Question
22 + x 3. For the function h(z) answer the following questions 212 + 6x[6]Note that 2x2 + 6r is zero when € =0 and when ~3- Use limits to describe the behavior of h near I = 0 and near x -3. Give the equations of all vertical asymptotes of h.[4] b) Use limits to describe the end behavior of h_ Give the equations of all horizontal asymptotes of h.
22 + x 3. For the function h(z) answer the following questions 212 + 6x [6] Note that 2x2 + 6r is zero when € =0 and when ~3- Use limits to describe the behavior of h near I = 0 and near x -3. Give the equations of all vertical asymptotes of h. [4] b) Use limits to describe the end behavior of h_ Give the equations of all horizontal asymptotes of h.


Answers
In Exercises $37-44$ , find the intercepts and asymptotes, use limits to describe the behavior at the vertical asymptotes, and analyze and draw the graph of the given rational function.
$$
h(x)=\frac{x^{2}+2 x-3}{x+2}
$$
We're considering function F of X equals two divided by X plus five Excuse me expose six and were trying to find the horizontal asthma totes of F if there are any. So we need to consider the limit as X goes to infinity of F of X and the other potential last maturity EVs the limits as X goes to negative infinity of ffx As X goes to infinity, it's clear that the denominator of the first limit goes to infinity. So we have two divided by infinity. That's zero. Similarly, for the second limit, therefore, F has one horizontal aspecto namely why equals zero.
For this item, we'll find the intercepts. Hasn't votes. Use limits to describe the behavior of the ASM totes, analyze and draw the graph of the rational function? Let's see for the X intercept, you want to know when the numerator will be zero. You take the expression, said it equal to zero and find that that's for X equals one. So the X intercept is 10 the Y intercept, because H of zero it was 1 12 that is the Y intercept looking for the vertical as and choked the expression in the denominator. It factors X minus four X plus three. Setting each of these expressions equal to zero in solving reveals the two vertical as in totes when we consider horizontal as in tote. Because the degree of the numerator is less than the degree of the denominator, we know that the horizontal isn't tote will be. Why equal zero. Let's take all this information and sketch a graph off this function. Let's re emphasize y equals zero. It's the horizontal, as in tow years wth e vertical as an Choate X equals negative three. My scale here is going to be none too perfect here is a vertical as a toad at X equals four. We have a Y intercept. It is 1/12 You We haven't X intercept that once all say it's here. So now when sketching out the curve, we have this piece. We know that we're going to cross through those intercepts and, of course, the final tail. So there is a sketch of the graph. I'm going to move to a second page to complete the analysis. Here, let's make a final record and said that the intercepts let's recall were 1 12 for the Y intercept 10 for the X domain. Recognizing that vertical, as in totes, isn't breaks or restrictions on the domain of the function, there's our Joe Mean range. Recalling that we had a continuous piece between the vertical as and totes. All real numbers are from negative affinity to positive infinity. If we look with an eye to continuity, it is everywhere except where X equals negative three and four. We look for a decreasing interval. Negative Infinity to negative three. Excuse me. I didn't mean to do that. These are all intervals, so we don't need to use a conjunction. Is thes three in journals. That's everywhere. It is not symmetric, it is unbounded. And let's make sure we address the behavior around the ass and totes the limit as X approaches. Negative three from the left of the function is negative. Infinity. The limit of function. This X approaches negative three from the right positive infinity limit of the function except H of X. This X approaches four from the right negative infinity. The limit of the function as X approaches for from the right is positive, infinity and finally, the limit of the function. As X approaches negative, infinity is equal to the limit of the function. It's X approaches, positive infinity, and that was equal to syrup.
Hello. So here we have our function H. Of X. And our denominator here is going to factor so we still have here an X plus six. But then over while this is going to factor as X squared is X times X 24 is in fact that's gonna be um what six times four? And then we have to get a plus two X. In the Middle East. We have to have an X. Plus six and then an X minus four. Right? So therefore we see we now have an X plus six over X plus six. So X is equal to negative six. We get 0/0. So therefore we have a whole at X. Is equal to negative six. And then if we cancel that holdout, cancel that 0/0 out just left with one over X minus four. And the denominator there is zero and X equal to four. Therefore we have a vertical ascent toe at the line X is equal 24.
So here. What we're gonna do is we're gonna try to find a vertical Assam totes or holes of irrational function. I'm gonna briefly give you out, um, a general strategy, and then we'll apply the general strategy to our specific problem. So in general, if we have this function rational function, it's a polynomial divided by Apollo. Now, the first thing we're gonna want to do is we're gonna want a factor factor, the numerator and the denominator. So let's imagine that we get like X minus a next month's be over X minus C times X minus D. Now what this means is that you see this X minus e in the denominator. Um, this means that at X equal, see, there is either so x equal. See is either a vertical Assen tote Dr Abbreviate v dot eh, or there is a hole, and the same thing applies for X minus. D X equals D is either and I'm getting this x equal. See, by just setting this equal to zero in solving same thing here. Explain Esty said it equal to zero in salt means X minus D is either a vertical Assam dude or a pull, and then depending on whether or not um anything simplifies that's going to determine whether or not we have a vertical ass into it are all. So, for example, if we had, um, X minus a X minus B over X minus a again and X minus D In this situation, there would be, Ah, hole at X equals A because there's this thing called the reduced forms of the reduced form would be X minus B over X minus d. And when we simplify, we have to say, hey, and he is technically not in the domains. We have to make that, um, extra step. And now this reduced form basically tells you, um, anything that's left in the denominator is gonna correspond to a vertical ascent. Oat. And if he simplified anything to get to this step, that's gonna correspond to a hole. So in general, factor the rational function as much as you can. Um, you'll have this factored form and then get the reduced form, have those sitting next to each other, and then you can compare those two and see where vertical Assam toots or our holes are. So let's try that for a particular problem in each of X equals X plus six over X squared plus two x minus 24. Step one is to factor, um, pro tip if you're, you know, finding holds and vertical ascent. Oates. If there's a there's only an explosive X in the numerator. Very likely there's going to be a factor of export. Six. And the denominator, that's how. Helps with the guessing and checking. Um, still, you know, kind of double check, but looks like exploits. Six X minus four is going to be we factor. Um, and then right so we can simplify there. There are two routes here. Um, note this Expo six simplified. So we have to make sure we add that additional domain restriction. Um, because here, we can't see that X can't be negative sinks. Or we can think about the reduced version one over X minus for Okay, So if there's a factor in the denominator of the reduced version, that's gonna correspond to a vertical ass in tow, So vertical ass into it and X equals four. Just set the sequel. That zero that gives you a vertical ass in tow. There's a hole. Um, when x equals negative. Six, right? So an explosive X was in the original function in the denominator. It's not in a reduced version or the simplified version. That means it's corresponds to a hole, Um, and coordinate. And then the X coordinate of the hole is the negative six. And to figure out the why. Cornet you input the X coordinate into the reduced version, right? If you try to input it here, you're gonna divide by zero. So that's no, that's no good. It's a negative six negative 1/10.