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Current time:0:00Total duration:10:29

let's see if we can tackle a slightly more difficult hyperbola graphing problem so let's say I have the hyperbola make this up on the fly X minus 1 squared over 16 minus y plus 1 squared over let's say 4 is equal to 1 so the first thing to recognize is that this is a hyperbola and well in a few videos do a bunch of problems where the first point is just to identify what type of conic section we have and then the second step is to actually graph the conic section so here when I already told you that we're going to be doing a hyperbola problem so you know it's a hyperbola but the way to recognize that is you have this minus of the Y squared term and then we actually have shifted right the classic or the standard non shifted form of a hyperbola or a hyperbola centered at 0 would look something like this especially if it has I guess the same asymptotes just shifted but Center to 0 would look like this x squared over 16 minus y squared over 4 is equal to 1 and the difference between this hyperbola and this hyperbola the center of this hyperbola is at let's see the center is that the point X is equal to 1 Y is equal to minus 1 and the way to think about it is x equals once makes this whole term 0 and so that's why it's the center and y equal to minus 1 makes this whole term 0 and on here of course the center is the origin Center is 0 0 so the easy way to graph this is to really graph this one but you shift it so you're at you're at the you use the center being 1 minus 1 instead of the center being 0 0 so let's do that so let's figure out more about let's figure out the slope of the two asymptotes here and then we could we can shift those two slopes so that it's appropriate for this hyperbola right here so if we go with this one let's just solve for y that's what we that's what I always like to do whenever I'm I'm graphing a hyperbola so we get let's see minus y squared over 4 subtracting x squared over 16 from both sides minus x squared over 16 plus 1 I'm working on this hyperbola right here not this one and then I'm just going to shift it later and then let's say I multiply both sides by minus 4 and you get Y squared is equal to see the minus cancels out with that and then 4 over 16 is x squared over 4 minus 4 and so y is equal to plus or minus square root of x squared over 4 minus 4 and to figure out the asymptotes you just have to think about well what happens is X approaches positive or negative infinity as X gets really positive or X gets really negative and we've done this a bunch of times already and I think this is important this is more important than just memorizing the formula cause it gives you an intuition of where those equations for the lines of the asymptotes actually come from because these are what this graph or this equation or this function approaches as X approaches positive or negative infinity so as X approaches X approaches positive or negative infinity as X approaches positive in regular infinity what is y approximately equal to in this case well once again this term is going to dominate this is just a 4 right here you could imagine when X is like a trillion or a negative trillion this is going to be a huge number and this is going to be just like you know you can almost fuse like the round off error you take the square root of that and so this is going to dominate so as you approach positive or negative infinity Y is going to be approximately equal to the square root the positive and negative square root of x squared over 4 which so Y would be approximately equal to positive or negative what's the square root of this x over 2 or 1/2 X so let's do that so let's draw our asymptotes and remember these are the asymptotes for this situation but now of course we're centered at 1 negative 1 so I'm going to draw two lines with these slopes at with positive 1/2 and negative 1/2 slopes but they're going to be centered at this point because I'm using I just got rid of the shift just so I could figure out the asymptotes but of course this is the real thing that we're trying to graph so let me do that see so this is my x-axis my y-axis sorry this is my x-axis and the center of this is that 1 negative 1 1 negative 1 so X is equal to 1 Y is equal to minus 1 fair enough and then the slopes of the asymptotes were positive and negative 1/2 right positive and negative 1/2 so let's do the positive 1/2 so that means for every 2 you run over so if you if you go positive in the positive x-direction - you move up 1 so you go to the right - and up 1 so that's the first one let me draw that asymptote it looks something like that and then let me draw it from this point to that point then I have a steady hand ok there you know I get the point and then the other asymptote is going to have a minus 1/2 slope so if I take if I remember this is our Center 1 - 1 so if I go down 1 and over so when I go over 2 I go down 1 so that'll be right there let me draw those asymptotes like that asymptote it's going to look like that and then just to continue it in the other direction it's going to look and I want to make it the lines overlap it looks something like that so we've drawn our asymptotes for this function and now we have to figure out it's going to be a kind of a vertical hyperbola or a horizontal hyperbola now the easy way to think about it is is to try to make and we can do it two ways I mean if you just look at this equation right here when you're taking the positive square route we're always going to be slightly below the asymptote right the asymptote is this thing but we're always going to be slightly below it so that tells us that we're always going to be slightly below the asymptote on the positive square root and we're always going to be slightly above the asymptote on the negative square root because it's going to be a little less and it's negative but I'll let you think about that so I already tell you know my intuition is going to be there and there it's more than intuition I know that we're going to be a little bit less than the negative square root but I'll do it the other way I'll do it the way I did in the last video so the other way to think about it is what happens when this term is here for this term to be 0 X has to be equal to 1 and does that ever happen can X be equal to 1 if X is equal to 1 here this term is 0 and then you have a situation where see if X is equal to 1 this is 0 and then you have a minus y squared over 4 would have to equal 1 or this would have to be a negative number so X could not be equal to 1 so Y could be equal to negative 1 let's try that out if Y is equal to negative 1 let's try out that point if Y is equal to negative 1 this this term right here disappears right when Y is equal to negative 1 you're negative 1 negative 1 plus 1 that's 0 0 4 so when y is equal to negative 1 you're just left with I'll do it down here well I don't want you're left with X minus 1 squared over 16 is equal to 1 right I just cancelled out this term come saying what happens when y is equal to negative 1 you multiply both sides by 16 you get let me do it over here these get messy X minus 1 squared is equal to 16 take the square root of both sides X minus 1 is equal to positive or negative 4 and so if X is equal to positive 4 if you add 1 to that X would be equal to 5 right and then if X if X minus 1 would be minus 4 and you add 1 to that you would have X is equal to 3 so our two points are I guess you can kind of our two points closest to our Center are the points five comma negative one and three comma negative one and let's plot those two so five one two three four five negative one and three negative one is that right so oh no minus three so because X minus 1 could be minus four that's what happens when you skip steps and if you have minus one plus let me write that so X minus one is either equal to 4 or X minus one is equal to minus 4 if you have the minus four situation then X is equal to minus three so you go you go one two three minus three minus one so those are both points on this hyperbola and then our intuition was correct or at least what I said that we're always going to be below the positive the positive square root is always going to be slightly below the asymptote so we get our curve is going to look something like this it's going to get closer and closer and then here it's going to get closer and closer in that direction it keeps getting closer and closer to that asymptote and here it's going to keep getting closer and closer to the asymptote on that side and then on that side and of course these asymptotes keep going on forever and forever and if you want you could try out some other points just to confirm you could plot that point there or that plot point there just to confirm that that's the case the hard part really is is just to identify the asymptotes and just to figure out do we sit kind of on the left and the right or do we sit on the top and the bottom and then you're done you you can graph your hyperbola see in the next video