Systems of quadratic equations
Solve the system of equations by graphing. Check your solution algebraically. Let's graph each of these, and let's start-- let me find a nice dark color to graph these with. Let me graph this top equation in blue, this parabola. The first thing to think about is this going to be an upward opening-- one, how did I know it's a parabola? That's because it's a quadratic function: we have an x squared term, a second degree term, here. Then we have to think about: it is going to be upward opening, or downward opening parabola? You see that it's a negative coefficient in front of the x squared, so it's going to be a downward opening parabola. What is going to be its maximum point? Let's think about that for a second. This whole term right here is always going to be negative, or it's always going to be non-positive. x squared will be non-negative when you multiply it by a negative, so it's going to be non-positive. So, the highest value that this thing can take on is when x is going to be equal to 0-- the vertex of this parabola is when x is equal to 0, and y is equal to 6. So, x is equal to 0, and y is 1, 2, 3, 4, 5, 6. So that right there is the highest point of our parabola. Then, if we want, we can a graph a couple of other points, just to see what happens. So let's see what happens when x is equal to-- let me just draw a little table here-- 2, what is y? It's negative x squared plus 6. So when x is 2, what is y? You have 2 squared, which is 4, but you have negative 2 squared, so it's negative 4 plus 6-- it is equal to 2. It's the same thing when x is negative 2. You put negative 2 there, you square it, then you have positive 4, but you have a negative there, so it's negative 4 plus 6 is 2. You have both of those points there, so 2 comma 2, and then you have a negative 2 comma 2. If I were to graph it, Let's try it with 3, as well-- if we put a 3 there, 3 squared is 9. It then becomes a negative 9 plus 3, it becomes negative 3, and negative 3 will also become a negative 3. Negative 3 squared is positive 9, you have a negative out front, it becomes negative 9 plus 6, which is negative 3. You have negative 3, negative 3, and then you have 3, negative 3. So those are all good points. Now we can graph our parabola. Our parabola will look something-- I was doing well until that second part --like that, and let me just do the second part. That second part is hard to draw-- let me do it from here. It looks something like that. We connect to this dot right here, and then let me connect this. So that it looks something like that. That's what our parabola looks like, and obviously it keeps going down in that direction. So that's that first graph. Let's graph this second one over here: y is equal to negative 2x minus 2. This is just going to be a line. It's a linear equation, and the highest degree here is 1. Our y-intercept is negative 2, so 0, 1, 2. Our y-intercept is negative 2. Our slope is negative 2. If we move 1 in the x direction, we're going to go 2 in the y-direction, and if we move 2 in the x direction, we're going to move down 4 in the y direction. If we move back 2, we're going to move up 2 in the y direction, and it looks like we found one of our points of intersection. Let's just draw that line, so that line will look something like-- It's hard for my hand to draw that, but let me try as best as I can. This is the hardest part. It will look something like that right there. The question is, where do they intersect? One point of intersection does immediately pop out at us, because they asked us to do it graphically. That point right there, which is the point negative 2, 2. It seems to pop out at us, so this is the point negative 2, 2. Let's see if that makes sense. When you have the point negative 2, when you put x is equal to negative 2 here, negative 2 times negative 2 is 4 minus 2, and y is equal to 2. When you put negative 2 here, y is also equal to 2, so that makes sense. There's going to be some other point way out here where they also intersect. There's also going to be some other point way out here if we keep making this parabola. When y is equal to positive 4, and you have negative 16 plus 6, you get negative 10. So, positive 1, 2, 3, 4, and then you go down 10. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. That looks like that might be our other point of intersection, so let me connect this right there. Our other point of intersection looks to be right there. If we just follow this red line it looks like we intersect there. Let's verify that it works out. So 4, negative 10. We know that that's on this blue line, so let's see if it's on this other line. So negative 2 times 4 minus 2, that is negative 8 minus 2, which is equal to negative 10. The point 4, negative 10, is on both of them. When x is equal to 4, y is negative 10 for both equations here, so they both definitely work out.