Graphing hyperbolas (old example)

in the last hyperbola video I didn't get a chance to do some concrete examples so I'll do that right now so let's say I had the hyperbola Y squared over 4 minus x squared over I don't know let me think of a good number let's say x squared over 9 is equal to 1 so the first thing to figure out about this hyperbola is what are its asymptotes and once again I always forget the formulas and I just try to solve for y and see what happens when X approaches positive or negative infinity so if you solve for y you can add x squared over 9 to both sides and you get Y squared over 4 is equal to x squared over 9 plus 1 now I can multiply 4 times both sides and you get Y squared is equal to 4 over 9 times x squared plus 4 right distribute the 4 take the positive and negative square root of both sides Y is equal to the plus or minus square root of 4 over 9x squared plus 4 and you can't really simplify this anymore but we can't think about what does this approach as X approaches positive or negative infinity so as X approaches plus or minus infinity what is this roughly equal what is this approximately what is the graph get a lot closer to well then Y is approximately equal to just the square root of this term because this becomes super huge and relative to this term this starts to matter less and less and less and that's why we get closer and closer to the asymptotes because when this number is like a trillion or a googol then this number is almost insignificant when you take the square root you're pretty much taking the square root of this and it'll you'll just be a little bit above the graph because you have this extra plus 4 there so as you approach positive or negative infinity this equation is approximately equal to the plus or minus square root of 4 over 9x squared and so that is so Y would be approximately equal to the plus or minus we can take the square root of this plus or minus square root of 4/9 is 2 over 3 right square root of 4 over square root of 9 times X so these are the asymptotes there's two lines here there's y is equal to 2/3 X and then there's y is equal to minus 2/3 X so let's draw those two lines let me draw my axes let's make that my Y axis so we cut the x axis let me switch some colors just to make things interesting so let me draw the first one see Y is equal to 2/3 X so you rise 2 for every 3 that you run so let me draw that so if this is 1 2 3 1 2 so that would be a point on the line and let me draw the line now it should go through the origin no that's not it let me draw it like this this way I can make sure it goes to the origin so it's going to go through like that and then I can go from here and then go like that so that's one asymptote and the other apps told is Y is going to be equal to minus 2/3 X right because plus or minus 2/3 X so minus 2/3 X you go down 2 for every 3 that you go out so that point will show up let's see if I do 1 2 so you get go down 2 for every 3 that you go out so it'll go there so if I draw that asymptote it looks something like go there go out there and then go from here go out there and we've drawn our asymptotes now the question is is it going to open up to the left on the right or up and down there's two ways we can think about it and I'll do it the way that might be more intuitive for you is can X what happens when X is equal to 0 well X is equal to 0 when X is equal to zero this disappears and we're just left with I'll do it here y squared over 4 is equal to 1 or Y squared is equal to 4 or Y is equal to plus or minus 2 so we know that the point 0 the points 0 plus or minus 2 is on this graph so X can be equal to 0 so 0 plus or minus 2 so 0 plus 2 is this point right here and 0 minus 2 is this point right there so that by itself actory is enough of a clue to know that it opens down here and up here because it will never a hyperbola will never cross the asymptote so it's not like it can go out here and cross this asymptote so we already know that the graph of this parabola you can try other points if you want just to verify it it's going to look something like this it's going to go and then good note I want to make it so it never touches so it's going to get really close I touched it it's going to get really close but never touch and then on this sides going to get really close but never touch and I don't want to touch it and then on the up on the top side is going to the same things get really close and as you approach infinity it's never going to touch it and if you get really close you get infinitely close but never touch it so that's what this parabola this hyperbola is going to look like and I did it by just trying to see if X could be equal to 0 and I encourage you to try what happens when y equals 0 and you you'll get no solution and that makes sense because this hyperbola never crosses y equals 0 right it never crosses the x axis and this should also be intuitive because if we saw here when we did this approximation is X approaches positive or negative infinity we saw that you know we we always did have this plus 4 sitting here but we said oh well as X gets super large or super you know super negative this starts to matter less and less but we will always be slightly larger than this number right especially in the positive quadrant right where R was going to be so the positive quatro is going to be slightly larger than the asymptote and even and E when we take the positive square root I guess it's the best way to say it when we take the positive square root will always be larger than either the asymptotes and likewise when you take the negative square root you're always going to be a little bit smaller than the easier to the asymptotes because this number is going to be a little bit bigger than this number then when you take the negative squared you're going to be a little bit small and that's why we're a little bit below I don't know which one's more intuitive for you maybe just the fun you know trying it when X is equal to zero and when y equals zero and see what points you get and say oh then I'm in the kind of a vertical hyperbola as opposed to a a horizontal one so let's see if I have time for I'll leave that video right there and then I'll do another video where I actually shift the hyperbola and shifting it at is actually no different than shifting an ellipse or a circle you just have you know Y minus something squared and X plus something or X minus something squared and that just tells you where you shift the origin this hyperbola of course is just centered at the origin anyway see you in the next video