So this video will show you how to deal with the rational function in general meaning that we are going to see how to find the vertical acetone, the horizontal acetone, the domain, the range, the x-intercept, the y-intercept, and then in the end we are going to find the graph for this rational function. So let's just go ahead and get to work. First off we see that there's nothing that we can reduce so this one I would say it's nothing that bad I would just say. Anyway go ahead get to the first one. For the vertical acetone We look at the denominator and we make it equal 0. So we get 8x minus 3. And of course, we put the equal 0. And solve for x.
You get x is equal to 3 over 8. And then that is it. So just say x is equal to 3 over 8. Next, we have the horizontal asymptote. So let's put the f of x as y. And then we have 2x plus 1 over 8x minus 3. And again, we care about this, care about that for the horizontal asymptote, right? the x cancel out and then 2 over 8 which is just 1 over 4 so that's it y is equal to 1 over 4. done deal cool now for the domain let's put it down right here again this right here we cannot reduce anything so nothing too tricky just go ahead and make the bottom not equal zero domain you have to state the restriction so you're going to say 8x minus 3 cannot be equal to zero well in that case we can see that x cannot be equal to 3 over 8 and based on this we can write our interval notation if you look at the graph of this real quick let's say here is 3 over 8 we have an open circle meaning that x can be any number but just not this right so x can be any number but just not that so look at this as the interval notation we go from negative infinity to this so let me write that down right here 3 over 8. And then be sure to use parentheses because you do not want to include 3 over 8. And then union the other part, which is again parentheses.
You go from 3 over 8 to infinity. yeah that's it and then you are supposed to write everything in one line but i cannot fit over there so that's that okay for the range this right here it's not bad but i would prefer to do that all the way at the end now let's look at e right let's look at e again usually do the green g at the end because that's the trickiest one so let's just have that habit for that for the x intercept we are going to make y which is this equals zero and solve it so we get zero is equal to 2x plus 1 over 8x minus 3. But you see, if the fraction is equal to 0, that means the top has to be 0. So we just have to look at 2x plus 1 is equal to 0. And then solve for x right here. We get x is equal to negative 1 over 2. So that's that. And because x intercept is a point, usually I like to write this as negative 1 over 2 comma 0. And again, this means the x value, and this means the y value. this is not the interval notation all right for the y intercept so let me put that down f for the y intercept we are going to just let x equal 0 and then plug in remember f of x is the y and then we have 2 times 0 and then plus 1 over 8 times 0 and then minus 3 so we see this is just negative 1 third right so that's pretty much the idea why it's equal to negative 1 third here And again, 1 intercept is a point, so we'll just write it as 0, negative 1 third.
Cool. Lastly, let's do a graph for this. And to do a sketch of the graph, this is what we're going to do. first off we have the horizontal we have the vertical as 3 over 8 so we are going to just put on a vertical dash line right here as 3 over 8 so let me just say this is 3 over 8 like so This is x equals 3 over 8. And we also have the horizontal isotope at 1 over 4. So let's say this right here is 1 over 4. y equals 1 over 4. Cool. Okay.
and we are also going to label the x and y intercept negative one half comma zero so let's say it's right here this is negative one half comma zero and we also have zero comma negative one third so let's say zero comma negative one third let's say this right here is negative one third cool and technically we should also be checking the sign of the function meaning that when does the qual go below the x-axis when does the graph go above the x-axis and things like that and after you finish all the graph by your hand you can just use a graphing calculator to double check and things like that let me just tell you guys how things are if X is less than this right negative one half you get positive it's actually like this just just follow along the graph will look like this Why? Because you see that I know the curve has to touch this and that point. And then when you go all the way to the left, the curve has to become flat.
And it has to go really close to 1 over 4. And when you have the curve right here, it has to go straight down. And if you pick a number... bigger than here you will see that the curve is actually going to be positive so let's go ahead and do that so it's actually going to be above this right here and again you can just double check with your graphing calculator now if If you look at this, once you have the graph, you see that x cannot be 3 over 8. We said that right here for the domain already, and for the range, you see y can be going from negative infinity to positive infinity except for this number.
The curve does not cross this number, namely 1 over 4. So for part. we have to make sure that y is not equal to 1 over 4. So let's go ahead and go back here, and then just say we have to go from negative infinity to 1 over 4, and we will have to put down a union, and then we go from 1 over 4 to infinity. So that's that.
Yeah, I just want to make sure that I write the interval notation on one line, so that looks much better. So, yeah, this is it. Here we are going to investigate this rational function.
We are going to find its vertical isotope, horizontal isotope, domain. range x intercept y intercept and then in the end we should have a sketch of the graph okay let's get to work with the verticals the first remember to check if we can reduce this or not on the top we can factor out a 3 that's good And then on the bottom, in fact, we can also factor this as x plus 2 times x plus 5, and we see we can actually cancel this and that. So the reduced version is 3 over x plus 5. For the vertical asymptote, let me just put down this right here.
We look at the reduced version and then look at the denominator, make it equal 0, and that's pretty much it. So x plus 5, make it equal 0, meaning x is equal to 0. equal to negative 5. Dan Dio. First answer right here. Very nice. Now, for the horizontal asymptote, you can take the reduced version or the original.
Doesn't matter. Let's do take the reduced version. Y is equal to 3 over x plus 5. Well, we see that on the top, we just have a number 3. And on the bottom, we do have x to the first power. If x goes to infinity, 3 over infinity is 0. So, So remember, whenever you have an x on the bottom right here, that will give you 0. So y is equal to 0. That will be the horizontal asymptote. Done.
Next, we are going to find the domain. Remember, remember, remember. For the domain of a rational function, you always look at the original before you reduce anything.
Because if the question was written like this, you have to prevent anything that will make the bottom equal 0. So look at that. which we have x squared plus 7x plus 10, go ahead and make it not equal 0, because we are trying to find the restriction. We can factor this, like what we did over there, which is x plus 2 times x plus 5, make it not equal 0, meaning that x cannot be equal to negative 2, x cannot be equal to negative 5. Okay, so these are the two numbers that we cannot use for the x. So, for the domain, let's go ahead and write down interval notation.
and a number line might help put down negative 5 right here and put down negative 2 right here we cannot have these two numbers but any other numbers are okay so the first piece is going to be from negative infinity to negative 5 do not include that union the other piece is negative 5 to negative 2 and then do not include that and last piece is negative 2 to infinity Okay, cool, so that will be what we have and perhaps let me just fix this a little bit cool Now for the range, let's save it until later and now let's look at the actual So I'm going to put that down right here. For the x-intercept, you have to look at the reduced version. You have to look at the reduced version.
So, if you look at this right here, you make it equal 0. 3 over x plus 5. Can we make that equal to 0? No, not possible. Because remember, the only chance for us to have a fraction to be 0 is when the top is equal to 0, but 3 is not equal to 0, right?
3 is not equal to 0. So this right here is not possible, meaning that we have no x-intercept. So I'll just say none. Okay, and now for the y-intercept, so this is going to be... Wait, this is supposed to be e, and here's f.
For the y-intercept, we are going to let x equal 0. Earlier, it was y equal 0, right? And then we are going to put 0 into the... This one doesn't matter.
You can put the original, or you can put it here. It doesn't matter, so let's put it here. y will be 3 over 0 plus 5, which is just 3 over 5. Okay, cool.
So here we have 0, 3 over 5. And now, let's take a look at the graph. Okay, be really careful with this. First off, let's label the vertical acetone, which is at x equals negative 5. So I'm just going to put it down right here. x is equal to negative 5. And we will also label the horizontal acetone, which is y is equal to 0. So let's put it down right here.
y is equal to 0. There's no x-intercept, so we don't put on any points on the x-axis. And we do have a y-intercept though, so 0,3 over 5. Let's say it's right here. This is 3 over 5. Good.
And if you take a look at the graph of this, you will get a picture like this on your graphing calculator. And you might be thinking like what's so tricky about this? This is indeed really tricky because we also have to account for The domain you see the domain says X cannot be negative 2 and X cannot be negative 5 so as you can see we already know x cannot be equal to negative 5 because we have a vertical asymptote.
However, right here it says x cannot be negative 2. So we will actually go to negative 2 that says right here, and you will have to go up, and you have to erase that point, and you have to put an open circle. And sometimes you will have to know the y value of the open circle. How do we do that?
Well, you have to utilize this. Do not plug in negative 2 into the original. Because if you do that, you get 0 over 0 because we have this factor. So, for this right here, we know this point. this right here is 3 over 5 because that's the y intercept but for this right here for the open circle we put negative 2 here so we get let me just write that down real quick we get y is equal to 3 over negative 2 plus 5 and if you take a look this is going to be 3 on the bottom and then 3 over 3 which is just 1 so this right here is at 1 the open circle is at negative 2 comma 1 and now why do we care about this it's because for the domain x cannot be negative 2 that means we have this open circle that tells you y cannot be 1 so i will come back here the notice from the graph we have this horizontal acetone the graph does not touch that so y cannot be zero and because of the open circle here y cannot be one so now let's put up into interval notation so zero right so we will just go from negative infinity to zero and the union 0 to 1 and then union and then 1 to infinity yeah just like that so this will be a picture your graphing calculator is not going to show you a little open circle so be really careful with the domain issue