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Find the horizontal and vertical asymptotes of ea…

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Problem 47 Medium Difficulty

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{5 + 4x}{x + 3} $

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Video Transcript

Okay here we have uh two function Y. X equals five plus four X over X plus three. I'd like to rewrite the top instead of five plus four X. Let's just rewrite it as four X plus five And then that is over X-plus three. Now, if we want to look for a vertical, ask them to, we want to look for when the denominator Uh might be zero. Okay, so to look for a vertical assam toe, when is the denominator? X plus three equals zero. While the denominator X plus three equals zero. When X equals negative three. So we're gonna have a vertical ascent. Okay, When X is negative three, we'll look at that on the graphing calculator in the moment. How about a horizontal asked them to horizontal, ask them to ah well to find a horizontal ascent. Okay, we need to see what happens to dysfunction as X goes towards positive infinity and as X goes towards negative infinity. The dominant term in the numerator is for ex as X gets very large toward positive infinity, four X plus five is dominated by four times X, adding five to a very large number is not going to change it. So four X is the dominating term in the numerator. And let's put a one in front of this ex uh in the denominator. Uh When X is getting very large for its positive infinity, one X plus three is going to be dominated by one X. And so as X is moving towards positive infinity. Uh This expression is tending towards is dominated by four X over one X. Now, four X over one X. We can cancel the exes and have 4/1. So his ex kids are very large towards positive infinity. Uh this expression is going to be headed towards 4/1. So why equals 4/1 or four is going to be a horizontal assume too. Now, how about his ex heads towards negative Anthony if X is heading towards negative infinity, that means it's very large and negative. While the dominating terms in the numerator and denominator will still be four X over one X. And even though X is headed towards negative infinity, meaning it's negative when we cancel out the X. Is X divided by X negative divided by negative from the a positive. So this expression is still heading towards 4/1. Uh As X approaches negative infinity. So as X approaches negative infinity, we're going to get the same horizontal aspect to Y equals support. So what I'm gonna do next is we're gonna grab our function Y of X along with the vertical ascent top line and the horizontal ass from tow line. Okay. Uh Two segments of uh the graph is our function Y of X. The green vertical line is our vertical ascent. Opec's equals negative three. And a purple horizontal line is our horizontal asked until Y equals support. Uh Let's look at the horizontal, asked until first as ex moves towards positive infinity. The graph gets closer and closer to the horizontal Assen, took mine as ex moves towards negative infinity. Uh The blue graph of our function gets closer and closer to the horizontal. Ask them to apply. Let's look at the vertical assam. Two of X equals negative three. Is green lying. Is our vertical ascent to As X approaches -3 from the right side, our graph gets closer and closer to the green assume top line as it moves down towards negative infinity. As X approaches -3 from the left side, our blue graph gets closer and closer to the vertical ascent. Oh, green line as it moves up towards possum fitting as X is moving towards negative three from the left, our graph is approaching, getting closer and closer to the vertical ascent. Oak line as our graph moves up towards positive infinity.

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