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2CQUESTION Find all vertical and horizontal asymptotes of the graph of the given function:61 f (x) 8 +4 [CLO-1] ~8; horizontal asymptote;y = - } Vertical asymptote,...

Question

2CQUESTION Find all vertical and horizontal asymptotes of the graph of the given function:61 f (x) 8 +4 [CLO-1] ~8; horizontal asymptote;y = - } Vertical asymptote, X= { horizontal asymptote, _ y =-8 Vertical asymptote, X = - X = 6; horizontal asymptote, y =-8 Vertical asymptote, asymptote, _ y=6 -8; horizontal Vertical asymptote, X

2C QUESTION Find all vertical and horizontal asymptotes of the graph of the given function: 61 f (x) 8 +4 [CLO-1] ~8; horizontal asymptote;y = - } Vertical asymptote, X= { horizontal asymptote, _ y =-8 Vertical asymptote, X = - X = 6; horizontal asymptote, y =-8 Vertical asymptote, asymptote, _ y=6 -8; horizontal Vertical asymptote, X



Answers

Asymptotes Find all horizontal and vertical asymptotes (if any).
$s(x)=\frac{8 x^{2}+1}{4 x^{2}+2 x-6}$

We need to find the or is a little and vertical ass and dotes for this function. Let's start by finding the horizontal Nassim thought. So we see that the power off the numerator is the same as the power off the denominator, that being the two over there in the numerator and the tool with their and the denominator. And when that's the case, we know that the horizontal Allison told, will just be no coefficients, the fraction of the coefficients for the leading terms. So that would be this eight and this four so and the horizontal ass and told, is going to be, Ah, why equals two 8/4, which is to now let's go ahead and find the political Assam totes to find the vertical Assam coats, we need to find what value off the off X makes this denominator zero. So let's go ahead and do that. So basically four x squared plus two x minus six is equal to zero. And when you look at this, then it should realize that we can use the quadratic formula to find out what value off X satisfies this bottom equation. The quadratic formula goes like this X is equal to minus be plus or minus square root off. Be square minus four a. C every to hey and over here, the A is our four. The baby is our two, and the sea is our six. So, basically, let's just go ahead and put in those values. So it's minus B, which is to plus or minus square root off be square, which is two square minus four times a, which is four times. See it just minus six. And this whole thing is under the square root, divided by to okay, which is four. And when you evaluate that, you should get minus to plus or minus 10/8. And this Let me just continue evaluating. This is idle. Ah, minus two plus 10 over it and minus two, minus 10 over it. Now, let's just find the value off it x for different values for these options over here, this side would be 8/8. Uh, the site would be find this 12 or eight. So 8/8 hours one. And we could just leave this as perhaps actually could simplify this. So this would be four to is a read four threes in 12 3/2 is negative. 1.5. So these are the vertical Assen tolls. So we have the horizontal Assam thought, which was two on the vertical Assen totes. No, it's just call these vertical hasem is that one, uh, and minus 1.5.

Okay. So um let's find uh horizontal a stem toad. Mhm. So mm. We know that an equals to two And m equals to two. Okay. From what we learned from the textbook, we know that if any goes to m mhm. Yeah. The lie 3/1. Or you can you have in the last three? Why you go to the three? Is the horizontal assembly thought. Yeah. Of mm hmm, affects Okay, So for part B Let's let's find the .1. H. F X crosses horizontal. A sum total. Okay, so Let H. FX equals 2 3. Okay, so we have three x squared plus. Yeah. Yeah. So we have okay, we multiply X uh square plus three to the both sides of the equation. Okay, so we have three X squared plus nine. Sure. So we have N. x. equals two 14. So actually goes to 14/8, Which it goes to 7/4. Okay. So the graph, so I think we should also plug this back into our our equation, our function here. So H off 7/4. It goes to. Okay. Okay. So this goes to This is a coastal three. Okay. Yeah. Telegraph across this the horizontal a sample tote at 7/4. 3

Okay, we are given a rational function and we're going to find its vertical and horizontal ascent tops and sketch the graph. So the first thing I'm gonna do is I'm going to look at my vertical ascent hopes. So when I do that I need to look at where my denominator equal zero. So for that I am going to have to factor so I'll factor into an excellent X. Looks like a four and a two and then I'll make my for the negative and my to the positive. That's why. Um so I'll have a negative four X plus two X. Get that negative two in the center. Okay, so now it's about the denominator equaling zero. So either of these factors can equal zero. So I'll put X plus two equals to zero. And I find I have a vertical ascent to Pet X equals negative two. I can also put X minus four equal to zero and I find I have one also at X equals four. So let's go ahead and draw those vertical ascent tops. Okay, so now I'm going to look at horizontal asientos. So that's all about comparing the degrees and my numerator and my denominator notice that I have a degree to in the denominator, so therefore my horizontal a sento will be Y equals zero. Okay, so um I'd like to figure out where I'm crossing the y axis. So I'm gonna put X equals zero in and I will get a one over negative eight. So a negative 1/8. So I can mark that. And then I like to look there's only a one on top. So really the bottom will tell me if I'm positive or negative. And my factored form is the easiest one to look at. So if I put any value between negative two and four into my factored form notice they're all going to give me negative values so I can draw this down below the X axis. And then I also know as I go towards my vertical ascent taub's, I have to go with unbound behavior. So now um with them being the factors being to the first power. What I know is I'm going if I went towards negative infinity on one side of my vertical ascent tobe, it's gonna go towards positive and fitting on the other. But I can also put those values in and see that I actually have positive values, which puts me up on the top.

Okay for problem 34. They give us a rational function. So a polynomial divided by a polynomial. Um, and in this case, it's a linear function divided by a quadratic. They want us to find the vertical assam totes. Now, vertical Assam totes occur when we have division by zero. Um, and there's one extra stipulation where, um, this does not occur. So where you cannot cancel any terms, Um, or I'm gonna say remove any variable factors. So what that essentially means is if I can factor the top and bottom and cancel a factor So something like X plus or minus the number divided by X plus or minus that same number. If I have the same factor divided by itself, I can cancel that out. Right? The division by itself always equals one which would make whatever this number is the plus or minus. That would be a removable dis continuity, otherwise known as a missing point or a whole dis continuity eso hole and or missing point, depending on what terminology you were taught. So I always referred to it as a whole. Um So what we need to do for this particular example? I'm going to factor. The top is already is factored is it could get but the bottom. I'm gonna I'm gonna factor um two x squared and see if there's any terms that will cancel. So two X squared minus six X minus eight. Well, first of all, I see that there's 26 and eight, so I can factor to out right away. Um, that's my greatest common factor. So I have two times x squared, minus three X minus four, and then I can see right away. I like to think about factoring. Is unf oiling? So this is the product of foiling the result. Right? So this is first times first, this is last times last. This is outer plus inner. So I need to figure out what numbers when they're multiplied together, they give me the last term. And when they're added together, they give me that middle term of negative three eso That's gonna be negative. Four and one. So negative for positive. One negative for a positive one. Um, which means that this factors into X minus four and X plus one right there is the minus four in the plus one and then don't forget about that to you. So I'm gonna rewrite f of X. Now I've factored the numerator, which nominator? I'm sorry. The top stays the same. It's already and fully factored form. So X plus two and then the numerator. The denominator is the two times X minus four times X plus one. Now I can see that there are no factors that will cancel, which means that both of these terms x minus four. These variable expressions and X plus one are non removable. So non removable tells me as, um tote vertical aspecto which hopefully we know that because of, uh, it's X and not why. So all I need to do now is solved for where I have division by zero. So where two times X minus four X plus one equals zero. The two doesn't really matter. I could divide to on both sides, and I you know, the cancel out. I just need to know where X minus four equals zero affordable sides. We get X equals four and where does X plus one equals zero? Subtract one. We get X equals negative one. So X equals for an X equals negative one. Our equations of vertical lines, those air the equations of our vertical Assam totes. And we know that for a negative one are not in the domain of this function because they give us division by zero at at those points, um, eso we know to exclude them from our domain. Um, but that is also what tells us that they're the vertical ass and totes of this function F of X.


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