in this video we're going to focus on graphing rational functions we're going to talk about how to graph the ASM tootes horizontal vertical even the slant and oblique ASM tootes how to find the holes and also how to write the domain range of the function so let's go ahead and begin let's start with the parent function yal 1 /x now this particular function has a vertical ASM toote at X Z which is on the Y AIS now it also has a horizontal ASM toote of y equals z which is the x-axis now this function the graph exists in the upper right corner and the lower left corner so it looks like this the domain for this function the domain represents the X values whenever you have whenever you need to write the domain of a function particularly a rational function remove the vertical asmp tootes and the x coordinate of any holes that you have so all the way to the left the lowest x value is negative Infinity the highest is positive infinity and X cannot equals z so the domain is from negative Infinity to 0 Union 0 to Infinity now for the range we need to remove the horizontal ASM toote and the Y value of any holes that we have which we don't have in this graph the lowest y value is negative infinity and the highest is positive infinity and Y cannot equal zero because that's the horizontal ASM toote so the range is going to be negative Infinity to0 Union 0 to Infinity so keep in mind the domain represents all of the Poss i x values that the function can have the range represents the possible y values that the function can have now what about the function y = 1 /x^2 what's the parent function for this how does it look like the vertical ASM toote is going to be the same anytime you have a function that is bottom heavy where the degree of the denominator is higher than the degree of the numerator the horizontal ASM toote will always be Y is equal to Z to find the vertical ASM toote set the denominate equal to zero so it's X is equal to zero so that's the vertical ASM toote and here's the horizontal ASM toote now the graph 1x^2 looks similar to 1/x the right side appears to be the same on the left side for 1/x the graph was here but whenever you square a number it can't be negative therefore it's going to be positive or above the x- axis so it's going to be in this uh vicinity and so that's the graph of 1/x^2 as you can see there's symmetry about the Y AIS for 1 /x there was symmetry about the origin now what about the domain and range of the function so let's start with the domain the lowest x value all the way to the left is negative infinity and the highest is positive infinity and X cannot equal zero the curve never touches the vertical ASM toote so the domain will be the same it's negative Infinity to 0 Union 0 to Infinity now what about the range because it's different this case notice that the lowest y value is not not negative Infinity but zero the graph starts from the horizontal asmt the highest y value is positive Infinity therefore the range is from zero to Infinity now let's try this one let's say f ofx is equal to 1 / X - 1 how can we graph this particular function well the first thing you want to do is find the vertical and the horizont Al Asm tootes now there's two ways you can find a vertical ASM toote if you compare it to the original function you could see that it's been shifted one unit to the right or what you can do is set the denom equal to zero and solve for x if you add one to both sides you can see that the vertical ASM toote is now at xal 1 or it's been shifted one unit to the right so that's the VA now what about the horizontal ASM toote so the function is still bottom heavy the degree of the denominator which is to the first power is greater than the degree of the numerator which since there's no X values the degree is zero anytime it's bottom heavy the horizontal ASM toope is y is equal to Zer if there's nothing else added to this rational function so right now we have enough information to graph the function so let's go ahead and do that so the vertical ASM toote is at xal 1 so here it is and the horizontal ASM toote is still on the x axis it's y equal 0 and we know that this particular function is going not sure what happened there it's going to look like this so that's a rough sketch of the graph that we have now if you have a picky teacher and you want to draw a graph accurately you may want to get some points so let's do that for this example so I'm going to alternate between an accurate graph and a rough sketch so what points can we plug in let's plug in two when X is two y Y is equal to 1 2 - 1 is 1 1/ 1 is 1 when X is equal to 3 Y is going to be 1 over two or 12 now if we continue to increase the X values we're going to get smaller fractions so we don't need to do that anymore now let's say if x is 1.5 what is the value of y for those of you who might be wondering why did I chose 1.5 I wanted to pick a number somewhere between 1 and two to see how the graph behaves when it's very close to the vertical ASM toote by choosing 1.5 notice that 1.5 minus 1 is .5 and 1 over5 is a nice number it's two and so that's why I chose it choose numbers that will give you a nice whole number for the value of y so let's plug in the numbers that we have so we have the 21 we have the3 comma 12 which is approximately there and 1.5 comma 2 and I think that's good enough now if you want to you can plug in 1.25 because 1.25 - 1 is 0.25 and 1 over 0.25 is 4 so that's a nice number so it should be over here so now we can get a very good graph now what about the left side chances are the left side is going to look like the right side but let's pick a few values that we can use to draw an accurate sketch so what values should we choose to the left of the vertical asmt the first value I will choose is z 1 0 - 1 is 1 over1 which is simply 1 so that's a nice point to pick now I'm going to choose negative 1 this is going to be 1 / -2 so that's negative2 so if I choose -2 I'm going to get a smaller fraction so I don't want that so I'm going to choose a number that's between zero and the vertical asmol 1 so let's choose negative2 I mean positive2 12 - 1 is2 or5 and 1 /5 is -2 now I'm going to choose not 1/4 but 34 I want to get closer to the vertical ASM 34 is about 75 75- 1 is .25 and we know that 1.25 is4 so we have the first point 011 12 12 and -2 34s and4 so this graph looks something like this so now what about the domain and a range of the function so let's start with the domain the lowest x value on the left side is negative Infinity the vertical ASM toote is x = 1 and the highest x value on the right side is positive Infinity whenever you write in the domain in interval notation always look at the X values from left to right so X cannot equal 1 since that's the vertical asmol so it's going to be from negative Infinity to1 union1 to Infinity now what about the range so now let's look at the Y values starting from the bottom to the top the lowest y value is negative Infinity the horizontal ASM toote is y equal 0 so the curve never touches the horizontal ASM toote in this particular graph and the highest y value is infinity so the range is from negative Infinity to 0 Union 0 to Infinity so now it's your turn try this one let's say Y is equal to 1 / x + 2 - 3 how would you graph this particular function well first let's begin by finding the vertical ASM toote to do that set the denominator equal to 0 so if x + 2 is equal 0 x = to -2 so we can see that it shifted two units to the left so that's the horizontal shift are there any vertical shifts what is the horizontal ASM toote now the function is bottom heavy so if we didn't have the3 the horizontal ASM toote would be Y is equal to Z but the horizontal ASM toote has been shifted down three so this would be Zer and then plus -3 is3 so that's the new horizontal ASM toote so now we can graph it now let's plot the asmp tootes so we have a vertical ASM toote at -2 and a horizontal ASM toote at -3 so if we just want to draw a rough sketch we know that oh wow that's terrible it's going to be something like this and then here's the other one so that's a rough sketch of the graph but let's get a few points at least one or two so we don't need actually we do need that let's get rid of this now let's plug in a point to the right of the vertical ASM toote let's try negative one if we replace x with1 what point will we get -1 + 2 is 1 1 over 1 is 1 1 - 3 is -2 so we have the point1 -2 which is in this region now let's try something closer to the vertical asmt let's say somewhere between1 and -2 so let's say 1.5 - 1.5 + 2 is positive .5 1 /5 is 2 2 - 3 is 1 so that's going to be somewhere over here now if we plug in zero it's going to be 12 minus 3 or5 - 3 which is uh -2.5 so this graph looks something like this which means this one is probably going to look similar to it now let's find the domain and range of the function so starting with the domain we can see that the lowest x value is negative infinity and X cannot equal the horizontal ASM toote -2 and the highest x value is infinity so it's going to be Infinity to -2 Union -2 to Infinity now let's move on to the range let's focus on the Y values the lowest y value is netive infinity and the horizontal ASM toote is at -3 so y cannot equal 3 the highest y value is infinity so it's going to be Infinity to -3 Union -3 to Infinity now we said that the parent function of the graph of 1x looks like this it's symmetric about the origin and apparent function of 1 / x^2 is symmetric about the Y AIS and so it looks like this so now what about the graph --1x and also -1 /x^2 if you put a negative sign in front of a function it's going to reflect over the x-axis so this particular curve it's above the xaxis now it's going to be below the x-axis and here it's below it now it's going to reflect above the x-axis now in this case both curves are above the x- axis so if we put a negative sign in front of the function they're going to both be below the x-axis so anytime you see a negative sign in front of the function it reflects over the xaxis try this one let's say Y is = -1 over x + 2 + 3 so the vertical ASM toote we can see it's been shifted two units to the left if we set the denominator equal to Z x is equal to -2 and for the horizontal ASM toote this portion alone is zero because it's bottom heavy and then plus three it's been shifted three units up so the horizontal ASM toote is yal 3 and because of the negative sign it's reflected over the x- axis so let's draw a rough sketch for this graph so here is the vertical ASM toote at x = -2 and the horizontal ASM toot is that y equal 3 now if the negative sign wasn't there the graph would be in these two quadrants but because the negative sign is there it's not going to be in this region it's going to reflect over the horizontal axis I mean the horizontal ASM toote so it's going to look something like this and it's going to be in this region too if you're not sure which region the graph is located plugin two two points plug in an x value that is to the right of the vertical ASM toote and one to the left you want to see if it's going to be above or below the horizontal ASM toote so let's plug in one point to the right of the vertical ASM toote which is1 and one point to the left which is -3 so if we plug in 1 this is going to be -1 + 2 which is posit 1 1 / POS 1 is1 + 3 that's 2 so we have the point-1 pos2 which is below the horizontal ASM toote so to draw a rough sketch is going to look something like that and now let's plug in -3 -3 + 2 is -1 -1 / 1 is 1 1 + 3 is 4 so at -3 the Y value is four so it's going to look something like that now let's find the domain and range so to find the domain let's analyze the graph from the left side to the right side on the left side the lowest x value is negative infinity and then we have the vertical Asal at -2 and the highest x value is positive Infinity so it's negative Infinity to -2 Union -2 to Infinity so the domain will always have the x value of the vertical ASM toote for rational functions the range is always going to contain the Y value of the horizontal ASM toote so for the horizontal asmt I mean for the range let's analyze the graph from the bottom to the top so the lowest y value is negative Infinity we have a horizontal asmode at y equal 3 and the highest y value is positive Infinity so it's negative Infinity to positive 3 Union pos3 to infinity and so that's the range of the function now what about this one 1 overx - 2^ 2 + 1 so let's focus on drawing a rough sketch so let's start with the vertical ASM toote if we set x - 2 = 0 we can see that that X will equal pos2 and now for the horizontal ASM toote this function is bottom heavy so it's zero but there's a plus one on the outside so the horizontal ASM toote is y is equal to+ one so now let's sketch the graph so notice that we have a square which means that this going to be symmetry on the left side and on the right side of the vertical asmt so it can be above the horizontal asmt like this or it can be below it like that now we don't have a positive one I mean we do have a positive one we don't have a negative one so because it's positive it's going to be above the horizontal asmt toe it's not going to reflect over the xaxis so therefore the graph looks something like this and this is just a rough sketch so now let's write the domain and range of the function so let's analyze the graph from the left side to the right side the lowest x value is negative infinity and then X cannot equal positive2 since that's the vertical astil and the highest x value is infinity so the domain is from negative Infinity to 2 Union 2 to infinity and now for the range let's analyze it from the bottom to the top the lowest y value starts at the horizontal ASM toote of one the highest is positive Infinity so it's from one to Infinity so what about this one -1 / x + 1^ 2 + 2 so let's start with a vertical ASM toote if we set x + 1 = to 0 x will equal 1 and for the horizontal ASM toote we have a function that's bottom heavy so that's zero and it's been shifted two units up so plus two now we do have a negative sign so it's going to reflect over the horizontal asmt so it's going to be below the horizontal asmt to now it's still squared we have an even exponent which means that this going to be symmetry about the Y AIS and not about the origin when I mean the origin the intersection between the vertical ASM and the horizontal Asm because the new origin that's where it's located now so here is the vertical ASM toote and here is the horizontal asmal at Y = 2 now because of the negative sign there's going to be symmetry across typically the y- AIS but now it's going to be across the vertical ASM toote and because of the negative sign it's going to be below the horizontal asot so it's going to look like this so because of the negative sign it's below the horizontal ASM toote and because of the square it's symmetric about the vertical cl to just want to clarify that in case I um mixed up my words now let's focus on the domain and the range the lowest x value is negative infinity and there's a discontinuity at1 and the highest x value is infinity so the domain is going to be negative Infinity to1 union1 to Infinity now for the range the lowest y value is negative infinity and the highest is at the horizontal ASM of two but it does not include two so it's from negative Infinity to two if it doesn't include two use parenthesis if it includes it which is probably not going to be the case for this video um then you would use brackets for infinity symbols you should always use parentheses never use a bracket for an infinity symbol now let's spend a few minutes talking about how to find a horizontal ASM toote for different functions we've considered functions that are bottom heavy such that the degree of the denominator is greater than that of the numerator and in situations like this the horizontal ASM toote is yal z now of course if you have a bottom heavy function and if there's a constant outside of it then this is going to change the horizontal ASM toote in this case the horizontal ASM toote is going to be yal 5 now what if we have a function that is not bottom heavy let's say if the degree of the numerator is the same as that of the denominator for example let's say if we have a 2X minus 4 / x + 3 the degree of the numerator is 1 the degree of the denominator is 1 whenever the top and the bottom have the same degree what you need to do is divide the coefficient the numbers front of the terms with the highest degree so 2 / 1 is 2 so the horizontal ASM toote is yal 2 now keep in mind if there's a constant outside of that it's going to shift up three units so it's going to be 2 + 3 and now the horizontal ASM toote is five so knowing that we go ahead and find the horizontal ASM toote for the following functions so for the first one it's going to be 3x^2 over 1x^2 or simply 3 over 1 which is 3 now for the second one we have A8 in front of the X2 and a 48 ID 4 is -2 but we need to add seven to it so the horizontal ASM toote is five here's another one you could try let's say if it's 3 * x + 1 * 2x - 2 / x - 3 * x + 2 + 4 so what's the horizontal ASM toote in this case now if if we foil the denominator x - 3 * x + 2 the first term is going to be x^2 plus we'll have some other stuff now on the top if we foil it this is going to be x * 2x which is 2x^2 * 3 the leading coefficient will be 6 so it's going to be 6x^2 plus a few other stuff so to find the horizontal asmt focus on the coefficients of the leading terms 6 / 1 is 6 + 4 it's going to equal 10 in this case so that's what you'll do if the degree of the numerator is the same as the degree of the denominator now what if the degree of the numerator is greater than that of the denominator let's say if we have this problem if if it's two units or more then there's no horizontal ASM tootes there's no other ASM tootes that you have to worry about but now let's say if the degree of the numerator exceeds the denominator only by one in that case you have a slant ASM toote to find a slant ASM toote you need to perform long division so I'll go over that later in the video and whatever ever function you get here that's going to be the equation of the slant or the oblique ASM toote so let's try this example let's say if Y is = to 2 * x - 2 / xus 3 go ahead and graph this function so let's find a vertical ASM toote so if we set x - 3 equal to 0 we can see that the vertical ASM toote is X is equal to 3 now for the horizontal ASM toote notice that the degree of the numerator is the same as that of the denominator so if we focus on the leading coefficients which is 2X and 1X 2 / 1 is equal to 2 so the horizontal ASM toote is y is equal to 2 now since we have an X variable in the numerator it might be advantageous to find the x intercept to find the x intercept you need to replace y with zero if the numerator is equal to zero the entire fraction will be equal to zero 0 divid anything is z so by setting the numerator equal to zero Y is zero and therefore you can get the X intercepts so we can set the factor x - 2 equal to Z so the X intercepts are two so we have the point 2 comma 0 so we have a vertical ASM toote at x = 3 and a horizontal asmt at Y = 2 so now let's plot it so we have the point2 which is here and we know this graph is going to look something like this and this graph is very similar to 1/x but there are some differences and this is going to look like this it's going to follow the ASM tootes now we can plug in another point if we want just to make the second curve a little bit more accurate let's replace x with four 4 - 3 is equal to 1 and 4 - 2 is 2 2 * 2 is 4 so we have the point 4 comma 4 so when X is four y will be equal to four if we plug in five 5 - 3 is 2 these cancel and 5 - 2 is 3 so when X is 5 Y is 3 so this graph looks something like this there's not much space to draw here but you get the point now what are the domains and the range of this particular function so starting with the domain the lowest x value is negative Infinity X cannot equal 3 and the highest is infinity so it's from negative Infinity to 3 Union 3 to infinity and for the range the lowest y value is negative infinity and there's a horizontal ASM toote at two and it goes to positive Infinity so the range is going to be negative Infinity to 2 Union 2 to Infinity try this one let's say that f X is equal to 3x^2 + 9 x - 12 / x^2 + x - 2 - 4 so how can we graph this particular rational function the first thing you should do if you get an expression that looks like this is you want to factor everything in the numerator notice that 3 9 and -12 are all divisible by 3 so let's take out the GCF first so this is what we're going to get and on the bottom we can Factor x^2 + x - 2 we have a trinomial where the leading coefficient is one what two numbers multiply to -2 but add to the middle coefficient of one that's going to be pos2 and 1 2 * 1 is -2 but 2 +1 is POS 1 so on the bottom we can Factor as x + 2 * x -1 now let's Factor this expression x^2 + 3x - 4 what two numbers multiply to -4 but add a positive three this is going to be positive4 and negative 1 so this is going to be x + 4 on top * x -1 and on the bottom x + 2 * X - one and let's not forget the4 so now let's start with the horizontal ASM toote what is the horizontal ASM toote of the function so looking at the original expression the leading coefficients of the highest terms are three and 1 3 / 1 is 3 and then - 4 that's going to be- 1 so the horizontal ASM toote is yal to1 now what about the vertical ASM toote how many vertical ASM tootes do we have here is it one two or none so we have a vertical ASM toote at -2 so therefore let's write it as X is equal to -2 but we do not have a vertical asmt at xal 1 notice that these two factors they cancel and when it cancels it's not a vertical ASM toot rather it's a hole it's still a point of discontinuity but it's not an infinite discontinuity it's not a vertical ASM to so the x value of the hole is at one to find the Y value of the hole plug it in into the surviving equation the surviving equation after you cancel it it's 3 * x + 4 / x + 2 - 4 we need the Y value of the whole because we need to take it out in the range in the domain we need to remove the x value of the hole so let's go ahead and plug it in so this is going to be 1 + 4 over 1 + 2 - 4 1 + 4 is 5 and 1 + 2 is 3 we can cancel the threes and so we let left with uh 5 - 4 which is 1 so the whole is at 1 comma 1 so now we have everything that we need let's get rid of this let's put this over here now there's something else that we can find and it is the x intercept so if you set the numerator equal to zero or basically X+ 4 equal to zero that's going to give you the x intercept so the x intercept we have it as -4 comma 0 we can't use the whole at xus one that's something else if you need to find a y intercept replace x with zero so it's going to be three * 0 + 4 these two cancel so we don't have to worry about it and then 0 + 2 - 4 so it's 3 * 4 / 2 - 4 3 * 4 is 12 12 divid 2 is 6 6 - 4 is 2 so the Y intercept is 0 comma 2 so now let's go ahead and graph it now there's one thing that we need to correct and that is the x intercept the x intercept would be -4 if this number wasn't here but since that number is there it's going to be a little different so what we're going to have to do is replace F ofx which is y with zero so zero is equal to 3 * x + 4 / x + 2 - 4 so if this4 wasn't here it would be A4 but because it's there it's not worth finding the x intercept it's too much work so let's just get rid of it so let's go ahead and plot it so we have a vertical ASM toote at -2 and we have a horizontal ASM toote at1 we have a hole at 1 comma 1 and we have a y intercept at 02 which is right there so we can see that this particular function looks like this now what about the left side let's plug in one point to the left of the vertical ASM toote so let's choose neg3 if we plug A3 into the surviving equation -3 + 4 is 1 -3 + 2 is 1 so this is -3 - 4 this is going to be all the way down at -7 so it's like way down here but you can see that the graph is going to look like this and so this is a rough sketch of the graph that we have so now what are the range and the domain of this function so let's start with the domain let's focus on the X values now let's analyze it from left to right the lowest x value is negative Infinity we have a vertical ASM toote at -2 so we got to take out -2 now the hole that's important too take out the x value of the hole the x value is positive one and the highest x value is in Infinity so it's going to be Infinity to -2 Union -2 to 1 Union 1 to Infinity so it's going to look like that now let's focus on the Range so let's analyze the graph from the bottom to the top the lowest y value is negative Infinity the horizontal asmo is1 so we got to take that out the Y value of the hole is positive one and the highest yvalue is infinity so it's going to be from negative Infinity to negative 1 Union -1 to 1 Union 1 to infinity and so that's how you can write the domain and range of a rational function whenever you have a horizontal ASM toote A vertical ASM toote and even a hole now let's work on an example that contains a slant or an oblique ASM toote so we have 2x^2 + 6x - 8 / x - 2 so notice that the degree of the numerator is greater than the degree of the denominator so there is no horizontal ASM toote but since the degree of the numerator is one unit higher than that of the denominator there is a slant ASM toote so let's use long division to find the SL ASM toote so we're going to put the uh numerator on the inside so let's divide 2x^2 / X is equal to 2X and now let's subtract actually before we subtract let's multiply 2x * X is 2x^2 and 2x * -2 is4 X so now let's subtract 2x^2 - 2x^2 is equal to 0 6x - -4x is the same as 6X + 4x so that's 10 x so now let's divide 10 x by x 10 x / X is simply 10 once you have a constant you don't need to finish the long division process you can stop here this is the equation of the slant ASM toote we don't need to finish the long division so I'm going to write sa for slant ASM toote so that's Y is equal to 2x + 10 so now let's go ahead and graph the function actually before we do that let's Factor the numerator let's take out a two so this is going to be x^2 + 3x - 4 and let's Factor x^2 + 3x - 4 two numbers that multiply to4 but add to three are pos4 and 1 so we can see that the vertical Asm toote is based on this factor and it's xal 2 now we don't have a number here so if we replace y with zero these two will give us the X intercepts so the X intercepts in this case are -4 comma 0 and 1 comma 0 we're going to need to go up to 10 so let's plot the vertical ASM toote which is at X is equal to 2 now to plot the select ASM toote it's all the way at 2x + 10 so what I'm going to do is instead of going by ones I'm going to go by twoos so every Mark is going to represent two so this is eight so here's the vertical ASM toote it's now at xals 2 so now to plot the SL ASM toote 2x + 10 the Y intercept is at 10 so if every Mark represents 2 10 is over here somewhere now the slope is two so every time we move one unit to the right we need to go up two units so the next point is going to be 1 comma 12 and then 2 comma 14 so if we go backwards we're going to be at1 comma 8 and then -2 comma 6 so here is the slent ASM toote so keep in mind this is just a rough sketch so now there's four regions in in which the graph can lie it can be here it can be here here or here so let's plug in a point to the right of one so we have let's use a three if we plug in three it's going to be 2 * 3 + 4 and then * 3 - 1 over 3 - 2 3 + 4 is 7 3 - 1 is 2 3 - 2 is 1 so that's going to be all the way up to uh 28 2 * 7 * 2 is 28 so we definitely know that it's going to be above the select ASM toote so the graph is going to be here now we have a lot of points to the left of the vertical ASM to those points are the X intercepts which we have at one Z keep in mind this is two and this is NE -2 so -4 is right here so this is ne40 and let's get the x intercept let's use I mean let's use this to get the Y intercept if we replace x with zero we can see that it's going to be this is zero that's zero so on top we're just going to have8 and on the bottom this is z so 0 - 2 is -28 / -2 is 4 so the Y intercept is 0a 4 which is right here so in this region we have a graph that looks like this so that's just a rough sketch of what we have so what's the domain of this function now this Curve will continue going down and towards the left so the lowest x value is going to be negative Infinity we have a vertical Asm toote at two and then this part will continue going up and to the right so it's going to go towards uh positive Infinity so all we need to do is take out the the vertical ASM toote so it's negative Infinity to 2 Union 2 to Infinity there's no holes to remove now for the range the lowest y value we could see is negative infinity and the highest is positive Infinity now we need to find how high this particular curve goes and what's the lowest point of that curve now we have a rough sketch so you can't exactly see where it stops so you may need to use a graphing calculator at this point using the graphing calculator I believe the highest point here is about 4.2 and the lowest point I wrote it down it's about 23.8 using the graphing calculator so once you find those points then you can write the range so the range viewing it from the bottom to the top the lowest is negative infinity and it stops at 4.2 and then the curve it begins again at 23.8 and then it goes to Infinity so to find a range for a graph that looks like this you need a graphing calculator if you don't have one just go to Google type in online graphing calculator put in the function and then you could probably zoom out until you see the whole graph and then you should find these points now these are rounded values I rounded it to the nearest 10th so now you know how to find the domain and range of rational functions you know how to identify the horizontal ASM toote the vertical ASM toote the slent ASM toote and also any holes in a graph so basically that's it for this video thanks for watching and have a great day