 Gang, check out this question. The question is this, okay, and this is Lorde, swords, and one posted it. Is there a good way to calculate the shortest distance between two functions that don't intersect? One of the functions being a quadratic, and it gives a quadratic equation, I'll write it down here, and the other one being a linear function. So let me write down the two functions first, okay? So the first function being a quadratic, by the way, I got new pens, I got new pens, it's going to be nice and bright. Equals negative x squared plus 5x. Okay, so that's our quadratic, quadratic is a parabola, right? This one opens down, it's negative in front. The other one is a linear function. Let's call it g of x is equal to negative 2x plus 15. Now keep in mind that it just doesn't have to be a quadratic and linear, it could be two linear lines that are parallel, and you want to find the shortest distance. It could be a linear line on a point, and what's the shortest distance? And you sort of follow the same technique that we're about to do here, okay? Let me get caught up with the chat, Speed of Gonzales style. Hi, Brando Hitz. MTL, I had one course last semester that covered a different branch of mathematics in each lecture. One week was linear programming, next was nonlinear programming, dynamic programming, on uncertainty analysis. That sounds pretty damn cool, man. Very cool. Yeah, Zahra, how's it going? Congrats on the graduates here. Yeah, indeed. Now, take a look at this thing. So let me draw your visual as to what it is that we want to do here, right? The linear function is easy to graph, and this one is easy to graph. You could do a completely square if you want to graph it, but what we'll do, we'll find the x-centers, find the average, and just draw an approximate graph of it. Okay, so let's draw a graph here. Okay, this one is a linear graph, right? Which follows the principle y equals the x plus b. That's the y-intercept, that's the slope, right? You go to the y-intercept and you do the slope. That's how you graph lines, right? So we've got to go up to 15. Let's take it up here. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen. We go to 15. That's the y-intercept. That's negative two. So it's negative two over one. So we go down two over to the right once, right? So you can think of it as negative two over one. Down two over one. So this is our linear graph. Okay? So this is G of x. And whenever you graph things on a Cartesian coordinate system, put the name of the function on there. That way you can easily reference it to know which graph it is, right? Now this one, let's bring this one down here. And instead of doing completely square, let's just find the two x-intercepts. And then take the average and find the coordinates of the vertex by plugging the average of the two x-intercepts, right? If you don't know what I just said, watch. This is what we're going to do. So we're going to find x-ins by setting f of x equal to zero, which f of x is your y, right? This is your x-axis. This is your y-axis, x and f of x, right? So what we're going to do, we're going to go negative x squared plus 5x is equal to zero. And then what we're going to do is we're going to factor out an x. And we've got negative x plus 5 is equal to zero. And we've put out videos regarding this. The power of zero, the thing that zero allows us to do is if we have two or more things that multiply together to give us zero, we can set each one equal to zero. So we've got two things multiply together to give you zero. And the only way that's possible if at least one of them is zero. So we're going to set each one equal to zero, split this, and go x is equal to zero. And negative x plus 5 is equal to zero. So x is equal to five, right? Those are our x-intercepts, okay? So x-intercept of zero, x-intercept of five. One, two, three, four, five. There's our x-intercept. This is a parabola. A parabola looks like this. This is opening down. We know it opens down because it's got a negative coefficient in front of the x square. So it does this, right? But even if we didn't know that, what we're going to do is we know it's symmetrical. So we're going to find the average of zero and five. Average of x-ints. Average of x-ints is x-average is equal to zero plus five divided by two, which is five divided by two, which is 2.5, right? So one, two, here it is. This is five over two. Now that's the x part of the vertex. It's the axis of symmetry, right? So we know that this guy, now this isn't part of the graph, it's just a mirror. The parabola is symmetrical along that mirror, right? Now what we need to do, we need to find the y associated with that axis of symmetry with the x part of the vertex. So all we do is just plug in five over two for x. So we're going to find f of five over two, which is equal to negative five over two squared plus five times five over two, which is going to be negative 25 over four plus 25 over two, which is going to be common denominators four. This is negative 25 plus, multiply that by two, multiply that by two, 50. So it's going to be 50 over four, right? That's the y part of the vertex. So the vertex for this parabola is vertex is five over two and 50 over four, which is 25 over two. I'm just going to reduce it, right? Which is equal to 25 over two, 25 over two, which is 12 and a half, right? So what we're going to do is, we're going to go up to 12 and a half. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, and a half. Oh, my graph really is touching it super close. What? 12 and a half. So let's say we're here, okay? And the parabola opens down, goes like this, and goes like this. Okay? Makes sense? So if we zoom into this area, right? It's unfortunate that they're so close together, right? Very unfortunate that they're so close together because it's hard to see this. But what we could do, as long as I got different colored pens, let's bring it out, MTL is a math tutor, even legit, if he doesn't have his own self-reflective drawing. So let's assume we're going to take this and zoom into here, right? So we're zooming into here. What you see here is this. Let me draw it in black as well. So this is our G of X, right? G of X. Really, it's too bad they gave the functions too close together. I would have given a function a little further away. So you can actually see where it is, right? And then here's the other function. The parabola looks like this. And you want to find the shortest distance between this line and this parabola. Okay? The shortest distance between this line and this parabola and this parabola is F of X. Okay? Okay, I'm going to skip over, slip, make your comment there. Just continue on this. Okay? All right? So what we're looking for is this. What's the shortest distance between this line and here? That's what we want. Okay. So how do we do this? Well, if you do this calculation, oh, actually, we want to gotta be... There's something we're missing. Slip, make. There's one thing you need from here, from this question. It should be... Oh, it should be 25 or 4. That's right. I did a boo-boo. This is... Thank you very much. If you find me making mistakes, please correct me. It should be 25 over 4. I did it in my mind. That's why I didn't reduce it. And then... So it should be 25 over 4. So 4 goes into 25, 6 times. So 6 and a quarter. That's way better. Let's erase this. So we don't even have to zoom in anymore. Let's kill this. Let's kill this. Let's kill this. But I think we need one bit of information, one more bit of info here. There it is. Let's say it's up there. So 6 and a half. One, two, three, four, five, six, six and a quarter. One, two, three, four, five, six and a quarter. So it would be here. Oh, that's way better, man. That's way better. So it goes like this. That's nice. Now we can see it. Right? That's also not my question. I think MLT asked. MLT asked would be the finding the... Yeah, it's the perpendicular. But we need a point on the line or a point on the parabola. We have to have a point. I'm pretty sure we have to have a point. Because the shortest distance... Oh, the shortest distance. Wait a second. The shortest distance. It would have to be at a point where this touches, right? Because the distance from this point to the parabola, the shortest distance. Wait a second. So check this out. This is my thinking, right? The shortest distance from a point is going to be the perpendicular. It's always going to be the perpendicular from the line, right? So if we have this, check this out. If we have this, right? The shortest distance from this point to the parabola is perpendicular. It's got to be at 90 degrees, right? The shortest distance from any point on the parabola is going to be perpendicular, right? So this question says what is the shortest distance, period, right? Now if they gave you the point, oh, by the way, I think they meant the shortest distance, y-wise. Oh, y-wise. Not just in any direction. The distance between the two functions are... The distance between the two functions are g of x minus f of x, okay? y-wise. Oh, that makes it easier. That makes it easier. I believe so, right? Why does it make it easier, y-wise? Because if it's y-wise, it would just be on the vertex, because that's the highest point that guy reaches. Is that correct? Is that correct? Is that correct? Would that be correct if it's the highest point? That's the highest point of the parabola, but if the line was going like this, I think it would just be to the vertex, correct me if I'm wrong, gang? And liquid source is saying the distance between g of x and f of x, g of x and f of x would just be g of x minus, I should write this down, f of x, right? So let's create some room for us to work. I'm going to take down all of this, okay? I'm going to take down all of this. No, I think it's a bit further than the vertex. Upper function minus lower function usually signifies a distance between them. Yeah, it does. You subtract them, right? So you're subtracting the y's. Let me erase all this. This would have been really easy if they give you a point on the line or a point on the parabola, but they're becoming very general, right? Dr. Meng Metten for small distances in x, 2x is greater. Yeah, that's the thing I'm thinking about, right? Because this is expanding at x square, but if it's a really short distance, 2x is going to be greater than x squared, right? But this is a longer distance. So let's, first of all, let's do this. Let's see where this takes us. This question is not, I thought it was obvious what the answer was, but it's not as obvious, okay? Since the slope is lower for a while after the vertex. Yeah, but this thing's expanding speedy Gonzales style, right? So if it's the shortest distance, it's going to be very close to the vertex if it's not the vertex. Like it won't be over here because that's expanding way too fast now, right? So it would have to be within this region, right? Within here and here. Otherwise you're already gone. It's past it, right? But the distance, distance between g of x and f of x, because g of x is your y because they want the vertical distance, right? So they want this, what's the shortest vertical distance? So technically speaking could just, or visually speaking could just do this, right? And you're going to pick the shortest line, right? That's what it means. That's what they're looking for, right? If it's just a y difference, then f of x and g of x are your y axes, right? And because g is higher up than f, we're just going to go g of x minus f of x, right? And then we're going to minimize that, right? Minimize what else? How do we go about minimizing? Let's do the subtraction first and then we'll figure out how to minimize it, right? To minimize it. Okay, let's draw it out. Let's see what we end up getting, right? Or write it out. So g of x is going to be negative 2x plus 15 minus negative x squared plus 5x. So this is going to be negative 2x plus 15 plus x squared minus 5x. So it's going to be x squared minus 7x plus 15, right? Did I have any brain farts there? Did I have any brain farts there? Before I continue, because I made one brain fart before with the vertex. The vertex here was, what was the vertex? We didn't write it down here. Let's write it down here. Vertex for parabola, vertex was 5 over 2 and 25 over 4, right? That's this point here. Okay. Was also thinking that creating a new function that is this, then using derivatives to find the critical point. Yeah, you could do that. And here's the other thing you could do. You want to find the minimum of this, right? If you want to find the minimum of this, would it work if we just found the vertex of this? Because the vertex of this function, because it's a combination of this function and this function, generating a new function, the minimum should be the minimum. So all you really need to do is find the vertex of this and you should have the minimum, okay? The y part of the vertex. Let's do both methods, okay? Let's do both methods. So the new function is the distance functions. We're getting into complex numbers, I think. Are we getting it? I don't know. No, I don't think it should be complex, is it? 49, oh crap. It is going to be complex. What the hell? This is going to be complex number. It is. The roots of this equation will be complex. Yeah, it is complex. So that doesn't make sense, right? If we take the derivative, but that should do it. How come it's not going to do it? Did we have a brain fart or did I have a brain fart here? Did I have a brain fart here? Was that supposed to be plus 5x? Or did I write it down properly? Yeah, it's plus 5x. Because if we take the derivative, not the derivative, if we try to find the factors of this, they're going to be complex. Yes, plus I think. Yeah, it is plus. I just scrolled up and it was plus. So that's not going to work. Okay, so let's call this deobags, the distance. So this function should give us the distance of the line and the parabola, if I'm not mistaken, right? Expired sandwich. You're such a shabby teacher. One of my late grandfather who was an amazing math tutor and carried me through high school math. And you also remind me of one of my favorite priests who passed away last year suddenly. I love watching your content. Much love. Much love right back, expired sandwich. And by the way, the way I teach is exactly the way you see here. If a question comes up that I don't know how to do right off the bat, we work on it together. Because I'm learning something, my students are learning something. That's one of the reasons I love what I do because I'm constantly kept on my toes, right? So from what I understand, this should give us a different distance between the y's, right? Oh, check this out. Check this out. This thing, here's the reason. Here's the reason that this is going to be complex. Have you guys figured out why it's going to be complex? I can tell you why it's going to be complex. Should we do it? Hold on, I'm going to give you a couple of seconds to figure out if it's going to be complex or not. So we can't find the x-intercepts. We need to find the vertex. Okay? It's a passion. It's a distance, but there's a reason why it won't work. Because when we find the vertex of this, it's not going to cross the x-axis, right? Because one of the reasons the one is very big, yeah, because it's the x squared taking in versus 2x, right? So what's happening is, it's trying to find the x-intercepts, the distance between the line and a problem that's going crazy. I believe so anyway, right? Or no, it should. We do not need the plus 15. Well, you know what? I'm going to go find the vertex of it. We don't need the plus 15. How come? So we can just take that out. I don't know if we can or not. Could we maybe look at the discriminant of d of x to find the line of symmetry? The problem with the discriminant of the v of x or d of x, it's going to be an imaginary number, it's a negative number, but let's find the vertex of this thing. Find the vertex, put those in there, brackets, divide this by 2. This is complaining the square, right? Square it. You get 49 over 4. Add and subtract that inside the brackets. x squared minus 7x plus 49 over 4 minus 49 over 4 plus 15. Grab this dookie, kick it out, and this factor is that guy. So x minus 7 over 2 squared. That guy comes out, becomes negative 49 over 4. So add those guys together. It's 4, 60 minus 49. So it's going to be 11 plus 11 over 49 plus 11 over 4. So that's the vertex of this parabola. It's just a displacement in the y direction. The vertex is going to stay at the same x value for any constant. The vertex is going to stay same value x value for any constant. Think about that. Is that correct? The two questions will never meet. Equations will never meet. So I think it's complex. But I think the vertex, the y part of the vertex, is the shortest distance. Is it not? Isn't this the shortest distance? And this is going to be the x part of it. So I believe, what's the problem? The problem is, find the shortest distance, shortest vertical distance between f of x and g of x. So if it's the shortest vertical, and we're not given any points to start off on the line or the parabola. If we were given a point, it would be easy. Then you just find the perpendicular, you find the shortest distance. If it was just the shortest distance. If it's the y, you just find the distance between them, right? I thought we searched for the x. I think we got the x. So the shortest distance between, not the shortest distance, the point where, would that be the shortest distance? It's the x part of it. So this would be 3.5. If I want to redraw this, I want to redraw this here. That way we get 1, 2, 3, 4, 5. So I'm getting 3.5. That's 7 over 2. You take the opposite sign of that. And 11 over 4 is 2 and 3 quarters, right? 1, 2, and 3 quarters. So that's the vertex. And that thing opens up, and we don't care about that. So that's the vertex of the distance, right? So it would be, I believe, it would be here for the x part, 7 over 2 and 11. Well, no, no, it wouldn't be 11 over 2. We don't know what the point is, right? But I think this is the distance. It would be 2 and 3 quarters, would it not? Am I mistaken on this? That is the x that gives the shortest distance in y. If it was the shortest distance, I'd use linear algebra. Yeah? Crafter? Does this channel cover differential? We've done a little bit of finding derivatives. A little bit. I mean, we'll find the derivative of this. What was the other method we're saying? Find the derivative and find the inflection point, right? Where it rolls over, where the slope is zero. And that would be, if you take the derivative here, let me do this in red, right? So if you take the derivative of D of x, so D prime of x is going to be 2x minus 7, right? So D, so D prime of us is 2x minus 7, which is this guy. If you set D, D prime of x equal to zero, then you get 2x minus 7 is equal to zero, so 2x is equal to 7. So x is equal to 7 over 2, which is also the same thing, right? This is essentially minimization over convex set, right? I don't know. I don't know if that's what they call it or not. It could be, right? So Dr. Heng, is that correct? So the x occurs at 7 over 2 and the distance, the shortest distance would be 11 over 4, which is this is 11 over 4. That's the shortest distance because our function was D of x was G of x minus f of x. Is that correct? This is actually really cool. I don't think, I haven't done a question like this teaching when they took it out. This complex of a question wasn't in the curriculum. They usually gave you a point and they said, find the shortest distance. So the solution is f prime of x equals f prime of G. No, I don't think so. I don't think so, Keaton. Should indeed be correct. Liquid source, 7 over 2 should indeed be correct. We have found the x value where the shortest distance occurs, but what is the shortest distance? So x is 7 over 2. Isn't this the shortest distance? Is this not the shortest distance because we're trying to find the minimum value because if this is the parabola, right? This is the parabola. It opens up so we know the minimum is this, which is 11 over 4, right? Okay, yeah, we have the same solution. True. I believe the answer is 11 over 4. Awesome. Great question. Fantastic. I love it. I've never done anything like this before. So cool. Okay, yeah, we have the same solution. So liquid source, were you guys doing this using, doing the derivative? So if you did the derivative, you got the x value. How do you get your 11 over 4? Do you kick this back up to here? Find the two y's. Oh, that's what you would do. Check this out. That's what you would do. You have the same solution at the point. So what you do is do this. Here, let's do this in green since we've got lots of colors going on. Shortest distance from, that's exactly it. That's what it would be, right? Slow, half right. Shortest distance between 3.5, 5.25 to the line that is the shortest is the line that passes through. I did use the derivative just now and plug it back in. Yeah, you plug it back in. Length of the line, segment of the line. Yeah, so basically what you would do is plug in 7 over 2 for f in x here. So let's do this here. So find f of 7 over 2, which is, let me erase that. Here, let's complete this again. It was actually a question someone asked me on a Swedish math form. Just wanted to share it. Awesome liquid source. Super cool. I like it. Negative 7 over 2 squared plus 507 over 2. So this would be negative 49 over 4 plus 35 over 2, which is going to be, that's going to be 70. Here, let's do it. 4 negative 49 plus 70. What is that? A brain fart. 21, right? 21 over 4. Is that correct? Yeah, 21 over 4. The guy that posed the question didn't even know what a derivative was. Yeah, you don't need derivatives to do this, by the way. We did it without derivatives, right? So what you would do, this would be 21 over 4. That's the y part of this point. So this point is 7 over 2 and 21 over 4. And then you plug in 7 over 2 for g of x. So g of 7 over 2 is going to be negative 2. 7 over 2 plus 15. 2 kills 2. So that's negative 7 plus 15 is equal to 8. Does that work? 8. Did I do a brain fart? Oh, yeah. That's not that. This point, because that was a parabola, this point is 7 over 2 and 21 over 4. And this point is 7 over 2 and 8, right? So if you want to find the shortest distance between this point and this point, you just subtract the y values. And that should give us the same answer as 11 over 4, right? Let's check it out. Let's do this in the green here. We're all over the place. I love it. This is the way math should be, chaos, right? As long as you've been following the work, you know exactly what it is that you're doing. And we've got different color pens. Rock and roll. So we got 8 minus 21 over 4, which is common denominators 4. 8 times 4 is 32 minus 21, which is equal to 11 over 4. Rock and roll. Same answer, right? Check it out. Awesome. Great question. Love it. Love it. Love it. Great question. So cool. So cool, right? I gotta remember this question. Give it to some of my students. They don't do this anymore though, so they took it out. Right. Super cool question, by the way. Thank you for that, Liquid Swords. You can go back and take a snapshot of this and here I'll step out of the way. Here's the solution. Follow the black color first and then the red color and then the green, right? Fun, fun, fun. I love these types of things that all of a sudden you start thinking about how you go about doing something that you haven't done before, right? Good for the brain. Good for the brain. Builds the connections a lot better, right? Builds the connections a lot better. By the way, gang, do you guys have snacks? I got my chocolates are back. They're, this time they're not melting. We're sitting outside and the sun was sitting on, so they melted. This is just vertical distance. Yeah, this is just vertical distance. Not the shortest. The shortest vertical distance. Yeah. Just vertical distance, indeed. Yeah. MR Sly or RM Sly. It's math. We're all used to following a complex mess of work. Yeah, I know. It's, man, if you can find, if you can, if you can follow this, your money, right? Okay. No spoon, no spoon required, but I brought it anyway. It was yesterday's chocolate melted and was so delicious eating it with a spoon later. And gang, don't forget free Assange, free Assange, free Assange. Julian Assange is a publisher and journalist that has been crucified for trying to bring transparency and accountability of capital as power to humanity. For more information see wikileaks.org, defend.wikileaks.org, or our Julian Assange and WikiLeaks playlist on censored too. Dr. Heng, this was super interesting. We could search for the shortest x distance now. Yeah, you could, you could do the same thing. Shortest x distance. How would you do the shortest x distance? How would you do the shortest x distance? Free Assange. Liquid source got beer and chips. Imperial stout has become a favorite. Nice, nice. I haven't had beer for so long. You don't know. I'm not sure either. I'm trying to think about it. How would you find the shortest x distance? You could. You could do this, right? Flip these around. Flip them around. Do a 90-degree rotation on this. Make the x, y, and the yx, and then find the shortest distance between those two points. I think you can invert functions, yeah, and do the same process. Yeah, that's what I was thinking, right? But it's been a while since I've done this. Yeah, I think you should be able to do that, right, Sly? Yeah, take the inverse of the functions, right? It might become tricky because now that's no longer quadratic. It's a, what do you call it, radical, right? I need to learn English as I did, not understand the class, and I was very interested in it. I hope to learn, don't say, yeah. Slowly. English is a hard language to learn. Choco milk, right? Keaton, yeah, invert and just use the right half of GeoVax. Yeah, I think so. I think that's what you would have to do, right? Oh yeah, that's the way to go. Yeah. It adds that one extra element, right? Pretty cool, though. And the board would look messier. And remember, we already had to do some work to be able to graph it, to get our visual, so we already did a little bit of racing. So this is a super cool question. Super cool question. I'm going to take these down, see if there's any other questions. If there isn't, we can definitely take a look at some ratios to confirm information from someone, right? Okay, so the total distance is minimized when F of X is at 7 over 2 and 21 over 4, and GeoVax is at 23 over 5 and 29 over 5. Ah, Keaton. I thought we got the points as, I erased it, as for GeoVax, it was 7 over 2 and 8. I believe that was the point of the vertical distance, right? Yeah. And then if you're going to make it perpendicular, so it was like this. This point here, this point was 7 over 2 and 8. This point was 7 over 2 and 21 over 4, 21 over 4, 21 over 4. That was the vertical distance, right? And if you take this and make it 90 degrees, is that the point you're talking about, which is 23 over 5, 23 over 5, and 29 over 5, 29 over 5. Is that what you're referencing? Your vertical distance seems fine. Yeah. This minimum total distance, not vertical. Yeah, I believe so, right? And that is easy to find. That would be easy to find, I believe, right? Because all you need to do is find an equation of a line. Here, we could do it. You need to find the equation of this line, right? Let's assume we don't know that yet, right? The equation of this line is going to be y is equal to mx plus b, right? Now, we need the m and the b for this line. Now, what was the g of x? g of x was negative 2x plus 15. I believe that was a function, right? So, if this line is perpendicular to this line, then the slope of the perpendicular, slope of the perpendicular is going to be negative reciprocal of that, right? Negative reciprocal of negative 2 is 1 over 2. You flip it and change the sign, right? So, 1 over 2. So, that means we have m. So, y is equal to 1 over 2x plus b. Now, we need to find the b, which is the y-intercept. The y-intercept was over here somewhere, right? The y-intercept would have been here somewhere, right? So, we need to find that point. That's your b. So, all you got to do is plug in a point you know in that's going to be on the line to find your b. So, you're going to take this guy and plug it into here for x and y. You're going to get 21 over 4 is equal to 1 over 2 times 7 over 2 plus b. Okay. So, this becomes 21 over 4 plus 7 over 4 plus b. Bring the 7 over. So, that's going to be 14 over 4 is b. So, you just found the equation of the line, this line, the perpendicular line to the other line, right? So, y is equal to a half x plus 14 over 4. So, now, all you need to do is find the intersection of this line and this line, right? And to find the intersection of that, you set them equal to each other, right? You want to find out at what point or what point exists on both this line and this line, right? Well, that means you want to find the y there, right? At the x and the y there. So, you set them both equal to each other. You set g of x, set g of x equal to y, right? That means you set that equal to this. So, you're going to get negative 2x plus 15 is equal to a half x plus 14 over 4. Multiply everything by 4 to get rid of the denominators, right? You're going to get negative 8x plus 60 is equal to 2x plus 14 and then solve for x. Here, I'm going to grab the negative 8, bring it over. So, that's 8x. I'm going to grab 14, bring it over. It becomes negative 14. So, you're going to get 10x is equal to 46. 46 divided by 10. So, x is equal to 23 over 5. Is that what we got? Yeah, the x is 23 over 5, right? So, that's correct. 23 over 5. And then if you want the y, all you got to do is just plug this in, either in for this x or this x. You're going to get the same y out. It doesn't make a difference, right? Because that's what you were doing. So, let's just plug it into, which one is easier to plug it into? If that one is easier to plug it into, obviously, right? So, let's find g of 23 over 5, which is going to be negative 2 times 23 over 5 plus 15, which is going to be negative 46 over 5 plus 15. Common denominator is 5. So, that's 60, which is going to be, what is that? 14. Oh, we didn't get 29. Did I have a brain fart? Did I have a brain fart? 15. Oh, it's 5, not 4. So, this is 75, right? 75. Okay. So, that should be 29. Here, 75 minus 46 is 6, 15, 29. So, that's 29. So, it's going to be 29 over 5, which is the y point. Once you got that, you have this point, and you have this point, and then you use your distance formula, which is Pythagorean theorem, which you're doing the triangle. You subtract the two y's. You subtract the two x's. You get your x and y distance there, and then you do a squared plus b squared equals c squared, right? Nice combination, right? Yeah. And the way you find that 7 over 2, 21 over 4, is the correct point to start from on f of x is by setting the derivative equal to each other, setting the derivatives equal to each other. The slope, okay. You can layer and layer and layer and layer. Keep on going with it, right? Super cool. And then you would just do a squared plus b squared, like I mentioned, right? Super cool question. Super cool question. Multi-layered. This type of question would be, if they gave you a point for parabola and a line, and they said it was the short assistance between the line and the parabola, or the parabola and the line, that would be more of a grade 11 question, grade 10, grade 11 question. If it was two lines, parallel lines, it was grade 10. They took it out 10 years ago. Don't teach it anymore. Too complicated, right? Don't want to make children too smart. They might question the system, right? They might figure out how to work around the system, right? So our centralized education system is more geared towards indoctrination, not education, right? But that was about 10 years ago they used to teach that in grade 10. Because this involved parabolas, where you had to find the equation of parabola, it would have been grade 11 that they would have a question like this. If it would involve derivatives, it would have been grade 12. I love this channel. Awesome styles, PG styles. Thanks, me too. Fun stuff. Great question. Great question. Super fun. Super fun. Good way to get the brain working. Cool, cool, cool.