Euclid 2001 8

hello hello hello ladies and gentlemen and welcome back to more of the grid 12 Euclid paper from the year 2001 put out by the University of Waterloo and we’re at question 8 and question eight nine ten these guys are all written every single part and so we’re gonna have to have our wits about us and question eight egg on the grid provided in the answer booklet sketch y equals x squared minus four so that that’s a parabola right there and y equals two times the absolute value of x now I don’t have a grid provided in the answer booklet because I don’t have an answer booklet but I do have a grid paper in my notebook so I’m just going to draw myself a grid and I’m gonna graph this and I’m gonna go over the key points of graphing these two okay so let’s that’s not a good at starting access there there’s oops I’m not bad throw the arrows on your axes and then we need to start marking off I’ll go every two here three four five six seven eight nine negative 1 and negative 2 negative 3 negative 4 negative 5 negative 6 negative 7 negative 8 okay so I will do each graph in a different color yes I will pick the really dark darker split now that might get confusing with the graph I look going with a lavender II purple for the parabola now although my parabola was why I use equal to x squared minus 4 unfortunately this is already in vertex form so I can see my vertex is going to be 0 negative 4 so negative 1 negative 2 negative 3 negative 4 it’s going to pass through this point right here and that’s going to be the vertex ok now if I want to what I can do is I need to plot more points and yeah I think I shall so at x equals 1 we’re going to be at minus 3 and at minus 1 will be minus 3 how about x equals 2 well 2 squared minus 4 that’s going to be 0 same with minus 2 this is going to be very symmetric about the graph how about at x equals 3 well 3 squared is 9 minus 4 is 5 so 1 2 3 4 5 and same with minus 3 we get an answer 5 and I think we will be off the graph paper with the next one 4 4 squared is 16 minus 4 is 12 volt I graph only went up to 9 so we’re sort of stuck here so no no it’s trying to play connect the points but we know it’s a parabola so it’s going to be kind of curvy and I’ll put a little arrows at the end of show that yes this still keeps going all right and then I guess I’ll use the green for our other equation y equals absolutely events there wasn’t a constant term right no just two times the absolute value of X okay so once again can just plot some points so at x equals zero to x 0 the absolute value of 0 of course being 0 3 and 0 here how about at x equals 1 well the absolute value of 1 is 1 so x 2 so we’re gonna have this point right here and then at 2 we’ll have 4 3 we’ll have 6 2 4 6 ok and we’ll be off the grid at that point for the next one so we’ll just

stop there and then how about that negative 1 well the absolute value of negative 1 is 1 and 2 times 1 is 2 so this one’s also going to be very symmetric about the y-axis and we get this V shape here okay so there we’ve graphed both of our functions that we were doing okay now I’m a little hesitant because in the as I mentioned this has a written symbol beside it so I would think that maybe they want some explanation but I you know I really don’t think there’s a whole lot of explanation that you can give you could probably give a table of values and when we plotted some points you could give the table of values there um a quick word that x squared minus 4 is in vertex form would probably not go amiss I said it verbally but you might want to write it down but other than that there’s not a whole lot that can be not me it’s written there just because you know they want a real real answer there although the question says sketched and I’m a little confused know what moving on to be part and I what have we got here we’ve got determined with justification all the values obtained for which what are equals x-squared much so the same parabolas before and y equals to absolute value of x plus K do not intersect and they do not intersect okay so hmm what that is where they don’t get your sentence well how about we draw a picture and see what that might look like okay we’ll just do we’ll do it even quicker sketch than we did before so here we have the example of K being zero but what about other other ones now parabola the Bible is always going to be the same so let’s all look like this oops okay now and we’re always gonna get that same v-shape it’s just whether or not it’s moved up or down now if we move it up we will eventually cross our parabola because parabolas always grow faster then after that his absolute values are essentially two linear functions meshed together they’re the two lines matched together and lines grow slower than parabolas so we’re going to get an intersection here so what we need to do is not Havilah so this is what J is positive and when K was zero we still got we still were going to get an intersection point equals how about if K is negative well if K is a very low negative or very small negative number we’re still going to get some intersections and we may get lower extent get a pair of intersections but if we may get low enough because parabolas grow faster the lines will never catch up to the parabola in any sense so we’ll expect some very low values some very negative values of check but how are we actually going to do that well it’s hard to manipulate to absolute value of x plus K in any equation it’s because of that pesky absolute value sign well the function absolute value is as I said sort of two lemons mashed together it’s negative x if X is less than zero and it’s X okay so what we can do is actually break our second function to absolute value of f2 plus K into two separate functions so we’ll have negative 2x plus K if X is less than 0

and 1/2 X plus K if X is greater than or equal to 0 ok so what we’re going to ask is that each of these lines here not intersect our parabola on their given ranges okay now I’m going to make a simplification here I’m going to say well we only really need to check the last case it’s both graphs are symmetric around the y-axis it suffices to show x squared minus 4 and 2x plus K don’t intersect when X is not negative okay because if there is if there were an intersection on the other side on the negative side then by symmetry there have to be an intersection on the positive side so we just need to check the positive case and I suppose there were such an interception well how do we set up any intersection between any equations we make the need for one another so x squared minus four must be equal to two X velocity okay and this is something that we can manipulate we could have setup by x squared minus four equals to absolute value of x plus KT but we wouldn’t easily be able to manipulate it because it’s really hard do eet can we add or subtract absolute value X to both sides that sort of thing and how does that really help us but now that we have no moral values we can very easily manipulate this x squared if we bring the minus 2x 2x onto the other side and we have minus 4 minus K equals 0 so we don’t want any intersections it’s double checking do not intersect so we never want this to happen so we want no values of X no positive values of X for this happens so since we can’t easily factor this because we don’t know what K is we’re going to use the quadratic formula okay and we’ll get negative so X is negative B so that’s 2 plus or minus square root of B squared so it’s negative 2 squared so that’s 4 minus 4 and a which is 1 C which is negative 4 minus okay all over 2 times a which is 2 okay so we can clean this up a little bit I’m never saying that we’ve got some factors of 4 there’s low C coming up the twos will cancel and we’re going to be left with 1 plus or minus 1 plus or plus hey which is just 1 a plus or minus square root of 5 okay now and we’re looking for intersections that are intersections that are negative are fine as we noted on the last page but a world is going as long as we can solve this square root here as long as we can square root what’s under the brackets will definitely be able to get a positive intersection this will be one plus a

positive number will be one of our roots so so since wine and root 5 plus the square root of 5 plus K is are both positive so now that way this can’t work is is if square root of five plus K must net it must not exist that means five plus K must be less than zero and rearranging that it means five must be less than K must be less than negative five okay if we wanted to if we had time what we could do is we could actually pick one say k equals negative six graph that and see for ourselves that we’re not going to get an intersection because we’ll graph it for a little bit around the one axis we’ll see there’s no intersection and we’ll be able to tell well the parabola is much berry far above our our absolute value function and so since linear terms will not grow fast not a faster than parabolas will definitely have an actual solution I’m not going to graph one right now but if we have time say we given up on nine and ten I would want to check our answers what we could do is pick a value of K where K is less than negative five and make sure that we don’t have any intersection so that’s something we could do okay so okay less than negative fired that should give us no interceptions and we have the math to prove it we’re definitely good on the written solution side of everything so on to see part state the values of K for which it’s the same parabola y equals x squared minus four and one equals to absolute value of x plus K so let’s same function as last time intersect in exactly two points justification isn’t required well we’re gonna try and get a little bit of justification here I guess it might have been difficult all right Karen with justification though before we have done intersect now on an intersection exactly two points so by symmetry again we we need to ensure that x squared minus 4 and I will take the positive side again intersect once for a positive x value okay so the reason we’re asking for that is if you have a if you intersect on the right of the on the right side of the graph by symmetry we’re gonna have an intersection on the left side of the graph so we want two intersections overall so one’s gotta be a one side one’s gonna be on the other side so we just ask that there’s only one intersection all the positive side of the graph okay so once again what does an intersection like that it looked like well 2x plus K and then we had x squared minus 2x minus 4 minus K and once again using the exact same quadratic formula we’ll get one on the plus and minus five plus K okay so we only want one a

non-negative root okay and this is where things get tricky because and I will just get it over here oops right now I’m asking about this parabola and sort of this line I want this intersection this intersection I don’t care about it might occur but it’s on the negative side and our real function actually goes up like this inside giving us the other interception so we’re talking about this specific black line but I don’t care about any of its negative intersections so I only care about how it behaves on the right side of breath so I’m only interested in this line on intersecting the parabola in one non negative okay any negative roots will be fine that doesn’t matter I don’t ticks and they’re not going to show up on the graph of two absolute value of x plus K okay so ailing like one non-negative root so we can either have exactly one root so we have two cases here so good I’d wonder it overall hold order we can have them one positive and one negative root okay so let’s clear that one plus root 5 plus K is going to be the positive root and so we’ll need 1 minus 5 plus root K 1 minus root 5 plus K to be the negative 3 okay so in our first situation to have one root over all we’ll need to be doing 1 plus or minus 0 so that means root 5 plus K is equal to 0 so squaring both sides we’ll get 5 plus K is equal to 0 or K is equal to negative 5 so that’s one situation and to get one negative root in order to make this ensure that this is a negative root we need to well we’ll bring things over to the other side we can square both sides so it’ll still be 1 less than 5 plus K and rearranging that bring the 5 on 4 the 5 onto the left-hand side we’ll have negative 4 is less than K so if your K equals negative I’m going to draw a picture or take greater than negative four we have exactly two intersections and so let’s just go out to pictures of what we got going on here I think we use the purple for the parabola so anything greater than negative four something about negative two or something we get two interceptions here and for the negative five one what will actually happen is it’ll touch it will just sort of just be tangent okay so these ones are outright intersections the lines will actually cross but for K equals negative five what we’ll get is sort of they’ll just they’ll just brush against each other the lines will be tangent to the parabola all right so we didn’t need to justify that we gave a bit of a justification and say equals negative five and ten greater than negative four okay that does it for question number eight and the last question with parts we did a B and C so we’ll move on to question number nine and then finally the question number ten but we’ll have to wait for those in the following videos so stay tuned and I’ll see you guys – question number nine on the 2001

you put paper next time