did cancel, there would be a hole instead of an asymptote. Also, you can take a look at the graph for confirmation.

[@joycexie8537]

Ohh.....I see it now! There would be a middle term if the numerator were to be factored into (x+2)(x+2) I was not looking carefully XD Thanks for replying.

[@BrentEuler]

How do you know how to graph these lines (functions) within the multiple vertical an horizontal asymptotes?

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[@user-bn6jr1mz8i]

We give values to hole equation or only x-3 for graph

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[@RDFspaniard]

Fyi x^2+1 can be factored, but it's factors are not on a real number graph.

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[@devendrautekar7305]

Why did you equate the denominator to zero in case of vertical asymptot?

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[@crislyngerolao9686]

I mean in 1:55 why is the curve on the Quadrant 1 and not on Q2? Then the curve on Q3 not on Q4??????

[@bonbonpony]

05:41 Actually, there's one more asymptote in this example: y = x² + x + 1. If you add it to your graph, you will see that the branches of the curve approach this parabola as `x` increases/decreases without bounds. 06:18 Well, at least not in the plane of the graph ;>

[@AadityaChawla-bc4ic]

I don’t understand can you please elaborate.

[@bonbonpony]

@@AadityaChawla-bc4ic Start with the rational function from the video: `y = (x³+2) / (x-1)`. Divide out the numerator `x³+2` through the denominator `x-1` (e.g. using polynomial long division) to get the polynomial part `x² + x + 1` and the remainder (which is still rational) part `3/(x-1)`. In other words, you get the following decomposition: y = (x³+2) / (x-1) = x² + x + 1 + 3/(x-1) Now observe what happens when `x` increases (or, for that matter, decreases) without bound: the denominator `x-1` in the last term (which is still rational) gets bigger and bigger, which means that you divide the numerator `3` into more and more parts, so each such part gets smaller and smaller, i.e. the entire fraction shrinks to 0, leaving just the polynomial part `y = x² + x + 1`, which is a formula of a parabola. This means that as `x` gets bigger and bigger (either positive or negative), the graph of this function starts looking more and more like a parabola `y = x² + x + 1`. Here's a graph that demonstrates it: www.desmos.com/calculator/kjs0e3ivya Notice how the blue curve approaches the red parabola as `x` gets more and more negative. But when `x` becomes more and more positive, approaching `x=1`, the curve gets farther from the parabola and starts approaching the vertical asymptote (which is a straight line) instead, `y` going towards `-∞`. Then it returns on the other side from `+∞` and starts diverging from the vertical asymptote and getting closer and closer to the parabola again as `x` increases even further (as well as `y`, going towards `+∞` again). You can also turn the green graph on or off to verify that these two formulas, `y = (x³+2) / (x-1)`, and `y = x² + x + 1 + 3/(x-1)`, are indeed the same function. Also try zooming the graph in/out with your mouse wheel to observe how the graph approaches the straight vertical line as well as the parabola as `x` gets bigger (disregarding its sign) without bounds.

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On 1:55 why is the curve on the upper right corner? Cant it be facing the left? And the other one to its right???

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i don''t understand one thing, my professor uses limits to find them and no matter how many videos i look up they all use this method, am i missing something?

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What about vertical and horizontal shifts

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[@selenadookaran4984]

thank you! I have a question like this 3x^2 + 17x + 10 / x^2 + 7x +10

[@hsmathsolutions]

The numerator (3x^2+17x+10) will factor as (3x+2)(x+5). The denominator (x^2+7x+10) will factor as (x+2)(x+5). The (x+5) factors will cancel between the numerator and denominator. However, the denominator still has the (x+2) factor. So there will be a vertical asymptote at x=-2 (x=-2 causes the denominator to be zero), and a hole in the graph at x=-5 because those factors canceled. Hope that helps!

[@youknownothingjonsnow2418]

what if the function doesn't have a denominator? fx= x+3x+7??

[@hsmathsolutions]

If the function doesn’t have a denominator, then there will be no asymptotes. Vertical asymptotes arise from division by zero. Without a denominator, there cannot be division by zero. Horizontal asymptotes arise from comparing the degrees of the numerator and denominator. Again, if there is no denominator, then there is no horizontal asymptote. Hope that helps!

[@jask320]

The second graph is not correct, there is a max point for the right side at (8, 0.0625) if you take a first derivative and then it goes down to zero inf.

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