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Find the horizontal and vertical asymptotes of each curve. If you have a graphing device, check your work by graphing the curve and estimating the asymptotes.

$ y = \dfrac{2x^2 + 1}{3x^2 + 2x -1} $

$x=\frac{1}{3}$

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Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Okay here we have Y. Is a function of X. And we want to see if we can find any vertical ascent oaks and any horizontal assam tops. Uh The first thing we're going to do to find vertical isotopes. Uh The best way to find vertical assam tops is uh find values of X. Where the denominator will equal zero. So let's go ahead and uh factor uh dysfunction now two X squared plus one is not factory bubble. So the numerator is still going to be written as two X squared plus one. But the denominator can be factored. Okay, so the denominator can be factored uh three X squared plus two X minus one, factors into three X minus one times X plus one. If we want to look for a vertical as um Toque. You wanna look to see when uh the denominator ah will be zero. So if we're looking for a vertical assam Toped which will be a vertical line uh That your graph of the function will get closer and closer to uh either heading up towards infinity or closer and closer to heading down towards negative infinity. Uh So vertical as um Toque let's look to see when the denominator might be zero. So we're looking to see when three X -1 times X plus one is zero. When will this denominator be 0? Well, if three x minus one times X plus one equals zero, then either three X minus one is zero or the X plus one could be zero. The three x minus one equals zero. That means X would be positive one third. So that is going to be one of our vertical ascent tops. We'll see that when we graph it, X plus one equals zero means X is negative one. So those are two values of X. That would make the denominator zero in the function. Uh So these are going to be two vertical assam. Tokes vertical line at X negative one. A vertical line at xs one third. We'll see that when we grab it, horizontal ascent tops. We want to look to see what happens as ex moves towards positive infinity and also his ex moves towards negative infinity. Now as uh let's look at the original expression of the function as X moves uh towards um Well, let me rewrite it real quick. Okay, so we're looking for horizontal ascent. Oops. Uh so to try to find a horizontal assam. Tope, we have to look at the behavior of our function as X moves far to the right towards positive infinity. And again, as ex moves far to the left towards negative infinity. No, the dominant term in the numerator here is two X squared. When X gets very large, two X squared is gonna be very large, adding one is not really going to change that much. Uh The dominant term in the in the denominator is going to be three X squared uh effects is very, very large. Um Three X squared is going to be very large adding to ex uh you know, makes it larger but really the three X squared is going is going to be the dominating term in the denominator. Uh So, as ex heads towards positive infinity. Uh This function is dominated by two X squared over three X squared. If we cancel out the x squares, basically what we have is as X gets very large towards positive infinity, why is going to approach two thirds? So why equals 2/3 is going to be a horizontal sm 2? How about as ex moves towards negative infinity? Well, his ex moves towards negative infinity because we have X squared here, X squared here as a dominating terms, even though X is moving towards negative infinity. Uh It's very large and negative. Once you square, it's very large and positive. And so even though when X is moving towards negative infinity, this function is still going to once again Be approaching 2/3. So we only expect why it was 2/3 to be our only horizontal s into. Okay, here we have our function Y of X uh indicated or represented by the three different portions of the three different green curves are all part of the graph of Y of X. Um and we have uh the vertical and horizontal uh isotopes Great. So the green curves or the actual function, Let's look at the vertical lines, x equals negative one vertical line, X equals negative, one is a vertical assam. Tope as X approaches negative one from the left, the graph is approaching the graph of Y of X is approaching the vertical ascent open as it moves towards positive infinity. As X approaches negative one from the right side. Uh The graph of Y of X approaches the vertical ascent topaz, it moves down towards negative infinity. You can see the same type of thing is happening for the other vertical ascent. Okay, At X is one third as ex moves towards one third from the left, the graph gets closer to the vertical ascent oak line as it moves down towards negative infinity. As X approaches one third from the right of the graph of Y fx gets closer to the vertical ascent. Oak line as it moves up towards positive infinity. Last but not least. Let's look at the horizontal as um toque y equals 2/3. So we don't need to look at this and this at the moment. Alright, so here we have the graph of Y is Y. Of X. And we have the horizontal. As some tope line, Y equals two thirds. Now, as ex moves towards positive infinity, you can see uh the green graph of Y of X gets closer and closer to the horizontal, assume top line Y equals two thirds. And uh as ex moves towards negative infinity, once again, uh the green graph of the Y of X function gets closer and closer to the horizontal line as ex moves towards negative infinity