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Graph the rational function f(x) by answering the following questions Answer each question and graph the function.x2 _ 4x - 5 f(x) = X-3Factor the function. What ar...

Question

Graph the rational function f(x) by answering the following questions Answer each question and graph the function.x2 _ 4x - 5 f(x) = X-3Factor the function. What are the domain restrictions?Simplify if possible: Find the vertical asymptote(s) ofthe function. Add this feature to the graph:Based on degree of the numerator and denominator; will this function have a horizontal asymptote? If so, what is it?Use long division to find the oblique asymptote of the function:Does the function cross the obl

Graph the rational function f(x) by answering the following questions Answer each question and graph the function. x2 _ 4x - 5 f(x) = X-3 Factor the function. What are the domain restrictions? Simplify if possible: Find the vertical asymptote(s) ofthe function. Add this feature to the graph: Based on degree of the numerator and denominator; will this function have a horizontal asymptote? If so, what is it? Use long division to find the oblique asymptote of the function: Does the function cross the oblique asymptote? If so, when? Place a point on the graph if this occurs. Find the x- and y-intercepts Place them On the graph: Look at the sections of the graph formed by the asymptotes Notice that we have already found important coordinates in each of the sections: The curves must stay very close to the asymptotes as x approaches infinity and negative infinity. In the left section of the coordinate plane, we can connect our curve through the intercepts and sketch it closely approaching each asymptote: In the right section of the coordinate plane, we know that the curve does not cross the oblique asymptote_ so it must g0 through the intercept and then closely approach the oblique and vertical asymptotes. Sketch the curves of your function



Answers

For the given rational function $f$ : $\bullet$Find the domain of $f$. $\bullet$Identify any vertical asymptotes of the graph of $y=f(x)$ $\bullet$Identify any holes in the graph. $\bullet$Find the horizontal asymptote, if it exists. $\bullet$Find the slant asymptote, if it exists. $\bullet$Graph the function using a graphing utility and describe the behavior near the asymptotes. $$f(x)=\frac{4 x}{x^{2}+4}$$

All right. So this is a rational function to start. We were going to actually simplify it. In fact, are the denominator. So, I will factor the denominator. Using difference of squares X plus two X minus two. All right, now, I can actually find my asientos. So, my vertical ascent oats come from the denominator. So I will set the two items down there even at zero. Been solved My too weird a glass and totes are X equals two and X equals negative two. Or my horizontal as I'm told, I'm looking at my degrees, my degree for the numerator is one. My degree of the denominator is too, since my top is less than my bottom. My rule states that my horizontal hospital is automatically at Y equals zero because nothing canceled. I have no holes. And because of my degrees, I have no slant asthma tote. Um So next I'm going to find my domain. So my domain will actually use uh my vertical as in terms of stopping points. So my domain will be negative Infinity to -2. And because it's a vertical ascent owed, it can't be on it. So, I'm using parentheses Union,- two. Union to infinity. Um Next, we're going to look at the graph which I have here. So, if you notice I have my vertical assam to its at -2 and two. And then my horizontal asthma toad is that wife was zero. It is okay to actually touch or go through horizontal assam tobe. Whereas you can't do that with vertical one. So that's why there is a point on 00 here. Um Next, I want to talk about the behavior near the ascent owes. So, if you look at my vertical ascent owes here, as uhh, the graph approaches As X approaches -2 from the left, why is going towards negative infinity keeps going down here as the graphic Approaches -2 from the right? Why is going up towards positive infinity? Now, we're gonna look over here as the graph approaches to from the left is going down towards negative infinity. Here is the graphic approaches to from the writers going up towards positive infinity. And lastly, for my horizontal Assen totes as X is approaching negative infinity. On the left, why is going toward your horizontal horizontal center, which is zero. You never hear on the right as X goes towards positive infinity, your graph is also going towards zero. Your horizontal aspecto

Okay, so for this rational function, first of all, we need to look to see if it can be simplified at all, which it cannot be because the denominator has x squared plus one, which is not difference of squares. What this means is there are no vertical ascent oats because they would be complex numbers. Also that means that nothing canceled. So there are no holes. Um looking at our degrees for a horizontal a sento uh to agree for the numerator is one to refer denominators to, since top is less than bottom, that means that is automatically at y equals zero because we have a horizontal assam toe. That means that we have no slant, awesome job. So, due to all this, our domain is going to be negative infinity to positive infinity all real numbers because there's nowhere to class and no hole to stop at in the domain. Now we're going to look at the graph of the function which is here. All right. Um So if you recall, you can actually pass through a horizontal a sento. Um so the graph does actually cross through here at 00. Um looking at what occurs as the graph nears your horizontal hasn't tip as X goes left towards negative infinity, the graph gets infinitely closer and closer to zero as X goes to the right towards positive infinity, that graph gets closer and closer to zero also, which is both are horizontal Asato

This rational function as we're identifying some characteristics, we always want to start off by looking to see, can we simplify it in this case? We can so this is these are both factor, but we have a sum of cubes on top and a difference of squares on bottom. So as we uh simplify that out, we end up getting this. And so that's the first thing to make sure that you note the reason that that's important is because because this cancels out or simplifies out, that's gonna tell us that we have a hole in the graph at negative one. And if we plug that negative one back and it would be a negative one, negative three halves. And then let's see our domain, we'll figure that out in a minute. Are vertical ascent. Oh, it's going to come from right here. So we have a vertical ascent toe at X equals one. And then now I think it's easier to go to the domain. So our domain, we have a break in the graph at one and a negative one from our vertical ascent to and our whole And so our domain is going to be from negative infinity to negative one, And then again from negative 1- one. And then again from 1 to infinity. Um as we take a look at the vertical assume toe, we can identify some behavior as X approaches one from the left side. So as it's coming in from the left side, our function f of X is approaching negative infinity. Oops looks like an eight negative infinity, you can see that going down right here And then as X approaches one from the right side, what is the graph doing? The graph is approaching positive infinity. So we can make note of that in this case when we take a look at the degree of the numerator, which is three compared to the degree of the denominator. The numerator is one more. Exactly one more. Which means we do not have a horizontal assam toe, but rather we have a slant awesome toe. And so we have to identify. What is that Selena sento. That is where we have to use long division or synthetic division to do such. So as we saw that out, you would have X. And this becomes X squared minus X. So subtracting that this would all cancel out. Bring down the one. But at that point that one doesn't really matter. That's our remainder. So we have a slant. As um tote at Y equals X. And we can see that here. Lastly, what is the end behavior of this graph When we use our calculator or dez most or graphing feature are in behavior is as X approaches negative infinity. Our graph is simply staying below the Y equals X line

So we're giving this function. And the first thing is we need to find out what makes the denominator, what makes the denominator on defined. So we want to factor this. So we have X-plus four, X -3 equals zero. So the roots X equals three and X equals negative four. Are going to cause a problem that's going to make it undefined. So my domain is going to be negative infinity and negative for -4, deposited three and then 3 to an affinity. That's my domain. Now the vertical esposito is the denominator study equal to zero. And that's what we just did. So we have two vertical ascent totes, X equals three and X equals negative four. 12341234. So I have one at X equals positive three And x equals -4. Yeah. Let's try to make this a little straighter. Okay, now are there holes in it? No, we can't cancel anything. There's nothing to cancel when I have this over this. So there are no removable discount annuities. The horizontal assam tote is bottom heavy. It's heavier on the bottom. So it's going to be approaching zero. There's my ass in. Tokyo. Come on. Uh This slant ascent tote. There is none because the top is not one degree higher than the bottom. The X intercepts means we're going to put um set the numerator equal to zero. So we have X equals zero. There's that. And yes, there can be points on the horizontal assam tote. And then the whiner cept is put X equals zero in there. And what do we get? We get zero again. So what this looks like is, oops, we are going to that market, we're going to get it to look like this because it has to go through and approach and what's happening on these sides. So this is three right here. Let's see what's our value equal when I put a foreign there. So I have a four over 16 plus, four is 20 so we have 4/8 or one half. So, because that's positive, we have about a one half right here, if that was one. So we know this looks like that. Now I'm gonna put in something over here, let's do X equals negative five. So, I end up with negative 5/25 minus five, negative 5/8. So that is going to be when I go right here, negative 5/8 is down there. So that's where a graph goes and it approaches the ascent oats and that's it.


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