In this video, we're going to focus on finding the horizontal and vertical asmtote of a rational function. In addition, we're also going to find the slant or oblique asmtote if it's there. So, let's begin. Let's go over our first function. Y is equal to 1 / xus 3. To find the vertical asmtote set the denominator equal to zero. So if we set x - 3 equal to zero and if we solve for x we can see that the equation of the vertical asmtote is x is equal to 3. That's all you need to do to find the vertical asmtote. To find the horizontal asmtote you need to see if the function is top heavy the same or bottom heavy. This function is bottom heavy. The degree of the denominator is higher than that of the numerator. Notice that the degree of the denominator is to the first power. We don't have any x variables on top. So we say it's bottom heavy. Whenever the degree of the denominator is higher than that of the numerator, the horizontal asmtote is simply zero. It's y is equal to z. What you really need to do is figure out the horizontal asmtote. You need to find the value of y as x approaches infinity. If you plug in a very large number for x, let's say a,000, y is going to approach zero. 1 over a,00 minus 3 is very close to zero. It might be like 01 approximately, but that's close to zero. So let's go ahead and graph the equation. So the vertical asmtote is x= 3 and the horizontal asmtote is y is equal to zero which means that it's the x axis. Now it might be good to plug in a few test points on the right side. The graph can be like this or like this to find out what it is for sure. Plug in a number for x that's greater than the vertical asmtote. So if we plug in four 1 over 4 - 3 is 1. So we have the point 4a 1 which is over here. Therefore the graph looks like this. Now for the left side the function could be like this or it can be like that. So let's plug in a test point to find out if it's above or below the horizontal asmtote. So let's plug in two for x. It's going to be 1 over 1 which is negative 1. So we have the point 2,1. So it's over here which means that it's below the horizontal asmtote. And so that is the graph for this function. By the way, what is the domain and range for the function? The domain represents all the possible values that x can be. The only value that x can't be is the vertical asmtote. It's three. So whenever you want to write the domain of a function, simply remove the vertical asmtote. The lowest x value is negative infinity. The highest is infinity. So therefore the domain is going to be negative infinity to 3 union 3 to infinity. It includes everything except three. Now what about the range? Whenever you want to write the range of a function simply remove the horizontal asmtote. Notice that the lowest y value is negative infinity and the highest is infinity. But the function never touches the x-axis where y is equal to zero. So we only need to remove zero from the range. So it's going to be negative infinity to 0 union 0 to infinity. Now what about the end behavior of this function? What happens as x approaches positive infinity? As the function approaches positive infinity, notice that the y value becomes zero. So y approaches zero. That's the right end behavior. For the left end behavior, as x approaches negative infinity, it still approaches the horizontal asmtote where y is still zero. So the end behavior of a function is usually the horizontal asmtote. So now let's move on to our next example. Consider this function. y is equal to 1 /x + 2 + 7. So what is the vertical asmtote for this function? So to find it just like before we're going to set the denominator equal to zero. So therefore the vertical asmtote if you subtract two from both sides you'll see that it's x is equal to -2. Now what about the horizontal asmtote? The horizontal asmtote for this portion of the function is y is equal to zero because we still have a rational function where it's bottom heavy. The degree of the denominator is higher than that of the numerator. But notice that we also have a plus seven on the outside. So the horizontal asmtote is going to be 0 + 7. It's shifted up seven units. So it's y is equal to 7. So the graph is going to look very similar to the one that we had in the last example. So we have a vertical asmtote at -2 and a horizontal asmtote. It's very high up. It's at seven. Now let's plug in a few test points to see where the graph is going to be. So the vertical asmtote is at -2. So let's plug in a number to the right of -2 to see if it's going to be above or below the horizontal asmtote. So let's substitute 1 for x. So 1 over -1 + 2 + 7 - 1 + 2 is 1 1 + 7 is 8. So we have the point -1a 8 which it looks like I plot the horizontal asmtote at six. So I need to fix that. It should be at seven. But we still have the point8 which is somewhere over here. Therefore we know that the curve is going to be above the horizontal asmtote on the right side. So it's going to look like that. Now let's plug in a number to the left of the vertical asmtote. So let's replace x with -3. So -3 + 2 that's -1. 1 + 7 is -6. So -3 -6 is the point that we have and that's below the horizontal asmtote. So the graph has to be in this region. So that's how you can graph a rational function. So what is the domain and range for the function? So keep in mind for the domain we simply need to remove the vertical asmtote. The lowest x value that we have isative infinity and the highest is infinity. So we just need to remove -2 from the domain. So x could be anything from -2 I mean fromative infinity to -2 union -2 to infinity. So that's the domain for this function. Now for the range all you need to do is remove the horizontal asmtote. The lowest y value that the function can be is negative infinity and the highest is infinity. However y can't be seven. It's not going to touch the horizontal asmtote. So therefore the range is going to beative infinity to 7 union 7 to infinity. Let's try another example. Consider the function y is equal to 6x - 18 / 2x + 4. So how can we find the horizontal and the vertical asmtote for this function? So notice that the degree of the numerator is the same as that of the denominator both is to the first power. So whenever you see that to find the horizontal asmtote ignore the 18 and the four because when x approaches infinity the 18 and 4 are insignificant. If you plug in a th00and for x it's going to be 6,000 - 18 / 2000 + 4 which is approximately 6,000 over 2000 which is approximately three. So therefore to quickly find the horizontal asmtote it's just going to be 6x / 2x which is 3. So when x becomes very large y approaches three. That's the end behavior of the function both the right side and the left side. So that's the horizontal asmtote for this function. It's y is equal to 3. Now to find the vertical asmtote, we need to set the denominator equal to zero. And let's solve for x. So 2x is equal to -4. If we subtract both sides by 4 and then if we divide both sides by 2, we can see that x is equal to -2. So that's the vertical asmtote for this function. And so let's graph it. By the way, how can we find the x intercept and the y intercept for this function? It might be helpful to do that before we graph it. So notice that we can factor out a six from the numerator. 6x / 6 is x. -18 / 6 is -3. And we can also take out a two. So we can see that because we have x + 2 in the bottom, the vertical asmtote is going to be -2. You could simply switch the sign from plus2 to minus2. Now notice that we have an xus 3 on top. That means that the x intercept is going to be three. To find the x intercept, you need to set the numerator equal to zero. So if you set 0 equal to 6 * x - 3, if you divide both sides by 6, 0 / 6 is 0. So therefore 0 = x - 3. If you add 3, x is 3. So that's the x intercept for the function. So we have the point 3 comma 0. Now let's also find the y intercept. To find the y intercept, replace x with zero and solve for y. So it's going to be 6 * 0 - 3 / 2 * 0 + 2. 6 * -3 is8 and 2 * 2 is 4 -8 over 4. We can make that -16 over 4 - 2 over 4 because -16 - 2 is8. 4 goes into 16 four times and 2 over 4 is basically a half or.5. So it's about4.5. So therefore we have the point 0 comma4.5. That's the y intercept for the function. So now let's graph it. So we have a vertical asmtote at -2 and the horizontal asmtote is y = 3. So now let's plot the intercepts. So the x intercept is 3 comma 0. Therefore we can see that the function is going to look like this. It's going to be below the horizontal asmtote on the right side. Now the y intercept is at 04.5. Maybe I should have plugged that in first before I graphed it. So this is a more accurate representation. Now we need to check the left side of the vertical asmtote. So let's plug in -3 into the equation to see if it's above or below the horizontal asmtote. So, it's going to be 6 * -3 - 3 over 2 * -3 + 2. 2 goes into 6 three times. -3 - 3, that's -6. -3 + 2, that's - 1. And 3 * 6 is 18. The two negatives will cancel. And so, it's going to be positive 18. So therefore, it's going to be way up here. Either case, we know that the graph is going to be in this section. So that's a rough sketch for our graph. Now, what do you think the domain and range of the function is going to be? How can we figure it out? Feel free to pause the video and see if you can get the answer. So for the domain all we need to do is remove the vertical asmtote -2. X could be anything except -2. So it's going to beative infinity to -2 union -2 to infinity. And for the range simply remove the horizontal asmtote. So the range for the function is going to beative infinity to 3 union 3 to infinity. Let's try this example. Y is equal to 2x^2 - 3x - 2 / x^2 + x - 6. So what do you think the horizontal and the vertical asmtote for this function is going to be? So to find the horizontal asmtote, let's look at the degree of the numerator and the denominator. Notice that it's the same. So therefore, we just need to look at the coefficients. We could ignore everything else. So it's going to be 2x^2 / 1x^2, which is approximately 2. And so the horizontal asmtote for this function is y is equal to 2. Now how can we find the vertical asmtote? What we need to do for this function is we need to factor the denominator and the numerator and we need to factor this expression completely. So let's focus on that at the moment. So how can we factor x^2 + x - 6? So notice that the leading coefficient is 1. In such a situation, what you need to do is find two numbers that multiply to -6 but add to the middle coefficient of 1. So what two numbers multiply to -6 but add to one? This is 3 and -2. 3 * -2 is -6, but 3 + -2 is pos 1. So the bottom part is simply going to be x + 3 * x - 2. Now what about the top part? 2x^2 - 3x - 2. How can we factor uh that particular tromial? So, it's going to be a little different. It's going to take more work because the leading coefficient is not one. It's a two. So, here's what you need to do. Multiply the leading coefficient and a constant term. 2 * -2 is -4. And then find two numbers that multiply to -4 but add to the middle term three. So this is -4 and 1. -4 * 1 is -4, but -4 + 1 is -3. So now we're going to rewrite the expression, but we're going to replace the middle term -3x with -4x + 1x. Notice that the value of the expression is still the same.4x + 1x is -3x. So the value hasn't changed. We simply change it into a form that will allow us to factor by grouping. To factor by grouping, factor out the GCF or the greatest common factor from the first two terms. So we can take out a 2x. I'm going to erase this and I put it back in later. If we take out a 2x, what's going to be inside? 2x^2 / 2x is x and -4x / 2x is -2. Now, for the last two terms, there's no GCF that we can take out. So, if you can't take out anything, take out a one. So, on the inside, it's just going to be x - 2. If these two terms are the same, then you're on the right track. So, we need to factor out x - 2. If we take out x - 2 from this term, what's left over is the 2x. And if we remove x - 2 from this term, what's left over is the plus one on the outside. And so that's how we can factor the numerator. So let's rewrite the equation. So y is equal to 2x + 1 * x - 2 / x + 3 * x - 2. So now that it's in factored form, we can now find the vertical asmtote. So what do you think the vertical asmtote is? Is it going to be 2, -3 or something else? Notice that we can cancel x - 2. So the vertical asmtote is not x= 2. In fact, x= 2, the part that's canled is a whole. The vertical asmtote is the part that remains in the surviving equation. So the VA is going to be x=3. You set x + 3 equal to zero. So now how can we graph the expression? So what we want to keep is the surviving equation which is the part that we have left over after we remove the hole. So we have the vertical asmtote at -3 and the horizontal asmtote is at y is equal to 2. Now the hole is on the right side of the vertical asmtote. So, we really don't need the x and y intercept for this function. Let's find the y-coordinate of the hole. So, you want to plug in two into the surviving equation. So, it's 2 * 2 + 1 over 2 + 3. 2 * 2 is 4 + 1 is 5. 2 + 3 is 5. 5 over 5 is 1. So, it's going to be uh 2, 1. So, it's somewhere over here. So the hole is also known as a point discontinuity. The vertical asmtote is an infinite discontinuity. Now we need to plug in a point less than or to the left side of the vertical asmtote. Since the vertical asmtote is at -3, let's uh plug in4. So 2 * -4 + 1 over -4 + 3. 2 *4 is8 + 1 that's -7. -4 + 3 is -1 which is pos 7. So we have the point -4 pos 7 which is probably going to be way up here. So at least we know the graph looks like this. Now, what is the domain and the range for this function? Be careful now because we do have a hole and you do have to take that into consideration. So, x could be anything fromative infinity to infinity except for two things. We need to remove the vertical asmtote because x cannot be -3. The curve never touches the vertical asmtote. And also we need to remove the x coordinate of the hole which is two because x will never be two since it's a point discontinuity. Therefore the domain for this function isative infinity to -3 since -3 comes before 2 and then union everything between -3 to pos2 and then union 2 to infinity. So we just need to remove the vertical asmtote and the x coordinate of the hole for the domain. Now for the range of the function y could be anything from negative infinity to positive infinity but we need to remove the horizontal asmtote and we need to remove the y-coordinate of the hole because y can't be one nor can it be two. So therefore the range is going to be negative infinity to one which is the y-coordinate of the hole union 1 to 2 where 2 is the horizontal asmtote union 2 to infinity. So this is going to be our final example for today. y is = 2x^2 - x + 1 / x - 2. So go ahead and find all the asmtotes for this function. So to find the vertical asmtote, we simply need to set the denominator x - 2 equal to zero. Therefore x will equal + 2. Now what about the horizontal asmtote? Notice that the degree of the numerator exceeds the degree of the denominator. So it's topheavy which means that there is no horizontal asmtote for this graph. Now because the numerator exceeds the denominator by one degree we do have a slant or an oblique asmtote. So in order to find that we need to use long division. 2x^2 / x is 2x. And then we're going to multiply. 2x * x is 2x^2 and 2x * -2 is -4x. And we have to subtract. 2x^2 - 2x^2 is 0. -x - -4x that's the same asx + 4x which is positive 3x and you can bring down the one. So now let's divide 3x / x is 3. 3 * x is 3x and 3 * -2 is -6. So we need to subtract 3x - 3x. Well, that cancels to zero. 1 - -6 is the same as 1 + 6. So the remainder is 7. Now we don't need to worry about the remainder. This is what we need. The equation of the slant or the oblique asmtote is y = 2x + 3. Now the next thing we need to do is we need to factor the numerator to see if x - 2 is a whole. We really didn't check that but we should because if it's a whole then the vertical asmtote won't be x um equals 2. So let's double check. So we have a tromial where the leading coefficient is 2. So we need to multiply the leading coefficient by the constant term 2 * 1 is 2. Now what two numbers multiply to two but add to the middle um coefficient which is negative 1. The only numbers that multiply to two is -1 and -2 or 1 and two. None of these add up to negative 1. So, we can't factor it, which means that x - 2 is likely um to not be a factor. So, let's go ahead and graph the equation. So we have a vertical asmtote at two and a slant asmtote at three. So this is a linear equation. It's in a form y is equal to mx + b. m is the slope. B is the y intercept. So the y intercept is three. So the graph is going to touch the y- ais at three and the slope is two which is the rise over run. So it's basically two over one. As you move one to the right, you need to go up two. One to the right, up two. Or you can go backwards. One to the left, down two. And so we can graph a rough sketch of our slant asmmptot. Now we need to plug in two points one to the left of the vertical asmtote and one to the right to see where the graph is going to be. So let's plug in zero for x because that's going to be the easiest. If we plug in 0, it's going to be 2 * 0^ 2 - 0 + 1 over 0 - 2, which is 1 over -2. So we have the y intercept, which is 02 or 0.5, which is somewhere in this area. So therefore, we could say that the graph is something it's going to look something like this. It's not going to be in this area. It's going to be below the SL asmtote. Now for the right side, let's plug in 3 into the equation. So 2 * 3^ 2 - 3 + 1 / 3 - 2. 3^2 is 9 and 2 * 9 is 18. -3 + 1 is -2. 3 - 2 is 1. So it's 16. So we have the point 316 which is probably way up here somewhere which tells us that the graph is going to be in this section and won't be in this section. So that's the rough sketch of our graph. So that is it for this video and thanks for watching and have a great day.