Graphing Simple Rational Functions

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Question

Graphing Simple Rational Functions

Graph the rational functions in Exercises 63–68. Include the graphs and equations of the asymptotes and dominant terms.

y = 2x/(x + 1)

Accepted Answer

Below there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Sketch the graph of the function Y equals 5 divided by X + 2, then write the equations for its asymptotes and identify its intercepts. Awesome. So it appears for this particular problem, we're ultimately trying to solve for 3 separate answers. So essentially our first answer that we're trying to solve for is we're trying to create a sketch of the graph for this function of Y, and we're also asked to find the equations for the asymptotes. So we're trying to determine whether or not this particular graph has and or horizontal or vertical asymptotes. It may have both, and may have one or the other. And identify. It's intercepts. And also when we talk about asymptotes, it might have an oblique asymptote too, but we will determine that as we go as we go along solving. So those are the three main things we're trying to solve what we're trying to solve for the, how to graph this particular function of Y. We're also trying to figure out if it has asymptotes. And then lastly, our third answer is we're trying to determine the intercept. So where does it intercept on our X or Y axis. So, with that said, now that we know what we're ultimately trying to solve for, let us start by trying to solve for, or rather, let's start by solving for our vertical asymptotes if there are any. So focusing our attention on solving for vertical asymptotes. So as we should recall, and note, vertical asymptotes occur where the denominator of the function equals 0, assuming that the numerator does not also equal 0 at that point. So we need to set our denominator equal to 0, so we need to take X + 2 and set it equal to 0, so when we isolate and solve for X, we will find that X is equal to -2. So that will mean, without a doubt that X equals -2 is our vertical asymptote. So that means there is a vertical asymptote at X equals -2, and that is one of our final answers. So let's put a box around it. So let us now solve for our horizontal asymptotes if there are any. And as we should recall a note, in order to find the horizontal asymptote or asymptotes, we need to examine the degrees of the numerator and denominator. So the numerator is a constant, which means It has, since the numerator is a constant, that means it has a degree of zero, since it's just the constant number 5. So it has a degree of 0. And because the denominator is a polynomial with a degree of 1 means you have this X plus 2, so it's X to the power of 1. And even though it doesn't show it's technically X to the power 1, so that means that since the denominator is a polynomial with the degree of 1, and the numerator is a constant with a degree of zero, that will mean undeniably that since the degree of the denominator is greater than the degree of the numerator, that means as a result, the horizontal asymptote without a doubt has to be. Y equals 0. So Y equals 0 is our horizontal asymptote. And that is also one of our final answers, so let's put a box around it. So now let's shift our focus to dominant terms. So as we should recall and note that for large values of absolute value of X, that will mean that the dominant behavior of Y will be dominated by the ratio of leading terms. In this case, as X approaches positive infinity. Or as X approaches negative infinity, that means that the value of Y approaches 0, and this is consistent with the horizontal asymptote that we determined earlier. So, with that said, we can now start to focus our attention on solving for the intercepts. So focusing our attention on the Y and X intercept, so starting with the Y intercept, so in order to find the Y intercept, as we should recall and note, we need to set X equal to 0 in our equation. So that will mean that Y is going to be equal to 5 divided by 0 + 2, which will be equal to. 5 halves. So that will mean that our Y intercept is going to be at parentheses 0.5 halves or 5 divided by 2. And for our X intercept, we, as we should recall, in order to find our X intercept, we need to set Y equal to 0, so that will mean that 0 is equal to 5 divided by X + 2, so we When we rearrange this particular expression to isolate sulfur X, that will mean that our final answer is going to be. Equal to 0. So as you can see, Since we have 0 is equal to 5 divided by X + 2, which will go to 5, which is equal to 0. So when we go to rearrange and solve, because if we multiply both sides for X + 2, that will be 0 is equal to 5. So that will give us a not possible answer. So that will mean that there is no X intercept, since when we try to isolate and solve for XL by itself, we will get An answer that doesn't exist, that's not possible. So with that in mind, our next step now is let us determine a few other points to help us graph our curve because it's important to figure out a couple different coordinate different coordinate points on our graph that we are going to use to connect them with a smooth curve, so it's a So our graph looks nicer. So, let us find a few random coordinate points. So let us say X is equal to -7, let us also say that X is equal to -3, let's also say that X is equal to -1. And let us also say that X is equal to 3. So that will mean that F of -7 is equal to 5 divided by -7 + 2, which when we solve for this particular. Expression, when we plug it into our calculator, we will find that it's equal to -1. So to save time, I'm not going to write out the whole equation every single time, so I'll just give you what the answer of F is. So F of -3 is going to be equal to -5. So once again, we're plugging in a value of -3 in for X. And for F of -1, so when we plug in a value of -1 in for X, that will mean that F of -1 is going to be equal to 5. And lastly, if we take F of 3, so F of 3, so we plug in a value of 3 in for X, we will find that it's equal to 1. So now, we can start to graph our function. So as we should recall a note for the asymptotes, we need to draw the vertical horizontal and or slant asymptotes as dashed lines, because these dashed or dotted lines will indicate that it is an asymptote and then it won't be mistaken for our curve because curves are drawn with solid lines. So, for the intercepts and other points, we need to connect these plotted points with a smooth curve, making sure that they approach our asymptotes that we've drawn, but they never cross or touch them. Because if you accidentally draw them and cross at the ascentdotes, you might lose points with your professor or teacher. So, with that said, now that we have enough information to graph, I've gone ahead and took our function and put it into my favorite graphing software choice. And when I did that, I produced the following graph. So we have our horizontal asymptote for Y equals 0 represented by a yellow dotted line. We also have our vertical asymptote, which is X equals -2 represented by a green dotted line. And then we have our curve for our graph of Y for our function of Y, and our curve is represented in red, and we have our intercepts. Drawn, which we have only one intercept, and of course, we determined our Y intercept to be at 0.2.5, and there is no X intercept. So that will mean Without a doubt, undeniably, that our final answers, without a doubt, will have to be once again to recap, we have our horizontal acetode at Y equals 0. We have our vertical ascentode at X equals -2. We have our X intercept, which once again is none, means there's no X intercept, and our Y intercept once again is 0.2.5. And as you can see, we have our other points that we found when we plugged in different values for X. We have -1.5, we have 3.1, we have 7.1 and -3.5. So it's usually good to plot at least 2 or 3 coordinate points to help us get a smooth curve, so like, you know, 2 to 3 minimum. To get an accurate curve, and as you can see, our curves get really, really close to our horizontal and vertical asymptotes, but they do not touch or intersect them. And that's it. We've officially solved for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Graphing Simple Rational Functions

Graph the rational functions in Exercises 63–68. Include the graphs and equations of the asymptotes and dominant terms.

y = 2x/(x + 1)

Key Concepts

Rational Functions

Asymptotes

Dominant Terms

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