 Hi everyone, let's take a look at an optimization problem that is very much geometry related. We have here a parabola given by the equation y equals x squared minus x plus 2, and we are trying to find the points on that parabola that are going to be closest to the point negative 7 comma 5. And you can see I've done this on my graph and calculator and copied over the image for you so we have a nice visual from which to work. If we are trying to find the points on this parabola that are closest to that point over there in the second quadrant, really what we are trying to minimize is the distance. And hopefully when you think of distance in geometry, what comes to mind is the distance formula. This becomes what we want to minimize. Let's substitute in the values that we know so far. We have that ordered pair negative 7 comma 5. So I'm going to substitute those coordinates in place of our x sub 2 and y sub 2. And I'm going to change the x sub 1 and y sub 1 to just regular x and y. Now we have an equation that we want to minimize but still we have the problem that it's in terms of two variables. We do need to eliminate one of these variables and the way in which we can do that is to make use of the equation we were given, the y equals x squared minus x plus 2. We're going to substitute that in place of our y right here and then we'll be able to simplify it down to one variable. If you substitute that equation in place of y, I'm going to go ahead and combine my constant term just so it's a little bit easier. So I end up with x squared minus x. I would have had a minus 5 plus 2 so that becomes minus 3 and we have that quantity squared. Now we have a primary equation that we can work from that is in terms of one single variable. We know that in order to minimize this equation we need to be finding the derivative, setting it equal to zero and solving it. This becomes a little bit tedious to find the derivative of this by hand though. So let me show you another way that we can maybe do this graphically instead. If you grab your graphing calculators I'm going to switch over to mine here. I'm going to go under y equals, I'll give you a moment to type this in yourself, and type in that distance equation we had under y1. Now remember what we want to be doing. We want to take the derivative of this and see where it equals zero. Graphically that means where is the derivative across the x-axis, what are the zeros, the roots, the x-intercepts. So I need to tell my calculator to do that. Remember we can do that by if you hit math in the far left column, number eight, and we're going to fill in our little boxes here so we want the derivative of what we had under y1, and we do want it to produce the graph of the derivative. So you're going to have where x equals x, so yours should look like this. I'm going to turn this original equation off. If you move your cursor, your black blinky cursor, so it's on top of the equal sign next to where it says y1 if you hit enter, that will turn that off. This way we're just going to be looking at the graph of the derivative. If you do zoom six, that should work good enough to get a graph. Now it might take a second to render because the calculator has a little bit of thinking to do to generate the graph of this derivative. And there we go. Looks like there's just the 1x-intercept. Let's go ahead and find that. If we hit trace, and I'm going to trace over to about where it crosses the x-axis, and I'll run my root finder, my zero finder. I'll need to have a left bound that has a negative y-value because as you look at this curve from the left to the right, the curve is underneath the x-axis first, so I need to have a negative y-value for my left bound hit enter. And then it will ask you for the right bound. And if you just go one click to the right, that should suffice because I now have a positive y-value hit enter there. There we go. So we have an x-value of approximately negative 1.625, it would round two. Now while we're looking at this derivative graph, remember that this negative 1.625 is our critical number. So if there is going to be a minimum for this original function, it will occur here. Remember though, you're looking at the derivative graph. Take a look at what's happening to the derivative graph. As you look at it from the left to the right, the derivative first is negative. Remember what that tells us? That tells us the original function would have been decreasing. It hits that x-intercept of negative 1.625 and then the derivative becomes positive, which means the original is increasing. So you have an original function that switches from decreasing to increasing, thereby creating a minimum at this x-value of negative 1.625. And that's your argument and your reasoning, your explanation as to why we do have a minimum at this value. If you'd like to write all that up, let's talk about how you would do that. So we set the derivative equal to zero. That gave us an x-value of negative 1.625. And that's pretty much all you would have to denote in terms of what you did. You specified, you took the derivative, you set it equal to zero and you solved it. That is what you need to be stating. This becomes our critical number. If you care to do a number line analysis, we've already talked about all the information that the derivative graph gives us, so this part's really easy. Remember how we said that derivative graph changed from negative? It hit the x-intercept, then it changed to positive. That's what created our relative minimum. We can then answer the question as to what point is going to be the closest. If we go back to the graph, we just figured out it has an x-value of negative 1.625. So that would be right around here. So if I trace that up, that's going to hit right around here on the curve. I need to find that y-value on the parabola. So I can substitute into the original equation, the x squared minus x plus 2. And you should come out with about a little bit more than 6. I'm kind of close. I'll leave you to do that number crunching to verify what it is. But it should be approximately 6.266. Please do verify that on your own. Then we can just write up the answer. We find that the point on the parabola closest to the point negative 7 comma 5 would be the point negative 1.625 comma 6.266. If you look at the graph, that makes sense because we were trying to minimize the distance to the parabola from this point out here. And it does look like that distance would be the closest. Because if the point were located anywhere else on the parabola, there's definitely going to be a farther distance getting to those points. So it very much makes sense that this is the point that we found as our final answer.