Understanding Coordinate Plane and Circles

Hello everyone, we're going to go ahead and go over section 1.9 today. This is on the coordinate plane, graphs of equations, and circles. Most of the sections should be pretty much review for everyone except for probably the last part at the end which is going to be about circles. But looking at the coordinate plane and graphs of equations should look somewhat familiar although there might be a little few new things here and there. So All right, starting off with the coordinate plane. The coordinate plane is where we graph things like lines, points, functions, equations, all that kind of stuff. Kind of looks like this picture here that we have. We see that points on a plane are identified with ordered pairs of numbers, and this forms a coordinate plane. It's also called the Cartesian plane, but we'll almost never call it that. If you do happen to see it named Cartesian plane, it's the coordinate plane. But so we see this horizontal line here. This is the x-axis. These are where we plot x values. So going up and down like this. The vertical plane is the y-axis. So the vertical line is the y-axis here. So when we plot points, we're kind of going side to side, going through that y-axis. And the plane is divided into four quadrants. We've got quadrant 1, quadrant 2, quadrant 3, and quadrant 4 here. The points on the axes are not actually assigned to any quadrant. They don't belong to one because they're right basically on the edge of two. So a point like this, point P, which is at the point A comma B, our first point there is that x value, point A, and the second is point B. And it's written within parentheses with a comma right there. That is our ordered pair. So whenever you plot things on... coordinate planes, you're most likely going to be plotting a lot of ordered pairs, drawing lines, equations, anything along those lines. So like we said, any point in the coordinate plane is located by its unique ordered pair of the numbers a comma b, where a, or in this case for this example, 5 comma negative 2, the x-coordinate is that first number, the y-coordinate is the second number. So that's how we write out our ordered pairs, like how we see here on this graph as well. We have a bunch of different ordered pairs here. Each time it's always the x value first, y value second, comma in between, parentheses around the whole thing. So these coordinates, these ordered pairs really just specify a specific location on our coordinate plane. We use them to graph other things as well as just graphing points in general. So it's really just identifying a specific location or potentially solutions that we're going to see kind of in the future. So. Moving on from there, we're going to learn about the distance and midpoint formulas. So the distance and midpoint formulas are going to be exactly how they sound. Distance formula is going to calculate the distance between two points, and the midpoint formula will calculate the point that lies directly between two points. So starting off with the distance formula, we want to recall, if you've ever seen it, if not, then this will be new, but the distance between points A and B on your number line would be written as... The distance between a and b, this first part, left hand side of the equal sign, is just stating what we're doing from the distance between two points. And this is equal to the absolute value of b minus a. Remember that absolute value, it's just going to take whatever's inside of it and make it positive in the end. So we take the larger number or the second number and we subtract the first or typically the smaller number. And if it ends up being negative, we add the absolute values on it, just make a positive. And this is really just to kind of run the idea home that distance is always positive. So since distance is always positive, you can't have a negative distance value, so when you subtract those values it always has to be positive, which is why we have the absolute values. So that's really just the reason why they're there. So since we know that that's the distance between two points on a number line, meaning just the typical number line that we know of, you know, 0, 1, 2, 3, 4, including negative numbers, things like that. We can do the same thing for horizontal and vertical distance as well. So that horizontal distance between any two given points are going to be two x values. So you would have a second x value minus a first x value. And whichever one you decided to be first and second really doesn't matter. because we have these absolute values. It's going to take whatever value you have on the inside and make it positive anyway. So it does not matter which one you pick to be second and which to be first. Then we can do the same thing with vertical distance. Any vertical distance between two given points is going to be two y values. So a second y value minus say the first y value, with once again absolute values around it. So just like with the x values, it doesn't matter which one you label to be first or second, but if you are picking points just make sure to be consistent. The second point will have the second x value and the second y value. Your first point we'll have the first x value and the first y value as well. So what we see in this picture here, we have point A and point B. Point A was said to be the first point. It has the little ones on the x's and y's. We call that x sub 1 or y sub 1. And B is the second point. It has the twos below the x and y. We call that x sub 2 and y sub 2. So we say that A is the first point. b is the second point and we want to find the distance between those two. Just like how we have it over here, we take the difference between those x values, the second x value minus the first x value, and we take the absolute value of that so we get a positive value. And same thing over here for the vertical distance. We take the second y value minus the first y value and apply those absolute values. I almost forgot the word. So that's how we would calculate distance with the specific formula. This is derived from the Pythagorean theorem. Just a fun fact, don't need to remember that. But here we go. This is the actual distance formula. So if you want to find the distance formula, this is what you're actually going to use. Everything on the last slide basically just gave you a background trying to explain kind of how we got here. But this is the bread and butter. This is what you're actually going to use. on things like your homework and exams and whatnot. So this is the important one. This is the distance formula. Distance formula between the points A and B, where we say A is that first point and B is the second, in the plane, which is any plane, the distance between points A and B, that's all that D with A and B in parentheses means, this just means distance between points A and B, is equal to the square root of x sub 2 minus x sub 1 squared plus y sub 2 minus y sub 1 squared. So this is our distance formula. We say we took the x value from that second point minus the x value from the first point, took that value, and we squared it. Then we took the difference between the second y value minus the first y value, squared that value, add those together. and take the square root of that result. That's how we know the distance between two points. So this is our distance formula. We're going to learn about the midpoint formula and then we're going to go ahead and do an example using both. So next is the midpoint formula. Just like we said before, the midpoint formula is going to calculate the point that lies directly between two other points for us. So we say at the midpoint m of x comma y just like how we had distance was d of x comma y, now we have m of x comma y, is the point that marks the exact halfway point on that line, this imaginary line most likely, that connects two points. So the horizontal distance of each half has to be equal. And this equation that we see here is just representing that it's equal. We don't have to particularly remember this. But if x is that x value that's right directly in the middle, we say that x minus x sub 1, the x value from that first point, is the same thing as having x sub 2, the x value from the second point, minus the x value of the midpoint. Just to say that it is exactly halfway. And same thing for the vertical distance with the y values are going to work the same way. The y value of the midpoint minus the y value of the first point has to be equal to the y value of the second point minus the y value of the midpoint. Just to say that they are. perfectly equal. They are right smack dab in the middle. That's all those are stating. So if you actually solve for x and y to get the midpoint formula, here's the bread and butter again. This is the one you need to remember. The midpoint formula says that the midpoint of the line segment from point a, which is our first point, to point b, which is our second, is this formula right here. Because remember, your midpoint is a point. It needs to be in the form of an ordered pair. distance is just a value. But since you have to have this ordered pair here, we see that the x value is going to be the sum of the x values of both points. x sub 1 plus x sub 2. and that number divided by 2. Then for the y value of your midpoint, same thing. You're going to take the sum of the y values of your two points, so y sub 1 plus y sub 2, and divide that by 2. And this result would end up being your midpoint. So this is the midpoint formula. These are the two really important formulas that you need to remember from this section, the distance and the midpoint formulas. Everything else was just mostly explanation. These are the ones you need to remember. So, all right, on to our first example. Example one, we want to find the distance between the points P and A, where P is at the values 1, negative 2 at the ordered pair, and A is at the ordered pair of 3, 2. So we want to find the distance between those points, and then we want to actually find the midpoint of the line PA. The line we're just labeling as PA is this line that just connects the two points. The midpoint formula suggests that there should be some line, but your line can be imaginary or real. Either way, you're still finding the midpoint. There doesn't actually have to be a line, although we will probably draw one in. So, all right, let's go ahead and head over to our notes so we can work out this problem. All right, so here is our problem here. We have a coordinate plane. I drew out, so excuse it. not being completely perfect, although it's pretty accurate. Here we go. So we have our coordinate plane. We have plotted point A and point P. We said P was at 1 comma negative 2, and A was at 3 comma 2. So here are the two points that we have, and we want to go ahead and we want to find the distance and the midpoint. So starting off with the distance, we said that the distance between two points in our case, we're just going to go ahead and do A and P or P, and whichever one you prefer. the distance is going to be x sub 2 minus x sub 1 squared plus y sub 2 minus y sub 1 squared. So if we know that this is the formula for the distance, we want to go ahead and figure out what is going to be our first point, x sub 1 and y sub 1, and what's going to be our second point, x sub 2 and y sub 2. So let's go ahead. We're going to label P as x sub 1 and y sub 1, and A will be x sub 2, y sub 2. So the distance between our points here, we can go ahead and fill in those values. So we've got our square root, we've got some parentheses, we're going to fill in all the important information here, we're going to add in some different colors, which is why I left it like that. All right, so our first point is... p, x sub 1 and y sub 1. So x sub 1 is going to be 1, since that is the x value of p, and y sub 1, the y value of p is negative 2. So we added in those values. Now let's go ahead and add in our values from a. We said a was the second point. So x sub 2 will be x value of a, which is 3, and y sub 2 will be the y value of a. which is 2. So from our values here, we're going to go ahead and we're going to simplify it. So we want to do whatever's in the parentheses first, following our order of operations. 3 minus 1 is going to be 2, so we've got 2 squared plus, and then 2 minus negative 2, we have a double negative there, so that becomes plus. So 2 plus 2 is 4. Ooh, my screen was wigging out. There we go. So we've got... Ooh, that was way too big. There we go. All right, we've got 2 squared plus 4 squared. Now we can go ahead and plug this into our calculators or reuse our brains, whichever one you prefer. 2 squared is going to be 4. And 4 squared is 16. We have the square root of 4 plus 16. We know 4 plus 16 is going to be 4 plus 16 is 20. So, all right, the distance between these two points is the square root of 20. Now, this answer is correct, but if you need your answer in decimal form for the homework or if you're doing any sort of comparisons and you want to know kind of an actual number instead of a square root number. you can go ahead and calculate that decimal. If you do that, you should get that the square root of 20 is about... 4.47. So this answer is also correct for your distance, whichever one you prefer. So all right, so we calculated the distance between those two lines, and this distance that we're talking about here is this distance here. Kind of drawing a little imaginary line. All right, so... That's kind of the distance that we're looking at. Now we want to find the midpoint. So our midpoint, we're going to scroll down just a little bit here. Our midpoint was this point M of X comma Y. And we said that the midpoint formula was going to be X sub 1 plus X sub 2 divided by 2. And Y sub 1 plus Y sub 2 divided by 2. Now we know what... x sub 1 and y sub 1 and x sub 2 and y sub 2 are, so we can just fill in this value. So we said x sub 1 was 1, x sub 2 was 3, so we have 1 plus 3, right over 2. And then we have y sub 1, which was negative 2, plus y sub 2, which was positive 2. and this is all divided by 2. So we can go ahead and calculate this. 1 plus 3 is going to be 4. So we've got 4 divided by 2, negative 2 plus 2. Well, that's just 0. So that's 0 divided by 2. So a final simplifying here, 4 divided by 2 is just 2. 0 divided by 2 is 0. So our midpoint is 2,0. So that is our midpoint there. We can go ahead fill in this on our graph here. Let's pick purple. That's the color between A and our point P here. Here is the-oh that was the wrong one. That's the highlighter. Here we go. So they picked purple. This is the midpoint. Okay so it's not exactly on the line that I drew but this was hand drawn anyway so forgive me. You should be going through that point. But you get it. So we're going to go ahead, we're going to show this on Desmos just to get more of an exact copy here. But this is our midpoint, which was at 2,0. So we'll get an exact version on Desmos here in a second. But this is how we would do this problem if we wanted to work it out by hand. So all right, let's just head over to Desmos real quick, just so we can get a real view of this. All right, so here we are with Desmos. This is going to be, you know, an extremely accurate representation unlike my slightly inaccurate representation with the hand-drawn coordinate plane. Whenever you hand-draw stuff, it doesn't have to be perfect, it just has to look somewhat correct. But all right, here we go. So this is Desmos. We have our point A, 3, 2, and our point P, 1, negative 2. Our midpoint, yes, lies directly on that line halfway in between both points at 2, 0. This is the equation of that line. I did restrict it between the two points just so we don't have that line running off here. But this is really the accurate kind of representation that we're looking at for example one. So all right, let's go ahead and head back to the lecture. All right, so we just finished up example one. So we can go ahead and move on. So now we're moving on to graphs of equations in two variables. Really all this means is that... We're graphing equations that have two variables like an x and a y, which should be something that we've seen before. So an equation in two variables, something like y is equal to x squared plus 1, this just expresses a relationship between two quantities, those quantities being x and y. A point will satisfy your equation if it makes the equation true when the values of x and y are plugged in. So if you have a point that lies on the line or lies on the equation of your function or your equation, those values of x and y would be solutions. So really all your graph is when you graph an equation, your graph is literally a picture of all the possible solutions of your equation. That's all a graph really is. So we have some, we have the fundamental principle of analytic geometry and then we have the official definition of the graph of an equation. So we're going to go through these but by no means do you have to. to memorize any definitions or anything like that. So the fundamental principle of analytic geometry from your book states that this tells us a point x comma y lies on the graph of an equation if and only if its coordinates satisfy the equation. A different version of the same statement is the definition of the graph of an equation. The graph of an equation in x and y is the set of all points, x comma y, in the coordinate plane that would satisfy the equation. A.k.a. your graph is a picture of all the possible solutions. That's all your graph is. So now that we've just stated that, let me go ahead and actually get into some specifics about graphs of equations. So first one being intercepts. The intercepts are points where your graph or your equation will intersect an axis. So for an x-intercept, that will be the point where the graph intersects the x-axis. And these are found by setting y is equal to zero in the equation of a graph, or if you actually have the physical, you know, graph on the coordinate plane, you have the picture, you just want to identify where that line It's going right through the x-axis. That would be your x-intercept. But if you wanted to find it algebraically, you would let y equal zero and solve for x. Then pretty much opposite for y-intercepts, the point where your graph intersects the y-axis is going to be the y-intercept. They also found algebraically by letting x equal zero and not solving for y in your equation. If you have the actual graph in front of you, that picture, that coordinate plane, you can go ahead and just identify wherever your equation or your graph that you drew goes through the y-axis. Those will be your intercept points. And those are also always going to be in the form of an ordered pair, with one of the numbers always being zero. x-intercepts always look like the ordered pair of x value comma zero. y-intercepts always look like the ordered pair of zero comma y value. So there's always a zero in there just knowing when which is which is important. So here we have a kind of a summed up version here. We have a chart where we have all the information. Our x intercepts coordinates of points where your graph of an equation intersects the x-axis. We have how to find them, set y equal to zero and solve for x, and where are they on the graph? What do they look like? So if you have your coordinate plane and you draw your equation or your function here in red, we see each point here where it crosses that x-axis, those are your x-intercepts. So in this case, we've got three, one, two, and three places where our graph crosses the x-axis. So three x-intercepts. Then for y-intercepts, we have the y-coordinates of points where the graph of an equation intersects the y-axis. So how to find them? we set x equal to zero and solve for y if we're solving algebraically and locating them on the graph we see we have a new function here drawn in and we see that it's going across the y-axis twice once here and once here so in this case we had two y-intercepts so using this information let's go ahead and sketch a graph so example two we're going to sketch the graph of the equation y is equal to x squared minus four And we're going to find the x and y-intercepts of the graph of the equation, both probably algebraically and by locating it on the actual graph. So, all right, let's go ahead and head over onto our notes to draw this up once again. All right, so here we have problem two. We've got y is equal to x squared minus four, and we have our coordinate plane here as well. So we want to go ahead and we want to graph. this equation we also have to identify the x and y intercepts. So in order to graph this equation by hand because that's really kind of what we're going for here we want to be able to graph them by hand and then just check on them with Desmos. So in order to actually graph this by hand we need to find points. So we need to figure out the ordered pairs that actually go along with this equation. So what we're going to do is we're going to create what's called a table of values. So we're going to start off by writing x and y. We're going to separate them out here. We're going to give more room to the y side because we're actually going to calculate y. So what happens here is that when you create a table of values is that you want to pick values of x. You're just going to pick some. Try to pick anywhere between five and seven. For us, we're going to go with 7, but anywhere between 5 and 7 is usually pretty good. If you feel like you really got the gist with 5, then you probably don't need to do 7. But if you feel like you need to look at it a little bit more, 7 is probably good. So you pick values of x. and then you calculate values of y. And you calculate values of y using your equation. You have that y is equal to x squared minus 4, so if you find values of x, you can pretty easily just find y. So let's go ahead and do that. We're going to pick some values of x, and then we'll calculate y. So if we want seven values, it's always best to try and get a spread around a specific number. So in our case, let's go ahead. We're going to pick the numbers negative 3 to 3, and all the numbers in between. This is just pretty a generic kind of spread of numbers that I typically choose for really any type of equation, but you can pick, you know, any type of spread of numbers that you like. But we're going to go from negative 3 to 3. So we're going to have negative 3, negative 2, negative 1, 0. 1, 2, and 3. So these are all the x values we want to calculate y for. So for calculating y, all we need to do is plug in that x value that we picked into our equation. So for example, to find the y value that corresponds to x being equal to negative 3, we're going to plug negative 3 in for x. So when we calculate y, it'll look like negative 3 squared minus 4. Now, Negative 3 squared, if we plug that into a calculator, if we just know it, that's going to be 9. So 9 minus 4, that's going to be 5. Let's extend this table out a little bit here. And now what we have is an ordered pair. So the ordered pair that now corresponds. is going to be the x value and the y value. The x value being negative 3, the y value being 5. So there we go. We have one ordered pair. We have one point that is going to be on this function now. All right, so let's go ahead. We'll do one more. We'll do the negative 2 value as well. So we're going to go ahead and plug in negative 2 for x. So we get negative 2 squared minus 4. Negative 2 squared is 4. minus 4, and that's just 0. So our ordered pair in this case is the x value comma y value. So we're going to have negative 2 comma 0. So there's another ordered pair. All right, I trust you all can most likely do exponents and subtraction. So I'm going to let you all go ahead, finish this out, and I'll write this out as well. I highly suggest pausing and calculating the values yourself, but if not... You're going to see them in about a second here. All right, and there we go. So we went ahead, calculated all those values of y, and then identified its corresponding ordered pair. So that is how we build a table of values, and then find all of the points that are going to go for our equation here when we graph it. And then we actually want to graph those values of the ordered pairs. We want to actually graph them. So if we go ahead and we start graphing our ordered... pairs. Let's go ahead. So we've got negative three and five. So that will be right about here. Then we have negative two and zero. That's going to go right about there. And we've got negative one, three, negative one, negative three. There we go. Negative one, negative three, zero, negative four. And then we've got 1, negative 3, 2, 0, and 3, 5. So there we go. We've plotted all of the points from our order pairs, and we can go ahead and actually draw the function now. So we're going to just connect all of our points, and we're always going to kind of start left to right. So we're going to connect our points in order from left to right. So... Starting over here, we're going to go ahead and connect all of these points. So now we have our equation here. We can imagine that if we pick, say, negative 4, negative 5, negative 10, or positive 4, positive 5, positive 10, then we'll get values for those as well. We just don't want to find a million values. So instead of actually calculating any more values, we're just going to tack on some arrows. on the ends of our graph here. So this is our graph of our equation y is equal to x squared minus 4. So there we have it. We have graphed an equation. Now the last thing that we had to do was find the x and y intercepts. So our x and y intercepts are going to be where our graph crosses each axis. So our x intercepts cross the x-axis right there, and our graph crosses the y-axis right there as well. Now the ordered pairs that these go along with are negative 2, 0, 0, negative 4, and 2, 0. The 2 for the negative 2, 0, and 2, 0. Ooh, wrong thing. There we go. These are the x-intercepts. Then the point 0, negative 4. is the y-intercept, because that is where our graph actually crosses those axes at those points. They're also pretty easily identifiable once you have your ordered pairs by looking at which one included the value of zero. That should tell you immediately that it is an intercept. You just have to figure out which one at that point. So remember x-intercepts have a number first, and then zero, and y-intercepts have a zero first, and then a number in their ordered pairs. So that's a pretty simple way to identify those in those cases. So all right, we're going to go ahead, just like last time, we're going to just verify this in Desmos really quick before we move on. All right, so here we are in Desmos. We have the function y equals x squared minus four is graphed, and then we see that it crosses the x-axis at the point, negative 2 comma 0 and positive 2 comma 0, and it crosses the y-axis at the point 0 comma negative 4. So the graph that we drew as well was also correct. Now if you are plotting this in Desmos, if you just plot the function right away, you'll see they have these gray kind of points on the intercepts, and you're actually just able to click those and see what the values are. So that'll be another pretty easy way to find intercepts. if you decide to use Desmos. But all right, let's go ahead and head back over to the lecture. All right, we just finished up example two. Now we can move on to our next concept here. So this is going to be where we enter in kind of newer concepts. So a concept of symmetry, and then we're also going to work with circles. So working with symmetry here, we say when a part of a graph looks like the mirror image of another part of the graph, your graph has symmetry. So if it kind of looks like if you ever fold a piece of paper in half, you draw a design on one half, and then maybe you copy it in front of like a window, you'll open it up and it looks like basically the same image on either side. That is the exact same idea with symmetry. You can have it across for kind of this vertically here, or up and down across. So you have symmetry in both cases, and even a third case. you can have symmetry diagonally as well. So a graph is symmetric to the x-axis, the y-axis, or the origin. If your graph is symmetric with respect to the x-axis, that means that it's symmetric across the x-axis. So that's kind of that vertical symmetry, the up and down symmetry. A graph being symmetric to the y-axis, since your y-axis is vertical, it means that it's symmetric across left and right. So that's more of a little bit of a horizontal symmetry. A graph that's symmetric about the origin is meaning that it is symmetric diagonally. So it means that in this case, you would fold that paper into fourths. So it means that it would be symmetric across that quadrant. So if you have the two halves in each one, you know, you have those quadrants. It means it's symmetric across quadrants and not. left or right or up and down. So it's that diagonal symmetry. We also have another chart here that is kind of wrapping up all of the different types of symmetry and its properties and how to find it. So with symmetry, if you have the graph in front of you, it should be a little easier to identify if it's symmetric. But if you don't have the graph in front of you, can't have it in front of you or anything along those lines, you can also test for symmetry using algebra. So starting off with symmetry to the x-axis. If you want to algebraically test if something is symmetric to the x-axis, you would replace y with negative y. If you can replace your function with negative y and the resulting equation is the same as the original, then it is symmetric. If it's not, then it's not symmetric. So here we are, here we have a graph and we can see this function is symmetric across the x-axis. We see that symmetry kind of in that up and down form here. The property is that the graph is unchanged or reflected. about the x-axis. So it looks the same if you folded it up and down. Then with symmetry to the y-axis, if you wanted to test if it's symmetric algebraically, you would replace x with negative x. And if that resulting equation is the same as your original, you have symmetry to the y-axis. If it's not, then you just don't. So here is an image as well with symmetry across the y-axis here. So we have these two sides. It looked like if you folded it sideways, left and right, it would look the same. So your graph is unchanged, reflected about the y-axis. It is symmetric, going from left to right. Then lastly, that symmetry with respect to the origin, that's that diagonal symmetry. So if you want to test this algebraically, you would combine the two previous symmetry tests. You would replace y with negative y and you replace x with negative x. If your resulting equation is the same as your original, you have symmetry. If it's not, you do not have symmetry. So here is a picture of kind of what that looks like. We saw before we saw symmetry up and down, symmetry left and right. Now we have symmetry diagonal. You can really only go in the two diagonal kind of forms here. One of these ways is depicted in the picture that we have here. So it means that your graph is unchanged if it's rotated about 180 degrees about the origin. That means flipping it in the opposite diagonal direction. So this is what symmetry looks like in each one of our different cases, how to find it algebraically, and mostly identifying it from a graph. All right, let's go ahead and test an equation for symmetry. Example three, we're testing the equation y is equal to x squared minus four. for symmetry. Now if you notice, this is the same problem that we had for example two. This is the same equation y is equal to x squared minus four. So if you remember what that graph looked like, you could probably already tell what it's symmetric to. We're going to algebraically test first and then we're obviously going to verify which one it's going to be in the end. So all right, let's go ahead and head over to our notes. All right, so here we go. We have y is equal to x squared minus 4. We want to test the x-axis symmetry, y-axis symmetry, and origin symmetry. So if we are testing all three, which is typically what you want to do when you test for symmetry, we're going to do each one of these things differently. So for the x-axis symmetry, we'll replace y with negative y. We obviously know our original equation is y is equal to x squared minus 4. So let's go ahead and change y to negative y. So this means that we now have negative y is equal to x squared minus 4. Okay, well, it obviously doesn't look the same. We have negative y instead of positive y. Let's go ahead. We're going to go ahead and try and get rid of the negative. So let's take the negative off of y, place it with the right hand side of our equation instead. To get rid of that negative, we can go ahead and multiply. multiply everything by a negative 1. So we're going to multiply a negative 1. This gives us y is equal to negative x squared plus 4. That is not the same. This does not match our original equation. Our original equation was y equals x squared minus 4. And the negatives are wrong in this case. We have y is equal to negative x squared plus 4. All the signs are wrong. So we do not have symmetry to the x-axis. Big X, not symmetric to x-axis. So that's what it looks like when it's... not symmetric. It's obviously going to look a little too different. It's not going to look the same. Let's go ahead and do y-axis symmetry. So instead of replacing y with negative y, now we're going to replace x with negative x. So we get y is equal to, and we see that x is squared. If we're just replacing x, that means that we would now have negative x that is squared. We don't just add a negative to the outside of x squared. You're replacing x. x squared is x times x. So if you're replacing x with negative x, you now have negative x times negative x. So the negative goes inside parentheses, and that squared goes on the outside. So that is the correct way to replace x with negative x. And we see that we have negative x squared. Well, if you ever square a number that is negative, and then square it, square that same number, but it's positive, you know that the answer is the same. The square root of negative 2 is going to be, sorry, I said square root, didn't I? The square of negative 2 is going to be 4, and the square of 2 is also 4. So even though we have negative x squared, it really doesn't affect anything. Negative x squared is essentially the same thing as x squared. So really, This is y equals x squared minus 4, which is the same thing as our original, which means that we have symmetry to the y-axis. We see how it kind of simplified in that case. So this is how we were able to figure out that it was symmetric. So we can write in here now, symmetric. I think I spelled that right. I think it just looks funny. You get it. All right, last one is origin symmetry. We know it's already going to be symmetric to the y-axis and not the x-axis, so we can kind of assume where this is going to go with the origin symmetry, but we're going to do it anyway. So with the origin symmetry, we do both. We're replacing y with negative y and we're replacing x with negative x. So our function in this case will look like negative y is equal to negative x squared. minus 4. Now, like we said before, the negative x that is squared minus 4 simplifies to x squared minus 4, just like how we did the last time. So we've got negative y is equal to x squared minus 4. But we still got that negative on the y. This is the same thing as we did when testing symmetry on the x-axis. We can get rid of the negative, multiply each side by a negative 1. but we see where this is going. We get y is equal to negative x squared plus 4. Once again, we know this is not symmetric. So our equation here is not symmetric to the origin. So, all right, this is how we tested for symmetry algebraically. We replaced each variable with the negative of itself instead, and then for the origin we did both. So this is how we kind of got these answers algebraically in our case. If you were to graph it like how we did before, we can see our graph is symmetric to the y-axis. It looks the same on the left and the right. If we folded our coordinate plane in half, it would look the same for both sides. So we see we have that left to right symmetry, which means we have symmetry to the y-axis since we had crafted before. And if we remember what it looked like, we could probably pretty easily be able to tell which answer was correct. But we showed it algebraically as well. So all right, let's go ahead and head on back over to our lecture. All right, so we just finished up example three. So we can move on to our last concept here, which is going to be circles. So we are learning about circles, equations of circles, how to find equations of circles, how to draw circles from equations of circles. So this is kind of what we're mostly covering in this case. So we have the equation of a circle. We say an equation of a circle with the center h comma k and a radius r is of the formula or of the equation here x minus h squared plus y minus k squared is equal to r squared. So we have x minus the x value of the center plus plus y minus the y value of the center squared is equal to the radius squared. The radius is the value between the center of the circle and the outer ring. This formula, we just saw this equation, this is called standard form for the equation of a circle. So if you're asked to find the equation of a circle in standard form, this is what you're looking for. If the center of your circle is at the origin zero zero, your equation is just going to be x squared plus y squared is equal to r squared. This is a result of standard form, which means that h and k were both zero, which essentially just simplified to x squared and y squared. So this is the formula you want to remember. Standard form. This is the important one right here. This is only if the center is at the origin, which is at zero, zero. This is the equation you want to remember pretty much all the time. So we have the equation of our circle. We have a picture of a circle here. We say that our circle has the center, hk, a point xy that lies on the edge of that circle, and the radius covers that distance between the center and the edge of your circle. So this is our equation, which is what we're going to be using a lot for specific questions from this section. So If we want to use standard form, a lot of the time we're going to have to know what completing the square is. Because we want to have these terms, x minus h squared, y minus k squared, they're not always going to be given to us. And we'll need to use completing the square in order to find them. So circles gives us the equation. This is the equation of a circle. We need completing the square in order to find... the equation of a circle in standard form. So completing the square is going to be a skill that we need in order to actually find the equation of a circle in standard form. So completing the square, you may have seen this before. If you have not, then this will be new, but it is this idea, this idea that we take some equation and we turn it into this perfect square. So Completing the square is a method to rewrite an expression into a perfect square by adding strategic constants. So let's say you've got some sort of expression, x squared plus bx. If you want to turn this into a perfect square, which a perfect square is x plus or minus a number squared. That's what the perfect square is. This is the result right here. But if we're starting with x squared plus bx. All we need to do is add a strategic constant. That strategic constant is b divided by 2 squared. That is always the constant we want to add. So we take this value b, whatever it is, we plug it into the strategic constant b over 2 squared, we get some new number, and this simplifies into that perfect square, x plus b. b divided by 2 squared. So this is how we turn something that is not a perfect square into a perfect square. And if you notice we have x plus some number squared. The plus doesn't have to be a plus, it can be minus instead, it'll work the same. But that is what we have in the equation of the circle. These are two perfect squares. So if you don't have it, we need to know how to create perfect squares and that is how we do that by completing the square. So this is what we need to use. So perfect square is also an expression of the form x plus a squared or x minus a squared and x or y doesn't matter it all works the same. So perfect square is one of these expressions when it is expanded meaning when it's actually multiplied out you It simplifies to one of these expressions. x plus a squared simplifies to x squared plus 2 times a times x plus a squared. This is what it multiplies out to be. Then we also have x minus a squared simplifies to x squared minus 2ax plus a squared. So these are just the forms and the formats that they use. Um, you can always solve for these using any sort of polynomial multiplication skills like FOIL or the square method or the triangle method. I see those used often. But this is basically a little way that you can kind of avoid the multiplication. If you know that it goes in these formats and you know what the value of a is, you can figure out how it looks when it's expanded just by using this formula here. But all right, the perfect square formula up here with completing the square, this is what's going to be most important to us because we're going to do our last example. We're going to show the following equation represents a circle by rewriting it in standard form. Then we just want to identify the center and the radius of the circle. We saw that in our equation of standard form of a circle here, we can easily identify where the center and the radius are. The center is at h and k. The radius is this r value here. So once we have it in standard form, identifying the center and the radius should not be too bad. But all right, here is our problem. We are showing that this equation represents a circle where we're writing it in its standard form, identifying the center and the radius. Our equation here is x squared plus y squared plus 10x minus 6y plus 33 is equal to 0. So let's go ahead and head on over to our notes to work this problem out. All right, so here's our example, example 4. We have x squared plus y squared plus 10x minus 6 squared plus 33 is equal to 0. And we have opted the side here, completing the square and how to do it really. So we want to make sure we identify some things. We want to know what is the standard form of a circle. So we'll write out standard form of equation of a circle. is x minus h squared plus y minus k squared is equal to r squared, where the center is h comma k and the radius is r. So we know that this is what we want. This is our goal. So we need to take the equation that we have and turn it into... our goal. So we're going to break this down in some steps. So step one, we're going to try to make our equation look as similar as possible to the standard form without doing any heavy lifting. So we're talking rearranging, throwing in some parentheses, and that's pretty much it. So we're going to rearrange our equation a little bit. We're going to group the x terms and we're going to group the y terms. And then we're going to send our term that doesn't have an extra y to the right side. So step one is going to be to group x and y terms together with parentheses and send. the constant number to the right hand side. All right, so what this means for our equation, we want to group the x terms together first. So we're going to start, we have x squared and 10x. We're going to use our highlighter here. We've got x squared, 10x. Those are the x terms. So we're going to write x squared plus 10x, and we're going to put those in parentheses. And those are our x terms. Next, we're going to group our y terms. Our y terms are y squared and negative 6. why we're always kind of trying to deal with positives here um as much so When we see that minus sign, we're just going to attribute that to negative instead. So for our y terms, we've got y squared minus 6y or plus negative 6y. And then this is equal to that constant number. That constant number we are sending to the right-hand side. The only way to do that would be to subtract 33 from both sides. If it were minus, you would add. So you're just putting it to the right-hand side. In our case, this means we subtracted, which means that we have a negative 33 on the end of our equation here. So if we look at our equation, we've got x terms plus y terms is equal to a number, which is... fairly similar to the standard form. We have x numbers that are squared, y terms that are squared, and then a radius squared at the end, a number at the end. So it looks fairly similar, but this is where the heavy lifting comes in. We need to have our x terms and our y terms look like x minus h squared and y minus k squared. We're not going to worry about radius squared or r squared because we're just going to hope that by the time we're done doing this it's just going to look correct so and that is pretty much the idea it will be correct in the end as long as you've done everything right remember your radius can just never be negative because you wouldn't have a circle with the center on the outside that's just really strange so your radius always has to be positive and greater than zero. If it were zero, it would just be a point, not a circle. So we're looking for positive numbers for the radius in the end. Not to worry about it now. We're going to hope that it just kind of works out. So now is when completing the square comes in. Step two, as that we need, we need to complete. the square for the x and y terms. So we need to complete the square. We said completing the square was if we had something of the form of x squared plus bx, which is what we have. In our case for the x terms, b is 10. For the y terms, b is negative 6. So what we need to do is add the strategic constant b over 2 squared. Then we can simplify. So that's what we're going to do. So we're going to scroll down a little bit here. So we have x squared plus 10x. And remember, we're completing the square. We're adding a strategic constant. We are adding b divided by 2 squared. And b is 10. So we're adding 10 divided by 2 squared. We are adding that value onto the end here. Then we're doing the same thing with our y values. We've got y squared. minus 6y. And we're adding our strategic constant. We are adding b, which is negative 6, divided by 2, squared. Now we can't forget that negative 33 is at the end here. But what we've just done, while it is completing the square, we kind of just broke an equation law. We just added things to one side without doing it to the other. So if you remember that rule when you probably lived in elementary school at some point, if you add or subtract something from one side, you have to do the same to the other. It has to stay equal. So since we added 10 divided by 2 squared and negative 6 divided by 2 squared to the left-hand side, we have to add it to the right hand side as well. So on the right hand side, we are adding those numbers. We are adding 10 divided by 2 squared and negative 6 divided by 2 squared, adding those numbers. Then what comes next is that since we added that strategic constant, if we look back at completing the square, we added that strategic constant. We are now kind of in this stage right there. We're in the second stage. We want this final stage. And all we have to do is write it in that form. We know what b is. We've added the strategic constant already. There's nothing stopping us from simplifying. So x squared plus 10x plus 10 divided by 2 squared is going to simplify without. really doing any real work, without doing any heavy lifting. We know this is going to simplify to x plus b over 2, all of that squared. That's what our completing the square states. We use completing the square as a formula to avoid doing hard work. We are avoiding doing lots of multiplication, lots of division. lots of factoring by knowing how to do completing the square. So we got to avoid a lot of work. We just followed the formula for completing the square. Completing the square is just a formula. And now we have what looks like x minus h squared. The minus can be a plus. It really doesn't matter what it is. It just means that your center has a negative value if it's plus instead of minus. But all right, so to do the same thing with the y terms, oh my gosh, I don't even know how I did that. All right, so completing the score with the y terms, we're doing the same thing. We've got y plus negative 6 over 2, and this whole thing is squared. So we're simplifying here in this case. And then on the left hand side, we can simplify those values. We've got negative 33. 10 divided by 2 is 5. So we've got plus 5 squared. Negative 6 divided by 2 is going to be negative 3. So plus negative 3 squared. From now on, we are simplifying. So step 3. simplify it. So we have x plus 10 over 2 squared. We know what 10 over 2 is. 10 over 2 is 5. So we've got x plus 5 squared plus, then we have y plus negative 6 over 2. Negative 6 over 2 is negative 3. So we'd have y plus negative 3 squared. If we have y plus negative 3, that's the same thing as saying minus 3. So if you like a specific way better, you can always choose to write it another way. We can just simplify that to be y minus 3 squared instead. Then lastly, on our left-hand side here, we've got negative 33 plus 5 squared. 5 squared is 25. So we've got 33 plus... 25 plus negative 3 squared is going to be 9. Whether or not we use a calculator or our brains, we're going to add up negative 33 plus 25 plus 9, and that leaves us with 1. Negative 33 plus 25 is negative 8, and then negative 8 plus 9 is positive 1. So here we go. This is our equation in standard form. So this right here, wrong thing, this right here is the equation of the circle. In standard form, The last thing we wanted to do is identify the center and the radius. We said the center, if our equation is of the form x minus h squared plus y minus k squared is equal to r squared, we said the center is at h and k. The only thing different about our result is that we have x plus 5. not x minus 5. But x plus 5 is the same thing as saying x minus negative 5 squared. So basically what we're trying to say here is that if you have plus instead of minus, it just means that that value is negative, it is opposite. So the center starts the x value is at negative 5. That was the only way that we got plus instead of minus. Then we have y minus 3, y minus k, so k is obviously 3. So our center is at negative 5, 3. And finally, the radius. We said that our formula has radius squared at the end, which means that whatever number that this equation is equal to, the radius is the square. root of that number. So in our case we just have 1, which means that radius, which is 1, so we have here radius, this number here is radius squared. The radius is just r, so we take the square root of that number. So that final number was 1, we're taking the square root of 1. which just so happens to be 1. If the radius, we said if this equation was equal to 4, instead we would take the square root of 4, the radius would then be 2. So we see that that's kind of how that works out. But all right, we found the center, we found the radius, we found the equation, and that was what we were looking for in this problem. So all right, let's go ahead and head back on over to our lecture 2. Wrap this up. All right, we just finished up example four, which was writing our equation in standard form and finding center in the radius. And that finishes up section 1.9. So in summary from section 1.9, we learned about the coordinate plane, the distance and midpoint formulas, graphing equations into variables, symmetry and circles, which included completing the square. So, all right. That finishes up section 1.9 and I'll see you in the next section.

But looking at the coordinate plane and graphs of equations should look somewhat familiar although there might be a little few new things here and there. So All right, starting off with the coordinate plane. The coordinate plane is where we graph things like lines, points, functions, equations, all that kind of stuff.

Kind of looks like this picture here that we have. We see that points on a plane are identified with ordered pairs of numbers, and this forms a coordinate plane. It's also called the Cartesian plane, but we'll almost never call it that. If you do happen to see it named Cartesian plane, it's the coordinate plane.

But so we see this horizontal line here. This is the x-axis. These are where we plot x values. So going up and down like this.

The vertical plane is the y-axis. So the vertical line is the y-axis here. So when we plot points, we're kind of going side to side, going through that y-axis. And the plane is divided into four quadrants.

We've got quadrant 1, quadrant 2, quadrant 3, and quadrant 4 here. The points on the axes are not actually assigned to any quadrant. They don't belong to one because they're right basically on the edge of two.

So a point like this, point P, which is at the point A comma B, our first point there is that x value, point A, and the second is point B. And it's written within parentheses with a comma right there. That is our ordered pair.

So whenever you plot things on... coordinate planes, you're most likely going to be plotting a lot of ordered pairs, drawing lines, equations, anything along those lines. So like we said, any point in the coordinate plane is located by its unique ordered pair of the numbers a comma b, where a, or in this case for this example, 5 comma negative 2, the x-coordinate is that first number, the y-coordinate is the second number.

So that's how we write out our ordered pairs, like how we see here on this graph as well. We have a bunch of different ordered pairs here. Each time it's always the x value first, y value second, comma in between, parentheses around the whole thing.

So these coordinates, these ordered pairs really just specify a specific location on our coordinate plane. We use them to graph other things as well as just graphing points in general. So it's really just identifying a specific location or potentially solutions that we're going to see kind of in the future. So. Moving on from there, we're going to learn about the distance and midpoint formulas.

So the distance and midpoint formulas are going to be exactly how they sound. Distance formula is going to calculate the distance between two points, and the midpoint formula will calculate the point that lies directly between two points. So starting off with the distance formula, we want to recall, if you've ever seen it, if not, then this will be new, but the distance between points A and B on your number line would be written as... The distance between a and b, this first part, left hand side of the equal sign, is just stating what we're doing from the distance between two points.

And this is equal to the absolute value of b minus a. Remember that absolute value, it's just going to take whatever's inside of it and make it positive in the end. So we take the larger number or the second number and we subtract the first or typically the smaller number.

And if it ends up being negative, we add the absolute values on it, just make a positive. And this is really just to kind of run the idea home that distance is always positive. So since distance is always positive, you can't have a negative distance value, so when you subtract those values it always has to be positive, which is why we have the absolute values. So that's really just the reason why they're there.

So since we know that that's the distance between two points on a number line, meaning just the typical number line that we know of, you know, 0, 1, 2, 3, 4, including negative numbers, things like that. We can do the same thing for horizontal and vertical distance as well. So that horizontal distance between any two given points are going to be two x values. So you would have a second x value minus a first x value. And whichever one you decided to be first and second really doesn't matter.

because we have these absolute values. It's going to take whatever value you have on the inside and make it positive anyway. So it does not matter which one you pick to be second and which to be first.

Then we can do the same thing with vertical distance. Any vertical distance between two given points is going to be two y values. So a second y value minus say the first y value, with once again absolute values around it.

So just like with the x values, it doesn't matter which one you label to be first or second, but if you are picking points just make sure to be consistent. The second point will have the second x value and the second y value. Your first point we'll have the first x value and the first y value as well.

So what we see in this picture here, we have point A and point B. Point A was said to be the first point. It has the little ones on the x's and y's. We call that x sub 1 or y sub 1. And B is the second point.

It has the twos below the x and y. We call that x sub 2 and y sub 2. So we say that A is the first point. b is the second point and we want to find the distance between those two.

Just like how we have it over here, we take the difference between those x values, the second x value minus the first x value, and we take the absolute value of that so we get a positive value. And same thing over here for the vertical distance. We take the second y value minus the first y value and apply those absolute values. I almost forgot the word. So that's how we would calculate distance with the specific formula.

This is derived from the Pythagorean theorem. Just a fun fact, don't need to remember that. But here we go. This is the actual distance formula. So if you want to find the distance formula, this is what you're actually going to use.

Everything on the last slide basically just gave you a background trying to explain kind of how we got here. But this is the bread and butter. This is what you're actually going to use.

on things like your homework and exams and whatnot. So this is the important one. This is the distance formula.

Distance formula between the points A and B, where we say A is that first point and B is the second, in the plane, which is any plane, the distance between points A and B, that's all that D with A and B in parentheses means, this just means distance between points A and B, is equal to the square root of x sub 2 minus x sub 1 squared plus y sub 2 minus y sub 1 squared. So this is our distance formula. We say we took the x value from that second point minus the x value from the first point, took that value, and we squared it.

Then we took the difference between the second y value minus the first y value, squared that value, add those together. and take the square root of that result. That's how we know the distance between two points.

So this is our distance formula. We're going to learn about the midpoint formula and then we're going to go ahead and do an example using both. So next is the midpoint formula.

Just like we said before, the midpoint formula is going to calculate the point that lies directly between two other points for us. So we say at the midpoint m of x comma y just like how we had distance was d of x comma y, now we have m of x comma y, is the point that marks the exact halfway point on that line, this imaginary line most likely, that connects two points. So the horizontal distance of each half has to be equal. And this equation that we see here is just representing that it's equal.

We don't have to particularly remember this. But if x is that x value that's right directly in the middle, we say that x minus x sub 1, the x value from that first point, is the same thing as having x sub 2, the x value from the second point, minus the x value of the midpoint. Just to say that it is exactly halfway.

And same thing for the vertical distance with the y values are going to work the same way. The y value of the midpoint minus the y value of the first point has to be equal to the y value of the second point minus the y value of the midpoint. Just to say that they are. perfectly equal.

They are right smack dab in the middle. That's all those are stating. So if you actually solve for x and y to get the midpoint formula, here's the bread and butter again.

This is the one you need to remember. The midpoint formula says that the midpoint of the line segment from point a, which is our first point, to point b, which is our second, is this formula right here. Because remember, your midpoint is a point. It needs to be in the form of an ordered pair.

distance is just a value. But since you have to have this ordered pair here, we see that the x value is going to be the sum of the x values of both points. x sub 1 plus x sub 2. and that number divided by 2. Then for the y value of your midpoint, same thing. You're going to take the sum of the y values of your two points, so y sub 1 plus y sub 2, and divide that by 2. And this result would end up being your midpoint.

So this is the midpoint formula. These are the two really important formulas that you need to remember from this section, the distance and the midpoint formulas. Everything else was just mostly explanation.

These are the ones you need to remember. So, all right, on to our first example. Example one, we want to find the distance between the points P and A, where P is at the values 1, negative 2 at the ordered pair, and A is at the ordered pair of 3, 2. So we want to find the distance between those points, and then we want to actually find the midpoint of the line PA. The line we're just labeling as PA is this line that just connects the two points.

The midpoint formula suggests that there should be some line, but your line can be imaginary or real. Either way, you're still finding the midpoint. There doesn't actually have to be a line, although we will probably draw one in.

So, all right, let's go ahead and head over to our notes so we can work out this problem. All right, so here is our problem here. We have a coordinate plane.

I drew out, so excuse it. not being completely perfect, although it's pretty accurate. Here we go. So we have our coordinate plane. We have plotted point A and point P.

We said P was at 1 comma negative 2, and A was at 3 comma 2. So here are the two points that we have, and we want to go ahead and we want to find the distance and the midpoint. So starting off with the distance, we said that the distance between two points in our case, we're just going to go ahead and do A and P or P, and whichever one you prefer. the distance is going to be x sub 2 minus x sub 1 squared plus y sub 2 minus y sub 1 squared.

So if we know that this is the formula for the distance, we want to go ahead and figure out what is going to be our first point, x sub 1 and y sub 1, and what's going to be our second point, x sub 2 and y sub 2. So let's go ahead. We're going to label P as x sub 1 and y sub 1, and A will be x sub 2, y sub 2. So the distance between our points here, we can go ahead and fill in those values. So we've got our square root, we've got some parentheses, we're going to fill in all the important information here, we're going to add in some different colors, which is why I left it like that. All right, so our first point is...

p, x sub 1 and y sub 1. So x sub 1 is going to be 1, since that is the x value of p, and y sub 1, the y value of p is negative 2. So we added in those values. Now let's go ahead and add in our values from a. We said a was the second point. So x sub 2 will be x value of a, which is 3, and y sub 2 will be the y value of a.

which is 2. So from our values here, we're going to go ahead and we're going to simplify it. So we want to do whatever's in the parentheses first, following our order of operations. 3 minus 1 is going to be 2, so we've got 2 squared plus, and then 2 minus negative 2, we have a double negative there, so that becomes plus.

So 2 plus 2 is 4. Ooh, my screen was wigging out. There we go. So we've got...

Ooh, that was way too big. There we go. All right, we've got 2 squared plus 4 squared. Now we can go ahead and plug this into our calculators or reuse our brains, whichever one you prefer.

2 squared is going to be 4. And 4 squared is 16. We have the square root of 4 plus 16. We know 4 plus 16 is going to be 4 plus 16 is 20. So, all right, the distance between these two points is the square root of 20. Now, this answer is correct, but if you need your answer in decimal form for the homework or if you're doing any sort of comparisons and you want to know kind of an actual number instead of a square root number. you can go ahead and calculate that decimal. If you do that, you should get that the square root of 20 is about...

4.47. So this answer is also correct for your distance, whichever one you prefer. So all right, so we calculated the distance between those two lines, and this distance that we're talking about here is this distance here.

Kind of drawing a little imaginary line. All right, so... That's kind of the distance that we're looking at. Now we want to find the midpoint. So our midpoint, we're going to scroll down just a little bit here.

Our midpoint was this point M of X comma Y. And we said that the midpoint formula was going to be X sub 1 plus X sub 2 divided by 2. And Y sub 1 plus Y sub 2 divided by 2. Now we know what... x sub 1 and y sub 1 and x sub 2 and y sub 2 are, so we can just fill in this value.

So we said x sub 1 was 1, x sub 2 was 3, so we have 1 plus 3, right over 2. And then we have y sub 1, which was negative 2, plus y sub 2, which was positive 2. and this is all divided by 2. So we can go ahead and calculate this. 1 plus 3 is going to be 4. So we've got 4 divided by 2, negative 2 plus 2. Well, that's just 0. So that's 0 divided by 2. So a final simplifying here, 4 divided by 2 is just 2. 0 divided by 2 is 0. So our midpoint is 2,0. So that is our midpoint there.

We can go ahead fill in this on our graph here. Let's pick purple. That's the color between A and our point P here.

Here is the-oh that was the wrong one. That's the highlighter. Here we go.

So they picked purple. This is the midpoint. Okay so it's not exactly on the line that I drew but this was hand drawn anyway so forgive me. You should be going through that point. But you get it.

So we're going to go ahead, we're going to show this on Desmos just to get more of an exact copy here. But this is our midpoint, which was at 2,0. So we'll get an exact version on Desmos here in a second.

But this is how we would do this problem if we wanted to work it out by hand. So all right, let's just head over to Desmos real quick, just so we can get a real view of this. All right, so here we are with Desmos. This is going to be, you know, an extremely accurate representation unlike my slightly inaccurate representation with the hand-drawn coordinate plane.

Whenever you hand-draw stuff, it doesn't have to be perfect, it just has to look somewhat correct. But all right, here we go. So this is Desmos. We have our point A, 3, 2, and our point P, 1, negative 2. Our midpoint, yes, lies directly on that line halfway in between both points at 2, 0. This is the equation of that line. I did restrict it between the two points just so we don't have that line running off here.

But this is really the accurate kind of representation that we're looking at for example one. So all right, let's go ahead and head back to the lecture. All right, so we just finished up example one. So we can go ahead and move on.

So now we're moving on to graphs of equations in two variables. Really all this means is that... We're graphing equations that have two variables like an x and a y, which should be something that we've seen before.

So an equation in two variables, something like y is equal to x squared plus 1, this just expresses a relationship between two quantities, those quantities being x and y. A point will satisfy your equation if it makes the equation true when the values of x and y are plugged in. So if you have a point that lies on the line or lies on the equation of your function or your equation, those values of x and y would be solutions. So really all your graph is when you graph an equation, your graph is literally a picture of all the possible solutions of your equation. That's all a graph really is.

So we have some, we have the fundamental principle of analytic geometry and then we have the official definition of the graph of an equation. So we're going to go through these but by no means do you have to. to memorize any definitions or anything like that. So the fundamental principle of analytic geometry from your book states that this tells us a point x comma y lies on the graph of an equation if and only if its coordinates satisfy the equation. A different version of the same statement is the definition of the graph of an equation.

The graph of an equation in x and y is the set of all points, x comma y, in the coordinate plane that would satisfy the equation. A.k.a. your graph is a picture of all the possible solutions. That's all your graph is. So now that we've just stated that, let me go ahead and actually get into some specifics about graphs of equations.

So first one being intercepts. The intercepts are points where your graph or your equation will intersect an axis. So for an x-intercept, that will be the point where the graph intersects the x-axis.

And these are found by setting y is equal to zero in the equation of a graph, or if you actually have the physical, you know, graph on the coordinate plane, you have the picture, you just want to identify where that line It's going right through the x-axis. That would be your x-intercept. But if you wanted to find it algebraically, you would let y equal zero and solve for x.

Then pretty much opposite for y-intercepts, the point where your graph intersects the y-axis is going to be the y-intercept. They also found algebraically by letting x equal zero and not solving for y in your equation. If you have the actual graph in front of you, that picture, that coordinate plane, you can go ahead and just identify wherever your equation or your graph that you drew goes through the y-axis. Those will be your intercept points. And those are also always going to be in the form of an ordered pair, with one of the numbers always being zero.

x-intercepts always look like the ordered pair of x value comma zero. y-intercepts always look like the ordered pair of zero comma y value. So there's always a zero in there just knowing when which is which is important.

So here we have a kind of a summed up version here. We have a chart where we have all the information. Our x intercepts coordinates of points where your graph of an equation intersects the x-axis. We have how to find them, set y equal to zero and solve for x, and where are they on the graph?

What do they look like? So if you have your coordinate plane and you draw your equation or your function here in red, we see each point here where it crosses that x-axis, those are your x-intercepts. So in this case, we've got three, one, two, and three places where our graph crosses the x-axis.

So three x-intercepts. Then for y-intercepts, we have the y-coordinates of points where the graph of an equation intersects the y-axis. So how to find them? we set x equal to zero and solve for y if we're solving algebraically and locating them on the graph we see we have a new function here drawn in and we see that it's going across the y-axis twice once here and once here so in this case we had two y-intercepts so using this information let's go ahead and sketch a graph so example two we're going to sketch the graph of the equation y is equal to x squared minus four And we're going to find the x and y-intercepts of the graph of the equation, both probably algebraically and by locating it on the actual graph.

So, all right, let's go ahead and head over onto our notes to draw this up once again. All right, so here we have problem two. We've got y is equal to x squared minus four, and we have our coordinate plane here as well.

So we want to go ahead and we want to graph. this equation we also have to identify the x and y intercepts. So in order to graph this equation by hand because that's really kind of what we're going for here we want to be able to graph them by hand and then just check on them with Desmos. So in order to actually graph this by hand we need to find points.

So we need to figure out the ordered pairs that actually go along with this equation. So what we're going to do is we're going to create what's called a table of values. So we're going to start off by writing x and y.

We're going to separate them out here. We're going to give more room to the y side because we're actually going to calculate y. So what happens here is that when you create a table of values is that you want to pick values of x. You're just going to pick some.

Try to pick anywhere between five and seven. For us, we're going to go with 7, but anywhere between 5 and 7 is usually pretty good. If you feel like you really got the gist with 5, then you probably don't need to do 7. But if you feel like you need to look at it a little bit more, 7 is probably good.

So you pick values of x. and then you calculate values of y. And you calculate values of y using your equation. You have that y is equal to x squared minus 4, so if you find values of x, you can pretty easily just find y. So let's go ahead and do that.

We're going to pick some values of x, and then we'll calculate y. So if we want seven values, it's always best to try and get a spread around a specific number. So in our case, let's go ahead.

We're going to pick the numbers negative 3 to 3, and all the numbers in between. This is just pretty a generic kind of spread of numbers that I typically choose for really any type of equation, but you can pick, you know, any type of spread of numbers that you like. But we're going to go from negative 3 to 3. So we're going to have negative 3, negative 2, negative 1, 0. 1, 2, and 3. So these are all the x values we want to calculate y for.

So for calculating y, all we need to do is plug in that x value that we picked into our equation. So for example, to find the y value that corresponds to x being equal to negative 3, we're going to plug negative 3 in for x. So when we calculate y, it'll look like negative 3 squared minus 4. Now, Negative 3 squared, if we plug that into a calculator, if we just know it, that's going to be 9. So 9 minus 4, that's going to be 5. Let's extend this table out a little bit here.

And now what we have is an ordered pair. So the ordered pair that now corresponds. is going to be the x value and the y value.

The x value being negative 3, the y value being 5. So there we go. We have one ordered pair. We have one point that is going to be on this function now. All right, so let's go ahead. We'll do one more.

We'll do the negative 2 value as well. So we're going to go ahead and plug in negative 2 for x. So we get negative 2 squared minus 4. Negative 2 squared is 4. minus 4, and that's just 0. So our ordered pair in this case is the x value comma y value.

So we're going to have negative 2 comma 0. So there's another ordered pair. All right, I trust you all can most likely do exponents and subtraction. So I'm going to let you all go ahead, finish this out, and I'll write this out as well.

I highly suggest pausing and calculating the values yourself, but if not... You're going to see them in about a second here. All right, and there we go.

So we went ahead, calculated all those values of y, and then identified its corresponding ordered pair. So that is how we build a table of values, and then find all of the points that are going to go for our equation here when we graph it. And then we actually want to graph those values of the ordered pairs. We want to actually graph them. So if we go ahead and we start graphing our ordered...

pairs. Let's go ahead. So we've got negative three and five. So that will be right about here.

Then we have negative two and zero. That's going to go right about there. And we've got negative one, three, negative one, negative three.

There we go. Negative one, negative three, zero, negative four. And then we've got 1, negative 3, 2, 0, and 3, 5. So there we go.

We've plotted all of the points from our order pairs, and we can go ahead and actually draw the function now. So we're going to just connect all of our points, and we're always going to kind of start left to right. So we're going to connect our points in order from left to right. So... Starting over here, we're going to go ahead and connect all of these points.

So now we have our equation here. We can imagine that if we pick, say, negative 4, negative 5, negative 10, or positive 4, positive 5, positive 10, then we'll get values for those as well. We just don't want to find a million values. So instead of actually calculating any more values, we're just going to tack on some arrows.

on the ends of our graph here. So this is our graph of our equation y is equal to x squared minus 4. So there we have it. We have graphed an equation. Now the last thing that we had to do was find the x and y intercepts.

So our x and y intercepts are going to be where our graph crosses each axis. So our x intercepts cross the x-axis right there, and our graph crosses the y-axis right there as well. Now the ordered pairs that these go along with are negative 2, 0, 0, negative 4, and 2, 0. The 2 for the negative 2, 0, and 2, 0. Ooh, wrong thing. There we go. These are the x-intercepts.

Then the point 0, negative 4. is the y-intercept, because that is where our graph actually crosses those axes at those points. They're also pretty easily identifiable once you have your ordered pairs by looking at which one included the value of zero. That should tell you immediately that it is an intercept. You just have to figure out which one at that point.

So remember x-intercepts have a number first, and then zero, and y-intercepts have a zero first, and then a number in their ordered pairs. So that's a pretty simple way to identify those in those cases. So all right, we're going to go ahead, just like last time, we're going to just verify this in Desmos really quick before we move on.

All right, so here we are in Desmos. We have the function y equals x squared minus four is graphed, and then we see that it crosses the x-axis at the point, negative 2 comma 0 and positive 2 comma 0, and it crosses the y-axis at the point 0 comma negative 4. So the graph that we drew as well was also correct. Now if you are plotting this in Desmos, if you just plot the function right away, you'll see they have these gray kind of points on the intercepts, and you're actually just able to click those and see what the values are. So that'll be another pretty easy way to find intercepts. if you decide to use Desmos.

But all right, let's go ahead and head back over to the lecture. All right, we just finished up example two. Now we can move on to our next concept here. So this is going to be where we enter in kind of newer concepts.

So a concept of symmetry, and then we're also going to work with circles. So working with symmetry here, we say when a part of a graph looks like the mirror image of another part of the graph, your graph has symmetry. So if it kind of looks like if you ever fold a piece of paper in half, you draw a design on one half, and then maybe you copy it in front of like a window, you'll open it up and it looks like basically the same image on either side. That is the exact same idea with symmetry. You can have it across for kind of this vertically here, or up and down across.

So you have symmetry in both cases, and even a third case. you can have symmetry diagonally as well. So a graph is symmetric to the x-axis, the y-axis, or the origin. If your graph is symmetric with respect to the x-axis, that means that it's symmetric across the x-axis.

So that's kind of that vertical symmetry, the up and down symmetry. A graph being symmetric to the y-axis, since your y-axis is vertical, it means that it's symmetric across left and right. So that's more of a little bit of a horizontal symmetry.

A graph that's symmetric about the origin is meaning that it is symmetric diagonally. So it means that in this case, you would fold that paper into fourths. So it means that it would be symmetric across that quadrant. So if you have the two halves in each one, you know, you have those quadrants.

It means it's symmetric across quadrants and not. left or right or up and down. So it's that diagonal symmetry.

We also have another chart here that is kind of wrapping up all of the different types of symmetry and its properties and how to find it. So with symmetry, if you have the graph in front of you, it should be a little easier to identify if it's symmetric. But if you don't have the graph in front of you, can't have it in front of you or anything along those lines, you can also test for symmetry using algebra. So starting off with symmetry to the x-axis.

If you want to algebraically test if something is symmetric to the x-axis, you would replace y with negative y. If you can replace your function with negative y and the resulting equation is the same as the original, then it is symmetric. If it's not, then it's not symmetric. So here we are, here we have a graph and we can see this function is symmetric across the x-axis. We see that symmetry kind of in that up and down form here.

The property is that the graph is unchanged or reflected. about the x-axis. So it looks the same if you folded it up and down.

Then with symmetry to the y-axis, if you wanted to test if it's symmetric algebraically, you would replace x with negative x. And if that resulting equation is the same as your original, you have symmetry to the y-axis. If it's not, then you just don't.

So here is an image as well with symmetry across the y-axis here. So we have these two sides. It looked like if you folded it sideways, left and right, it would look the same.

So your graph is unchanged, reflected about the y-axis. It is symmetric, going from left to right. Then lastly, that symmetry with respect to the origin, that's that diagonal symmetry. So if you want to test this algebraically, you would combine the two previous symmetry tests.

You would replace y with negative y and you replace x with negative x. If your resulting equation is the same as your original, you have symmetry. If it's not, you do not have symmetry. So here is a picture of kind of what that looks like. We saw before we saw symmetry up and down, symmetry left and right.

Now we have symmetry diagonal. You can really only go in the two diagonal kind of forms here. One of these ways is depicted in the picture that we have here. So it means that your graph is unchanged if it's rotated about 180 degrees about the origin.

That means flipping it in the opposite diagonal direction. So this is what symmetry looks like in each one of our different cases, how to find it algebraically, and mostly identifying it from a graph. All right, let's go ahead and test an equation for symmetry. Example three, we're testing the equation y is equal to x squared minus four. for symmetry.

Now if you notice, this is the same problem that we had for example two. This is the same equation y is equal to x squared minus four. So if you remember what that graph looked like, you could probably already tell what it's symmetric to. We're going to algebraically test first and then we're obviously going to verify which one it's going to be in the end.

So all right, let's go ahead and head over to our notes. All right, so here we go. We have y is equal to x squared minus 4. We want to test the x-axis symmetry, y-axis symmetry, and origin symmetry. So if we are testing all three, which is typically what you want to do when you test for symmetry, we're going to do each one of these things differently. So for the x-axis symmetry, we'll replace y with negative y.

We obviously know our original equation is y is equal to x squared minus 4. So let's go ahead and change y to negative y. So this means that we now have negative y is equal to x squared minus 4. Okay, well, it obviously doesn't look the same. We have negative y instead of positive y.

Let's go ahead. We're going to go ahead and try and get rid of the negative. So let's take the negative off of y, place it with the right hand side of our equation instead. To get rid of that negative, we can go ahead and multiply.

multiply everything by a negative 1. So we're going to multiply a negative 1. This gives us y is equal to negative x squared plus 4. That is not the same. This does not match our original equation. Our original equation was y equals x squared minus 4. And the negatives are wrong in this case. We have y is equal to negative x squared plus 4. All the signs are wrong. So we do not have symmetry to the x-axis.

Big X, not symmetric to x-axis. So that's what it looks like when it's... not symmetric. It's obviously going to look a little too different. It's not going to look the same.

Let's go ahead and do y-axis symmetry. So instead of replacing y with negative y, now we're going to replace x with negative x. So we get y is equal to, and we see that x is squared. If we're just replacing x, that means that we would now have negative x that is squared.

We don't just add a negative to the outside of x squared. You're replacing x. x squared is x times x. So if you're replacing x with negative x, you now have negative x times negative x. So the negative goes inside parentheses, and that squared goes on the outside.

So that is the correct way to replace x with negative x. And we see that we have negative x squared. Well, if you ever square a number that is negative, and then square it, square that same number, but it's positive, you know that the answer is the same. The square root of negative 2 is going to be, sorry, I said square root, didn't I? The square of negative 2 is going to be 4, and the square of 2 is also 4. So even though we have negative x squared, it really doesn't affect anything.

Negative x squared is essentially the same thing as x squared. So really, This is y equals x squared minus 4, which is the same thing as our original, which means that we have symmetry to the y-axis. We see how it kind of simplified in that case.

So this is how we were able to figure out that it was symmetric. So we can write in here now, symmetric. I think I spelled that right.

I think it just looks funny. You get it. All right, last one is origin symmetry. We know it's already going to be symmetric to the y-axis and not the x-axis, so we can kind of assume where this is going to go with the origin symmetry, but we're going to do it anyway.

So with the origin symmetry, we do both. We're replacing y with negative y and we're replacing x with negative x. So our function in this case will look like negative y is equal to negative x squared. minus 4. Now, like we said before, the negative x that is squared minus 4 simplifies to x squared minus 4, just like how we did the last time.

So we've got negative y is equal to x squared minus 4. But we still got that negative on the y. This is the same thing as we did when testing symmetry on the x-axis. We can get rid of the negative, multiply each side by a negative 1. but we see where this is going.

We get y is equal to negative x squared plus 4. Once again, we know this is not symmetric. So our equation here is not symmetric to the origin. So, all right, this is how we tested for symmetry algebraically. We replaced each variable with the negative of itself instead, and then for the origin we did both.

So this is how we kind of got these answers algebraically in our case. If you were to graph it like how we did before, we can see our graph is symmetric to the y-axis. It looks the same on the left and the right. If we folded our coordinate plane in half, it would look the same for both sides.

So we see we have that left to right symmetry, which means we have symmetry to the y-axis since we had crafted before. And if we remember what it looked like, we could probably pretty easily be able to tell which answer was correct. But we showed it algebraically as well. So all right, let's go ahead and head on back over to our lecture. All right, so we just finished up example three.

So we can move on to our last concept here, which is going to be circles. So we are learning about circles, equations of circles, how to find equations of circles, how to draw circles from equations of circles. So this is kind of what we're mostly covering in this case.

So we have the equation of a circle. We say an equation of a circle with the center h comma k and a radius r is of the formula or of the equation here x minus h squared plus y minus k squared is equal to r squared. So we have x minus the x value of the center plus plus y minus the y value of the center squared is equal to the radius squared. The radius is the value between the center of the circle and the outer ring. This formula, we just saw this equation, this is called standard form for the equation of a circle.

So if you're asked to find the equation of a circle in standard form, this is what you're looking for. If the center of your circle is at the origin zero zero, your equation is just going to be x squared plus y squared is equal to r squared. This is a result of standard form, which means that h and k were both zero, which essentially just simplified to x squared and y squared. So this is the formula you want to remember.

Standard form. This is the important one right here. This is only if the center is at the origin, which is at zero, zero. This is the equation you want to remember pretty much all the time.

So we have the equation of our circle. We have a picture of a circle here. We say that our circle has the center, hk, a point xy that lies on the edge of that circle, and the radius covers that distance between the center and the edge of your circle. So this is our equation, which is what we're going to be using a lot for specific questions from this section.

So If we want to use standard form, a lot of the time we're going to have to know what completing the square is. Because we want to have these terms, x minus h squared, y minus k squared, they're not always going to be given to us. And we'll need to use completing the square in order to find them. So circles gives us the equation.

This is the equation of a circle. We need completing the square in order to find... the equation of a circle in standard form.

So completing the square is going to be a skill that we need in order to actually find the equation of a circle in standard form. So completing the square, you may have seen this before. If you have not, then this will be new, but it is this idea, this idea that we take some equation and we turn it into this perfect square. So Completing the square is a method to rewrite an expression into a perfect square by adding strategic constants. So let's say you've got some sort of expression, x squared plus bx.

If you want to turn this into a perfect square, which a perfect square is x plus or minus a number squared. That's what the perfect square is. This is the result right here. But if we're starting with x squared plus bx. All we need to do is add a strategic constant.

That strategic constant is b divided by 2 squared. That is always the constant we want to add. So we take this value b, whatever it is, we plug it into the strategic constant b over 2 squared, we get some new number, and this simplifies into that perfect square, x plus b.

b divided by 2 squared. So this is how we turn something that is not a perfect square into a perfect square. And if you notice we have x plus some number squared.

The plus doesn't have to be a plus, it can be minus instead, it'll work the same. But that is what we have in the equation of the circle. These are two perfect squares.

So if you don't have it, we need to know how to create perfect squares and that is how we do that by completing the square. So this is what we need to use. So perfect square is also an expression of the form x plus a squared or x minus a squared and x or y doesn't matter it all works the same. So perfect square is one of these expressions when it is expanded meaning when it's actually multiplied out you It simplifies to one of these expressions.

x plus a squared simplifies to x squared plus 2 times a times x plus a squared. This is what it multiplies out to be. Then we also have x minus a squared simplifies to x squared minus 2ax plus a squared.

So these are just the forms and the formats that they use. Um, you can always solve for these using any sort of polynomial multiplication skills like FOIL or the square method or the triangle method. I see those used often. But this is basically a little way that you can kind of avoid the multiplication. If you know that it goes in these formats and you know what the value of a is, you can figure out how it looks when it's expanded just by using this formula here.

But all right, the perfect square formula up here with completing the square, this is what's going to be most important to us because we're going to do our last example. We're going to show the following equation represents a circle by rewriting it in standard form. Then we just want to identify the center and the radius of the circle.

We saw that in our equation of standard form of a circle here, we can easily identify where the center and the radius are. The center is at h and k. The radius is this r value here.

So once we have it in standard form, identifying the center and the radius should not be too bad. But all right, here is our problem. We are showing that this equation represents a circle where we're writing it in its standard form, identifying the center and the radius. Our equation here is x squared plus y squared plus 10x minus 6y plus 33 is equal to 0. So let's go ahead and head on over to our notes to work this problem out.

All right, so here's our example, example 4. We have x squared plus y squared plus 10x minus 6 squared plus 33 is equal to 0. And we have opted the side here, completing the square and how to do it really. So we want to make sure we identify some things. We want to know what is the standard form of a circle. So we'll write out standard form of equation of a circle. is x minus h squared plus y minus k squared is equal to r squared, where the center is h comma k and the radius is r.

So we know that this is what we want. This is our goal. So we need to take the equation that we have and turn it into...

our goal. So we're going to break this down in some steps. So step one, we're going to try to make our equation look as similar as possible to the standard form without doing any heavy lifting. So we're talking rearranging, throwing in some parentheses, and that's pretty much it. So we're going to rearrange our equation a little bit.

We're going to group the x terms and we're going to group the y terms. And then we're going to send our term that doesn't have an extra y to the right side. So step one is going to be to group x and y terms together with parentheses and send. the constant number to the right hand side.

All right, so what this means for our equation, we want to group the x terms together first. So we're going to start, we have x squared and 10x. We're going to use our highlighter here.

We've got x squared, 10x. Those are the x terms. So we're going to write x squared plus 10x, and we're going to put those in parentheses. And those are our x terms. Next, we're going to group our y terms.

Our y terms are y squared and negative 6. why we're always kind of trying to deal with positives here um as much so When we see that minus sign, we're just going to attribute that to negative instead. So for our y terms, we've got y squared minus 6y or plus negative 6y. And then this is equal to that constant number.

That constant number we are sending to the right-hand side. The only way to do that would be to subtract 33 from both sides. If it were minus, you would add.

So you're just putting it to the right-hand side. In our case, this means we subtracted, which means that we have a negative 33 on the end of our equation here. So if we look at our equation, we've got x terms plus y terms is equal to a number, which is... fairly similar to the standard form. We have x numbers that are squared, y terms that are squared, and then a radius squared at the end, a number at the end.

So it looks fairly similar, but this is where the heavy lifting comes in. We need to have our x terms and our y terms look like x minus h squared and y minus k squared. We're not going to worry about radius squared or r squared because we're just going to hope that by the time we're done doing this it's just going to look correct so and that is pretty much the idea it will be correct in the end as long as you've done everything right remember your radius can just never be negative because you wouldn't have a circle with the center on the outside that's just really strange so your radius always has to be positive and greater than zero. If it were zero, it would just be a point, not a circle.

So we're looking for positive numbers for the radius in the end. Not to worry about it now. We're going to hope that it just kind of works out.

So now is when completing the square comes in. Step two, as that we need, we need to complete. the square for the x and y terms. So we need to complete the square. We said completing the square was if we had something of the form of x squared plus bx, which is what we have.

In our case for the x terms, b is 10. For the y terms, b is negative 6. So what we need to do is add the strategic constant b over 2 squared. Then we can simplify. So that's what we're going to do.

So we're going to scroll down a little bit here. So we have x squared plus 10x. And remember, we're completing the square.

We're adding a strategic constant. We are adding b divided by 2 squared. And b is 10. So we're adding 10 divided by 2 squared. We are adding that value onto the end here. Then we're doing the same thing with our y values.

We've got y squared. minus 6y. And we're adding our strategic constant.

We are adding b, which is negative 6, divided by 2, squared. Now we can't forget that negative 33 is at the end here. But what we've just done, while it is completing the square, we kind of just broke an equation law. We just added things to one side without doing it to the other.

So if you remember that rule when you probably lived in elementary school at some point, if you add or subtract something from one side, you have to do the same to the other. It has to stay equal. So since we added 10 divided by 2 squared and negative 6 divided by 2 squared to the left-hand side, we have to add it to the right hand side as well. So on the right hand side, we are adding those numbers.

We are adding 10 divided by 2 squared and negative 6 divided by 2 squared, adding those numbers. Then what comes next is that since we added that strategic constant, if we look back at completing the square, we added that strategic constant. We are now kind of in this stage right there. We're in the second stage.

We want this final stage. And all we have to do is write it in that form. We know what b is. We've added the strategic constant already.

There's nothing stopping us from simplifying. So x squared plus 10x plus 10 divided by 2 squared is going to simplify without. really doing any real work, without doing any heavy lifting.

We know this is going to simplify to x plus b over 2, all of that squared. That's what our completing the square states. We use completing the square as a formula to avoid doing hard work.

We are avoiding doing lots of multiplication, lots of division. lots of factoring by knowing how to do completing the square. So we got to avoid a lot of work. We just followed the formula for completing the square. Completing the square is just a formula.

And now we have what looks like x minus h squared. The minus can be a plus. It really doesn't matter what it is.

It just means that your center has a negative value if it's plus instead of minus. But all right, so to do the same thing with the y terms, oh my gosh, I don't even know how I did that. All right, so completing the score with the y terms, we're doing the same thing.

We've got y plus negative 6 over 2, and this whole thing is squared. So we're simplifying here in this case. And then on the left hand side, we can simplify those values.

We've got negative 33. 10 divided by 2 is 5. So we've got plus 5 squared. Negative 6 divided by 2 is going to be negative 3. So plus negative 3 squared. From now on, we are simplifying.

So step 3. simplify it. So we have x plus 10 over 2 squared. We know what 10 over 2 is. 10 over 2 is 5. So we've got x plus 5 squared plus, then we have y plus negative 6 over 2. Negative 6 over 2 is negative 3. So we'd have y plus negative 3 squared. If we have y plus negative 3, that's the same thing as saying minus 3. So if you like a specific way better, you can always choose to write it another way. We can just simplify that to be y minus 3 squared instead.

Then lastly, on our left-hand side here, we've got negative 33 plus 5 squared. 5 squared is 25. So we've got 33 plus... 25 plus negative 3 squared is going to be 9. Whether or not we use a calculator or our brains, we're going to add up negative 33 plus 25 plus 9, and that leaves us with 1. Negative 33 plus 25 is negative 8, and then negative 8 plus 9 is positive 1. So here we go. This is our equation in standard form.

So this right here, wrong thing, this right here is the equation of the circle. In standard form, The last thing we wanted to do is identify the center and the radius. We said the center, if our equation is of the form x minus h squared plus y minus k squared is equal to r squared, we said the center is at h and k. The only thing different about our result is that we have x plus 5. not x minus 5. But x plus 5 is the same thing as saying x minus negative 5 squared. So basically what we're trying to say here is that if you have plus instead of minus, it just means that that value is negative, it is opposite.

So the center starts the x value is at negative 5. That was the only way that we got plus instead of minus. Then we have y minus 3, y minus k, so k is obviously 3. So our center is at negative 5, 3. And finally, the radius. We said that our formula has radius squared at the end, which means that whatever number that this equation is equal to, the radius is the square. root of that number. So in our case we just have 1, which means that radius, which is 1, so we have here radius, this number here is radius squared.

The radius is just r, so we take the square root of that number. So that final number was 1, we're taking the square root of 1. which just so happens to be 1. If the radius, we said if this equation was equal to 4, instead we would take the square root of 4, the radius would then be 2. So we see that that's kind of how that works out. But all right, we found the center, we found the radius, we found the equation, and that was what we were looking for in this problem.

So all right, let's go ahead and head back on over to our lecture 2. Wrap this up. All right, we just finished up example four, which was writing our equation in standard form and finding center in the radius. And that finishes up section 1.9. So in summary from section 1.9, we learned about the coordinate plane, the distance and midpoint formulas, graphing equations into variables, symmetry and circles, which included completing the square.

So, all right. That finishes up section 1.9 and I'll see you in the next section.

Transcript for:Understanding Coordinate Plane and Circles

Transcript for:
Understanding Coordinate Plane and Circles